Q1: Find the 20^th and n^th terms of the G.P. 5/2, 5/4, 5/8, …

Given: The G.P. is 5/2, 5/4, 5/8, … To Find: The 20^th term ( a_20 ) and the n^th term ( a_n ). Solution: The first term is a = 5/2. The common ratio r is found by dividing the second term by the first term: r = 5/4/5/2 = 5/4 × 2/5 = 1/2 Formula for the n^th term of a G.P.: a_n = ar^n-1 Finding the…

Q2: Find the 12^th term of a G.P. whose 8^th term is 192 and the common ratio is 2.

Given: The 8^th term of a G.P. is a_8 = 192. The common ratio is r = 2. To Find: The 12^th term ( a_12 ). Formula for the n^th term of a G.P.: a_n = ar^n-1 Solution: First, we use the given information to find the first term a. For the 8^th term: a_8 = ar^8-1 = ar^7 192 = a(2)^7 192 = a(128) a =…

Q5: Which term of the following sequences: (a) 2,2 √(2), 4, … is 128? (b) √(3), 3,3 √(3), … is 729? (c) 1/3, 1/9, 1/27, … is 1/19683?

Solution: We need to find the position 'n' of the given term in each sequence. (a) Sequence: 2, 2√(2), 4, … is 128? Given: First term a = 2. Common ratio r = frac2√(2)2 = √(2). Let the n^th term be a_n = 128. Formula: a_n = ar^n-1 Calculation: 128 = 2(√(2))^n-1 128/2 = (√(2))^n-1 64 = (2^1/2)^n-1…

Sequences and Series Class 11 Mathematics NCERT Solutions

Q1EXERCISE 8.1

Write the first five terms of each of the sequences in Exercises 1 to 6 whose $n^{\text {th }}$ terms are: $a_{n}=n(n+2)$

Solution

Given: The

n^{\text{th}}

term of the sequence is

a_{n}=n(n+2)

.

To Find: The first five terms of the sequence.

Solution: We substitute

n = 1, 2, 3, 4, 5

into the formula for

a_n

.

For

n=1

:

a_1 = 1(1+2) = 1 \times 3 = 3

For

n=2

:

a_2 = 2(2+2) = 2 \times 4 = 8

For

n=3

:

a_3 = 3(3+2) = 3 \times 5 = 15

For

n=4

:

a_4 = 4(4+2) = 4 \times 6 = 24

For

n=5

:

a_5 = 5(5+2) = 5 \times 7 = 35

Final Answer: The first five terms of the sequence are 3, 8, 15, 24, and 35.

Q2EXERCISE 8.1

Write the first five terms of each of the sequences in Exercises 1 to 6 whose $n^{\text {th }}$ terms are: 2. $a_{n}=\frac{n}{n+1}$

Solution

Given: The

n^{\text{th}}

term of the sequence is

a_{n}=\frac{n}{n+1}

.

To Find: The first five terms of the sequence.

Solution: We substitute

n = 1, 2, 3, 4, 5

into the formula for

a_n

.

For

n=1

:

a_1 = \frac{1}{1+1} = \frac{1}{2}

For

n=2

:

a_2 = \frac{2}{2+1} = \frac{2}{3}

For

n=3

:

a_3 = \frac{3}{3+1} = \frac{3}{4}

For

n=4

:

a_4 = \frac{4}{4+1} = \frac{4}{5}

For

n=5

:

a_5 = \frac{5}{5+1} = \frac{5}{6}

Final Answer: The first five terms of the sequence are

\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}

, and

\frac{5}{6}

.

Q3EXERCISE 8.1

Write the first five terms of each of the sequences in Exercises 1 to 6 whose $n^{\text {th }}$ terms are: 3. $a_{n}=2^{n}$

Solution

Given: The

n^{\text{th}}

term of the sequence is

a_{n}=2^{n}

.

To Find: The first five terms of the sequence.

Solution: We substitute

n = 1, 2, 3, 4, 5

into the formula for

a_n

.

For

n=1

:

a_1 = 2^1 = 2

For

n=2

:

a_2 = 2^2 = 4

For

n=3

:

a_3 = 2^3 = 8

For

n=4

:

a_4 = 2^4 = 16

For

n=5

:

a_5 = 2^5 = 32

Final Answer: The first five terms of the sequence are 2, 4, 8, 16, and 32.

Q4EXERCISE 8.1

Write the first five terms of each of the sequences in Exercises 1 to 6 whose $n^{\text {th }}$ terms are: 4. $a_{n}=\frac{2 n-3}{6}$

Solution

Given: The

n^{\text{th}}

term of the sequence is

a_{n}=\frac{2n-3}{6}

.

To Find: The first five terms of the sequence.

Solution: We substitute

n = 1, 2, 3, 4, 5

into the formula for

a_n

.

For

n=1

:

a_1 = \frac{2(1)-3}{6} = \frac{2-3}{6} = -\frac{1}{6}

For

n=2

:

a_2 = \frac{2(2)-3}{6} = \frac{4-3}{6} = \frac{1}{6}

For

n=3

:

a_3 = \frac{2(3)-3}{6} = \frac{6-3}{6} = \frac{3}{6} = \frac{1}{2}

For

n=4

:

a_4 = \frac{2(4)-3}{6} = \frac{8-3}{6} = \frac{5}{6}

For

n=5

:

a_5 = \frac{2(5)-3}{6} = \frac{10-3}{6} = \frac{7}{6}

Final Answer: The first five terms of the sequence are

-\frac{1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}

, and

\frac{7}{6}

.

Q5EXERCISE 8.1

Write the first five terms of each of the sequences in Exercises 1 to 6 whose $n^{\text {th }}$ terms are: 5. $a_{n}=(-1)^{n-1} 5^{n+1}$

Solution

Given: The

n^{\text{th}}

term of the sequence is

a_{n}=(-1)^{n-1} 5^{n+1}

.

To Find: The first five terms of the sequence.

Solution: We substitute

n = 1, 2, 3, 4, 5

into the formula for

a_n

.

For

n=1

:

a_1 = (-1)^{1-1} 5^{1+1} = (-1)^0 5^2 = 1 \times 25 = 25

For

n=2

:

a_2 = (-1)^{2-1} 5^{2+1} = (-1)^1 5^3 = -1 \times 125 = -125

For

n=3

:

a_3 = (-1)^{3-1} 5^{3+1} = (-1)^2 5^4 = 1 \times 625 = 625

For

n=4

:

a_4 = (-1)^{4-1} 5^{4+1} = (-1)^3 5^5 = -1 \times 3125 = -3125

For

n=5

:

a_5 = (-1)^{5-1} 5^{5+1} = (-1)^4 5^6 = 1 \times 15625 = 15625

Final Answer: The first five terms of the sequence are 25, -125, 625, -3125, and 15625.

Q6EXERCISE 8.1

Write the first five terms of each of the sequences in Exercises 1 to 6 whose $n^{\text {th }}$ terms are: 6. $a_{n}=n \frac{n^{2}+5}{4}$

Solution

Given: The

n^{\text{th}}

term of the sequence is

a_{n}=n \frac{n^{2}+5}{4}

.

To Find: The first five terms of the sequence.

Solution: We substitute

n = 1, 2, 3, 4, 5

into the formula for

a_n

.

For

n=1

:

a_1 = 1 \times \frac{1^2+5}{4} = 1 \times \frac{1+5}{4} = \frac{6}{4} = \frac{3}{2}

For

n=2

:

a_2 = 2 \times \frac{2^2+5}{4} = 2 \times \frac{4+5}{4} = 2 \times \frac{9}{4} = \frac{9}{2}

For

n=3

:

a_3 = 3 \times \frac{3^2+5}{4} = 3 \times \frac{9+5}{4} = 3 \times \frac{14}{4} = \frac{21}{2}

For

n=4

:

a_4 = 4 \times \frac{4^2+5}{4} = 4 \times \frac{16+5}{4} = 21

For

n=5

:

a_5 = 5 \times \frac{5^2+5}{4} = 5 \times \frac{25+5}{4} = 5 \times \frac{30}{4} = \frac{75}{2}

Final Answer: The first five terms of the sequence are

\frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21

, and

\frac{75}{2}

.

Q7EXERCISE 8.1

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose $n^{\text {th }}$ terms are: 7. $a_{n}=4 n-3 ; a_{17}, a_{24}$

Solution

Given: The

n^{\text{th}}

term of the sequence is

a_{n}=4n-3

.

To Find: The terms

a_{17}

and

a_{24}

.

Solution: To find

a_{17}

, we substitute

n=17

into the formula for

a_n

.

a_{17} = 4(17) - 3 = 68 - 3 = 65

To find

a_{24}

, we substitute

n=24

into the formula for

a_n

.

a_{24} = 4(24) - 3 = 96 - 3 = 93

Final Answer:

a_{17} = 65

and

a_{24} = 93

.

Q8EXERCISE 8.1

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose $n^{\text {th }}$ terms are: 8. $a_{n}=\frac{n^{2}}{2^{n}} ; a_{7}$

Solution

Given: The

n^{\text{th}}

term of the sequence is

a_{n}=\frac{n^{2}}{2^{n}}

.

To Find: The term

a_7

.

Solution: To find

a_7

, we substitute

n=7

into the formula for

a_n

.

a_7 = \frac{7^2}{2^7} = \frac{49}{128}

Final Answer:

a_7 = \frac{49}{128}

.

Q9EXERCISE 8.1

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose $n^{\text {th }}$ terms are: 9. $a_{n}=(-1)^{n-1} n^{3} ; a_{9}$

Solution

Given: The

n^{\text{th}}

term of the sequence is

a_{n}=(-1)^{n-1} n^{3}

.

To Find: The term

a_9

.

Solution: To find

a_9

, we substitute

n=9

into the formula for

a_n

.

a_9 = (-1)^{9-1} 9^3 = (-1)^8 \times 729 = 1 \times 729 = 729

Final Answer:

a_9 = 729

.

Q10EXERCISE 8.1

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose $n^{\text {th }}$ terms are: 10. $a_{n}=\frac{n(n-2)}{n+3} ; a_{20}$

Solution

Given: The

n^{\text{th}}

term of the sequence is

a_{n}=\frac{n(n-2)}{n+3}

.

To Find: The term

a_{20}

.

Solution: To find

a_{20}

, we substitute

n=20

into the formula for

a_n

.

a_{20} = \frac{20(20-2)}{20+3} = \frac{20 \times 18}{23} = \frac{360}{23}

Final Answer:

a_{20} = \frac{360}{23}

.

Q11EXERCISE 8.1

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: 11. $a_{1}=3, a_{n}=3 a_{n-1}+2$ for all $n>1$

Solution

Given: A sequence defined by the recurrence relation

a_{1}=3

and

a_{n}=3 a_{n-1}+2

for all

n>1

.

To Find: The first five terms of the sequence and the corresponding series.

Solution: We are given the first term

a_1 = 3

. We use the recurrence relation to find the next four terms.

For

n=2

:

a_2 = 3a_{2-1} + 2 = 3a_1 + 2 = 3(3) + 2 = 9 + 2 = 11

For

n=3

:

a_3 = 3a_{3-1} + 2 = 3a_2 + 2 = 3(11) + 2 = 33 + 2 = 35

For

n=4

:

a_4 = 3a_{4-1} + 2 = 3a_3 + 2 = 3(35) + 2 = 105 + 2 = 107

For

n=5

:

a_5 = 3a_{5-1} + 2 = 3a_4 + 2 = 3(107) + 2 = 321 + 2 = 323

The first five terms of the sequence are 3, 11, 35, 107, and 323.

The corresponding series is the sum of these terms. Series:

3 + 11 + 35 + 107 + 323 + \dots

Final Answer: The first five terms are 3, 11, 35, 107, 323. The corresponding series is

3 + 11 + 35 + 107 + 323 + \dots

.

Q12EXERCISE 8.1

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: 12. $a_{1}=-1, a_{n}=\frac{a_{n-1}}{n}, n \geq 2$

Solution

Given: A sequence defined by the recurrence relation

a_{1}=-1

and

a_{n}=\frac{a_{n-1}}{n}

for

n \geq 2

.

To Find: The first five terms of the sequence and the corresponding series.

Solution: We are given the first term

a_1 = -1

. We use the recurrence relation to find the next four terms.

For

n=2

:

a_2 = \frac{a_{2-1}}{2} = \frac{a_1}{2} = -\frac{1}{2}

For

n=3

:

a_3 = \frac{a_{3-1}}{3} = \frac{a_2}{3} = \frac{-1/2}{3} = -\frac{1}{6}

For

n=4

:

a_4 = \frac{a_{4-1}}{4} = \frac{a_3}{4} = \frac{-1/6}{4} = -\frac{1}{24}

For

n=5

:

a_5 = \frac{a_{5-1}}{5} = \frac{a_4}{5} = \frac{-1/24}{5} = -\frac{1}{120}

The first five terms of the sequence are

-1, -\frac{1}{2}, -\frac{1}{6}, -\frac{1}{24}

, and

-\frac{1}{120}

.

The corresponding series is the sum of these terms. Series:

(-1) + (-\frac{1}{2}) + (-\frac{1}{6}) + (-\frac{1}{24}) + (-\frac{1}{120}) + \dots

Final Answer: The first five terms are

-1, -\frac{1}{2}, -\frac{1}{6}, -\frac{1}{24}, -\frac{1}{120}

. The corresponding series is

-1 - \frac{1}{2} - \frac{1}{6} - \frac{1}{24} - \frac{1}{120} - \dots

.

Q13EXERCISE 8.1

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: 13. $a_{1}=a_{2}=2, a_{n}=a_{n-1}-1, n>2$

Solution

Given: A sequence defined by

a_{1}=a_{2}=2

and the recurrence relation

a_{n}=a_{n-1}-1

for

n>2

.

To Find: The first five terms of the sequence and the corresponding series.

Solution: We are given the first two terms

a_1 = 2

and

a_2 = 2

. We use the recurrence relation for

n>2

to find the next three terms.

For

n=3

:

a_3 = a_{3-1} - 1 = a_2 - 1 = 2 - 1 = 1

For

n=4

:

a_4 = a_{4-1} - 1 = a_3 - 1 = 1 - 1 = 0

For

n=5

:

a_5 = a_{5-1} - 1 = a_4 - 1 = 0 - 1 = -1

The first five terms of the sequence are 2, 2, 1, 0, and -1.

The corresponding series is the sum of these terms. Series:

2 + 2 + 1 + 0 + (-1) + \dots

Final Answer: The first five terms are 2, 2, 1, 0, -1. The corresponding series is

2 + 2 + 1 + 0 - 1 + \dots

.

Q14EXERCISE 8.1

The Fibonacci sequence is defined by $1=a_{1}=a_{2}$ and $a_{n}=a_{n-1}+a_{n-2}, n>2$ Find $\frac{a_{n+1}}{a_{n}}$ , for $n=1,2,3,4,5$

Solution

Given: The Fibonacci sequence defined by

a_1=1, a_2=1

and

a_n = a_{n-1} + a_{n-2}

for

n>2

.

To Find: The value of the ratio

\frac{a_{n+1}}{a_n}

for

n=1, 2, 3, 4, 5

.

Solution: First, we need to find the first six terms of the Fibonacci sequence to calculate the required ratios.

a_1 = 1

a_2 = 1

a_3 = a_2 + a_1 = 1 + 1 = 2

a_4 = a_3 + a_2 = 2 + 1 = 3

a_5 = a_4 + a_3 = 3 + 2 = 5

a_6 = a_5 + a_4 = 5 + 3 = 8

Now, we can calculate the ratios for

n=1, 2, 3, 4, 5

.

For

n=1

:

\frac{a_{1+1}}{a_1} = \frac{a_2}{a_1} = \frac{1}{1} = 1

For

n=2

:

\frac{a_{2+1}}{a_2} = \frac{a_3}{a_2} = \frac{2}{1} = 2

For

n=3

:

\frac{a_{3+1}}{a_3} = \frac{a_4}{a_3} = \frac{3}{2}

For

n=4

:

\frac{a_{4+1}}{a_4} = \frac{a_5}{a_4} = \frac{5}{3}

For

n=5

:

\frac{a_{5+1}}{a_5} = \frac{a_6}{a_5} = \frac{8}{5}

Final Answer: The values of the ratio

\frac{a_{n+1}}{a_n}

for

n=1, 2, 3, 4, 5

are

1, 2, \frac{3}{2}, \frac{5}{3}

, and

\frac{8}{5}

respectively.

Q1EXERCISE 8.2

Find the $20^{\text {th }}$ and $n^{\text {th }}$ terms of the G.P. $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots$

Solution

Given: The G.P. is

\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots

To Find: The

20^{\text{th}}

term (

a_{20}

) and the

n^{\text{th}}

term (

a_n

).

Solution: The first term is

a = \frac{5}{2}

. The common ratio

r

is found by dividing the second term by the first term:

r = \frac{5/4}{5/2} = \frac{5}{4} \times \frac{2}{5} = \frac{1}{2}

Formula for the $n^{\text{th}}$ term of a G.P.:

a_n = ar^{n-1}

Finding the $n^{\text{th}}$ term: Substituting the values of

a

and

r

:

a_n = \left(\frac{5}{2}\right) \left(\frac{1}{2}\right)^{n-1} = 5 \times \frac{1}{2} \times \frac{1}{2^{n-1}} = \frac{5}{2^n}

Finding the $20^{\text{th}}$ term: Substitute

n=20

into the formula for

a_n

:

a_{20} = \frac{5}{2^{20}}

Final Answer: The

n^{\text{th}}

term is

a_n = \frac{5}{2^n}

and the

20^{\text{th}}

term is

a_{20} = \frac{5}{2^{20}}

.

Q2EXERCISE 8.2

Find the $12^{\text {th }}$ term of a G.P. whose $8^{\text {th }}$ term is 192 and the common ratio is 2.

Solution

Given: The

8^{\text{th}}

term of a G.P. is

a_8 = 192

. The common ratio is

r = 2

.

To Find: The

12^{\text{th}}

term (

a_{12}

).

Formula for the $n^{\text{th}}$ term of a G.P.:

a_n = ar^{n-1}

Solution: First, we use the given information to find the first term

a

. For the

8^{\text{th}}

term:

a_8 = ar^{8-1} = ar^7

192 = a(2)^7

192 = a(128)

a = \frac{192}{128} = \frac{3 \times 64}{2 \times 64} = \frac{3}{2}

Now, we can find the

12^{\text{th}}

term using the formula for

a_{12}

:

a_{12} = ar^{12-1} = ar^{11}

a_{12} = \left(\frac{3}{2}\right) (2)^{11} = 3 \times 2^{11-1} = 3 \times 2^{10}

a_{12} = 3 \times 1024 = 3072

Final Answer: The

12^{\text{th}}

term of the G.P. is 3072.

Q3EXERCISE 8.2

The $5^{\text {th }}, 8^{\text {th }}$ and $11^{\text {th }}$ terms of a G.P. are $p, q$ and $s$ , respectively. Show that $q^{2}=p s$ .

Solution

Given: In a G.P., the

5^{\text{th}}

term is

p

, the

8^{\text{th}}

term is

q

, and the

11^{\text{th}}

term is

s

.

To Show:

q^2 = ps

.

Proof: Let the first term of the G.P. be

a

and the common ratio be

r

.

The formula for the

n^{\text{th}}

term is

a_n = ar^{n-1}

.

According to the given information:

a_5 = ar^{5-1} = ar^4 = p \quad ...(1)

a_8 = ar^{8-1} = ar^7 = q \quad ...(2)

a_{11} = ar^{11-1} = ar^{10} = s \quad ...(3)

Now, let's evaluate the Left Hand Side (LHS) and Right Hand Side (RHS) of the equation to be proved.

LHS =

q^2

From equation (2),

q = ar^7

. So, LHS =

(ar^7)^2 = a^2 r^{14}

RHS =

ps

From equations (1) and (3),

p = ar^4

and

s = ar^{10}

. So, RHS =

(ar^4)(ar^{10}) = a^{1+1} r^{4+10} = a^2 r^{14}

Since LHS =

a^2 r^{14}

and RHS =

a^2 r^{14}

, we have LHS = RHS.

Hence Proved.

Q4EXERCISE 8.2

The $4^{\text {th }}$ term of a G.P. is square of its second term, and the first term is -3 . Determine its $7^{\text {th }}$ term.

Solution

Given: The first term of a G.P. is

a = -3

. The

4^{\text{th}}

term is the square of the

2^{\text{nd}}

term, i.e.,

a_4 = (a_2)^2

.

To Find: The

7^{\text{th}}

term (

a_7

).

Formula for the $n^{\text{th}}$ term of a G.P.:

a_n = ar^{n-1}

Solution: First, we use the given condition to find the common ratio

r

.

a_4 = ar^{4-1} = ar^3

a_2 = ar^{2-1} = ar

Given condition:

a_4 = (a_2)^2

ar^3 = (ar)^2

ar^3 = a^2 r^2

Since

a=-3 \neq 0

and we assume

r \neq 0

for a G.P., we can divide both sides by

ar^2

:

r = a

We are given

a = -3

, so the common ratio is

r = -3

.

Now, we can find the

7^{\text{th}}

term:

a_7 = ar^{7-1} = ar^6

a_7 = (-3)(-3)^6 = (-3)^1 (-3)^6 = (-3)^{1+6} = (-3)^7

a_7 = -2187

Final Answer: The

7^{\text{th}}

term of the G.P. is -2187.

Q5EXERCISE 8.2

Which term of the following sequences:

(a)

2,2 \sqrt{2}, 4, \ldots

is 128 ?

(b)

\sqrt{3}, 3,3 \sqrt{3}, \ldots

is 729 ?

(c)

\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots

is

\frac{1}{19683}

?

Solution

Solution: We need to find the position 'n' of the given term in each sequence.

(a) Sequence: $2, 2\sqrt{2}, 4, \ldots$ is 128?

Given: First term

a = 2

. Common ratio

r = \frac{2\sqrt{2}}{2} = \sqrt{2}

. Let the

n^{\text{th}}

term be

a_n = 128

.

Formula:

a_n = ar^{n-1}

Calculation:

128 = 2(\sqrt{2})^{n-1}

\frac{128}{2} = (\sqrt{2})^{n-1}

64 = (2^{1/2})^{n-1}

2^6 = 2^{(n-1)/2}

Equating the exponents:

6 = \frac{n-1}{2}

12 = n-1

n = 13

Answer (a): 128 is the

13^{\text{th}}

term of the sequence.

(b) Sequence: $\sqrt{3}, 3, 3\sqrt{3}, \ldots$ is 729?

Given: First term

a = \sqrt{3}

. Common ratio

r = \frac{3}{\sqrt{3}} = \sqrt{3}

. Let the

n^{\text{th}}

term be

a_n = 729

.

Formula:

a_n = ar^{n-1}

Calculation:

729 = \sqrt{3}(\sqrt{3})^{n-1}

729 = (\sqrt{3})^{1 + n-1} = (\sqrt{3})^n

729 = (3^{1/2})^n = 3^{n/2}

Since

729 = 3^6

, we have:

3^6 = 3^{n/2}

Equating the exponents:

6 = \frac{n}{2}

n = 12

Answer (b): 729 is the

12^{\text{th}}

term of the sequence.

(c) Sequence: $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ is $\frac{1}{19683}$ ?

Given: First term

a = \frac{1}{3}

. Common ratio

r = \frac{1/9}{1/3} = \frac{1}{9} \times 3 = \frac{1}{3}

. Let the

n^{\text{th}}

term be

a_n = \frac{1}{19683}

.

Formula:

a_n = ar^{n-1}

Calculation:

\frac{1}{19683} = \left(\frac{1}{3}\right) \left(\frac{1}{3}\right)^{n-1}

\frac{1}{19683} = \left(\frac{1}{3}\right)^n

Since

19683 = 3^9

, we have:

\frac{1}{3^9} = \left(\frac{1}{3}\right)^n

\left(\frac{1}{3}\right)^9 = \left(\frac{1}{3}\right)^n

Equating the exponents:

n = 9

Answer (c):

\frac{1}{19683}

is the

9^{\text{th}}

term of the sequence.

Q6EXERCISE 8.2

For what values of $x$ , the numbers $-\frac{2}{7}, x,-\frac{7}{2}$ are in G.P.?

Solution

Given: The numbers

-\frac{2}{7}, x, -\frac{7}{2}

are in G.P.

To Find: The value(s) of

x

.

Concept: If three numbers

a, b, c

are in G.P., then the ratio of consecutive terms is constant, i.e.,

\frac{b}{a} = \frac{c}{b}

. This implies

b^2 = ac

.

Solution: Here,

a = -\frac{2}{7}

,

b = x

, and

c = -\frac{7}{2}

. Using the property of G.P., we have:

x^2 = \left(-\frac{2}{7}\right) \times \left(-\frac{7}{2}\right)

x^2 = 1

Taking the square root of both sides:

x = \pm \sqrt{1}

x = \pm 1

So, the possible values for

x

are 1 and -1.

Final Answer: The values of

x

for which the numbers are in G.P. are

x = 1

and

x = -1

.

Q7EXERCISE 8.2

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 7. $0.15,0.015,0.0015, \ldots 20$ terms.

Solution

Given: The G.P. is

0.15, 0.015, 0.0015, \ldots

Number of terms

n=20

.

To Find: The sum of the first 20 terms (

S_{20}

).

Solution: First term

a = 0.15

. Common ratio

r = \frac{0.015}{0.15} = \frac{15/1000}{15/100} = \frac{15}{1000} \times \frac{100}{15} = \frac{1}{10} = 0.1

. Since

|r| < 1

, we use the formula for the sum of a G.P.:

S_n = \frac{a(1-r^n)}{1-r}

Substitute the values

a=0.15

,

r=0.1

, and

n=20

:

S_{20} = \frac{0.15(1 - (0.1)^{20})}{1 - 0.1}

S_{20} = \frac{0.15(1 - (0.1)^{20})}{0.9}

S_{20} = \frac{15}{90}(1 - (0.1)^{20})

S_{20} = \frac{1}{6}(1 - (0.1)^{20})

Final Answer: The sum of the first 20 terms is

\frac{1}{6}(1 - (0.1)^{20})

.

Q8EXERCISE 8.2

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 8. $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots n$ terms.

Solution

Given: The G.P. is

\sqrt{7}, \sqrt{21}, 3\sqrt{7}, \ldots

Number of terms is

n

.

To Find: The sum of the first

n

terms (

S_n

).

Solution: First term

a = \sqrt{7}

. Common ratio

r = \frac{\sqrt{21}}{\sqrt{7}} = \sqrt{\frac{21}{7}} = \sqrt{3}

. Since

|r| > 1

, we use the formula for the sum of a G.P.:

S_n = \frac{a(r^n-1)}{r-1}

Substitute the values

a=\sqrt{7}

and

r=\sqrt{3}

:

S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)}{\sqrt{3} - 1}

To rationalize the denominator, we multiply the numerator and denominator by

(\sqrt{3}+1)

:

S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)}{\sqrt{3} - 1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}

S_n = \frac{\sqrt{7}(\sqrt{3}+1)((\sqrt{3})^n - 1)}{(\sqrt{3})^2 - 1^2}

S_n = \frac{\sqrt{7}(\sqrt{3}+1)(3^{n/2} - 1)}{3 - 1}

S_n = \frac{\sqrt{7}(\sqrt{3}+1)(3^{n/2} - 1)}{2}

Final Answer: The sum of the first

n

terms is

\frac{\sqrt{7}(\sqrt{3}+1)(3^{n/2} - 1)}{2}

.

Q9EXERCISE 8.2

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 9. $1,-a, a^{2},-a^{3}, \ldots n$ terms (if $a \neq-1$ ).

Solution

Given: The G.P. is

1, -a, a^2, -a^3, \ldots

Number of terms is

n

. Condition:

a \neq -1

.

To Find: The sum of the first

n

terms (

S_n

).

Solution: First term is

A = 1

. Common ratio

r = \frac{-a}{1} = -a

.

Since

a \neq -1

, the common ratio

r = -a \neq 1

. We use the formula for the sum of a G.P.:

S_n = \frac{A(1-r^n)}{1-r}

Substitute the values

A=1

and

r=-a

:

S_n = \frac{1(1 - (-a)^n)}{1 - (-a)}

S_n = \frac{1 - (-a)^n}{1 + a}

Final Answer: The sum of the first

n

terms is

\frac{1 - (-a)^n}{1 + a}

.

Q10EXERCISE 8.2

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 10. $x^{3}, x^{5}, x^{7}, \ldots n$ terms (if $x \neq \pm 1$ ).

Solution

Given: The G.P. is

x^3, x^5, x^7, \ldots

Number of terms is

n

. Condition:

x \neq \pm 1

.

To Find: The sum of the first

n

terms (

S_n

).

Solution: First term

a = x^3

. Common ratio

r = \frac{x^5}{x^3} = x^2

.

Since

x \neq \pm 1

, the common ratio

r = x^2 \neq 1

. We use the formula for the sum of a G.P.:

S_n = \frac{a(r^n-1)}{r-1}

Substitute the values

a=x^3

and

r=x^2

:

S_n = \frac{x^3((x^2)^n - 1)}{x^2 - 1}

S_n = \frac{x^3(x^{2n} - 1)}{x^2 - 1}

Final Answer: The sum of the first

n

terms is

\frac{x^3(x^{2n} - 1)}{x^2 - 1}

.

Q11EXERCISE 8.2

Evaluate $\sum_{k=1}^{11}\left(2+3^{k}\right)$ .

Solution

Given: The summation

\sum_{k=1}^{11} (2+3^k)

.

To Find: The value of the summation.

Solution: We can split the summation into two parts using the properties of sigma notation:

\sum_{k=1}^{11} (2+3^k) = \sum_{k=1}^{11} 2 + \sum_{k=1}^{11} 3^k

Part 1: $\sum_{k=1}^{11} 2$ This is the sum of the constant 2, repeated 11 times.

\sum_{k=1}^{11} 2 = 2 \times 11 = 22

Part 2: $\sum_{k=1}^{11} 3^k$ This is the sum of a geometric series:

3^1 + 3^2 + 3^3 + \ldots + 3^{11}

. This is a G.P. with: First term

a = 3

. Common ratio

r = 3

. Number of terms

n = 11

.

We use the formula for the sum of a G.P.,

S_n = \frac{a(r^n-1)}{r-1}

:

S_{11} = \frac{3(3^{11}-1)}{3-1} = \frac{3(3^{11}-1)}{2}

Combining the parts:

\sum_{k=1}^{11} (2+3^k) = 22 + \frac{3(3^{11}-1)}{2}

Let's calculate

3^{11}

:

3^5 = 243

3^{10} = (3^5)^2 = 243^2 = 59049

3^{11} = 3^{10} \times 3 = 59049 \times 3 = 177147

Now substitute this value back:

= 22 + \frac{3(177147-1)}{2}

= 22 + \frac{3(177146)}{2}

= 22 + 3(88573)

= 22 + 265719

= 265741

Final Answer: The value of the summation is

22 + \frac{3}{2}(3^{11}-1) = 265741

.

Q12EXERCISE 8.2

The sum of first three terms of a G.P. is $\frac{39}{10}$ and their product is 1 . Find the common ratio and the terms.

Solution

Given: Sum of the first three terms of a G.P. is

\frac{39}{10}

. Product of the first three terms is 1.

To Find: The common ratio and the terms of the G.P.

Solution: Let the three terms of the G.P. be

\frac{a}{r}, a, ar

. This choice simplifies the calculation for the product.

From the product:

(\frac{a}{r}) \times (a) \times (ar) = 1

a^3 = 1

a = 1

From the sum:

\frac{a}{r} + a + ar = \frac{39}{10}

Substitute

a=1

into the sum equation:

\frac{1}{r} + 1 + r = \frac{39}{10}

To solve for

r

, we first get a common denominator:

\frac{1 + r + r^2}{r} = \frac{39}{10}

Cross-multiply:

10(1 + r + r^2) = 39r

10 + 10r + 10r^2 = 39r

10r^2 + 10r - 39r + 10 = 0

10r^2 - 29r + 10 = 0

This is a quadratic equation in

r

. We can solve it by factorization.

10r^2 - 25r - 4r + 10 = 0

5r(2r - 5) - 2(2r - 5) = 0

(5r - 2)(2r - 5) = 0

This gives two possible values for

r

:

5r - 2 = 0 \implies r = \frac{2}{5}

2r - 5 = 0 \implies r = \frac{5}{2}

Case 1: $r = \frac{2}{5}$ The terms are

\frac{a}{r}, a, ar

. With

a=1

and

r=\frac{2}{5}

:

\frac{1}{2/5}, 1, 1(\frac{2}{5}) \implies \frac{5}{2}, 1, \frac{2}{5}

Case 2: $r = \frac{5}{2}$ The terms are

\frac{a}{r}, a, ar

. With

a=1

and

r=\frac{5}{2}

:

\frac{1}{5/2}, 1, 1(\frac{5}{2}) \implies \frac{2}{5}, 1, \frac{5}{2}

In both cases, the set of terms is the same.

Final Answer: The common ratio can be

\frac{2}{5}

or

\frac{5}{2}

. The terms are

\frac{2}{5}, 1, \frac{5}{2}

.

Q13EXERCISE 8.2

How many terms of G.P. $3,3^{2}, 3^{3}, \ldots$ are needed to give the sum 120 ?

Solution

Given: The G.P. is

3, 3^2, 3^3, \ldots

The sum of

n

terms is

S_n = 120

.

To Find: The number of terms,

n

.

Solution: From the G.P., we can identify: First term

a = 3

. Common ratio

r = \frac{3^2}{3} = 3

.

Since

r > 1

, we use the formula for the sum of a G.P.:

S_n = \frac{a(r^n-1)}{r-1}

Substitute the given values into the formula:

120 = \frac{3(3^n - 1)}{3 - 1}

120 = \frac{3(3^n - 1)}{2}

Now, solve for

n

:

120 \times 2 = 3(3^n - 1)

240 = 3(3^n - 1)

\frac{240}{3} = 3^n - 1

80 = 3^n - 1

81 = 3^n

Since

81 = 3^4

, we have:

3^4 = 3^n

Equating the exponents:

n = 4

Final Answer: 4 terms are needed to give the sum 120.

Q14EXERCISE 8.2

The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to $n$ terms of the G.P.

Solution

Given: Sum of the first three terms of a G.P. is 16. Sum of the next three terms is 128.

To Find: The first term (

a

), the common ratio (

r

), and the sum to

n

terms (

S_n

).

Solution: Let the G.P. be

a, ar, ar^2, ar^3, ar^4, ar^5, \ldots

Sum of the first three terms:

a + ar + ar^2 = 16

a(1 + r + r^2) = 16 \quad ...(1)

Sum of the next three terms (i.e., 4th, 5th, and 6th terms):

ar^3 + ar^4 + ar^5 = 128

ar^3(1 + r + r^2) = 128 \quad ...(2)

Now we have a system of two equations. Divide equation (2) by equation (1):

\frac{ar^3(1 + r + r^2)}{a(1 + r + r^2)} = \frac{128}{16}

r^3 = 8

r = 2

Now, substitute

r=2

back into equation (1) to find

a

:

a(1 + 2 + 2^2) = 16

a(1 + 2 + 4) = 16

a(7) = 16

a = \frac{16}{7}

Now we need to find the sum to

n

terms,

S_n

. We use the formula:

S_n = \frac{a(r^n-1)}{r-1}

Substitute the values of

a

and

r

we found:

S_n = \frac{\frac{16}{7}(2^n - 1)}{2 - 1}

S_n = \frac{\frac{16}{7}(2^n - 1)}{1}

S_n = \frac{16}{7}(2^n - 1)

Final Answer: The first term is

a = \frac{16}{7}

. The common ratio is

r = 2

. The sum to

n

terms is

S_n = \frac{16}{7}(2^n - 1)

.

Q15EXERCISE 8.2

Given a G.P. with $a=729$ and $7^{\text {th }}$ term 64, determine $\mathrm{S}_{7}$ .

Solution

Given: A G.P. with the first term

a = 729

. The

7^{\text{th}}

term is

a_7 = 64

.

To Find: The sum of the first 7 terms,

S_7

.

Solution: First, we need to find the common ratio

r

.

Formula for the $n^{\text{th}}$ term:

a_n = ar^{n-1}

For the

7^{\text{th}}

term:

a_7 = ar^{7-1} = ar^6

64 = 729 \times r^6

r^6 = \frac{64}{729}

We can write 64 as

2^6

and 729 as

3^6

.

r^6 = \frac{2^6}{3^6} = \left(\frac{2}{3}\right)^6

This gives

r = \frac{2}{3}

.

Now, we can find the sum of the first 7 terms,

S_7

. Since

|r| = \frac{2}{3} < 1

, we use the formula:

S_n = \frac{a(1-r^n)}{1-r}

Substitute

a=729

,

r=\frac{2}{3}

, and

n=7

:

S_7 = \frac{729 \left(1 - \left(\frac{2}{3}\right)^7\right)}{1 - \frac{2}{3}}

S_7 = \frac{729 \left(1 - \frac{128}{2187}\right)}{\frac{1}{3}}

S_7 = 3 \times 729 \left(\frac{2187 - 128}{2187}\right)

S_7 = 2187 \left(\frac{2059}{2187}\right)

S_7 = 2059

Final Answer: The sum of the first 7 terms,

S_7

, is 2059.

Q16EXERCISE 8.2

Find a G.P. for which sum of the first two terms is -4 and the fifth term is 4 times the third term.

Solution

Given: Sum of the first two terms is -4. Fifth term is 4 times the third term.

To Find: The G.P.

Solution: Let the first term be

a

and the common ratio be

r

.

From the second condition:

a_5 = 4 \times a_3

ar^{5-1} = 4 \times ar^{3-1}

ar^4 = 4ar^2

Assuming

a \neq 0

and

r \neq 0

, we can divide both sides by

ar^2

:

r^2 = 4

r = \pm 2

Now we use the first condition: Sum of the first two terms is

a + ar = -4

.

a(1+r) = -4

We have two cases for

r

.

Case 1: $r = 2$ Substitute

r=2

into

a(1+r) = -4

:

a(1+2) = -4

3a = -4

a = -\frac{4}{3}

The G.P. is

a, ar, ar^2, \ldots

-\frac{4}{3}, -\frac{4}{3}(2), -\frac{4}{3}(2)^2, \ldots

-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \ldots

Case 2: $r = -2$ Substitute

r=-2

into

a(1+r) = -4

:

a(1 + (-2)) = -4

a(-1) = -4

a = 4

The G.P. is

a, ar, ar^2, \ldots

4, 4(-2), 4(-2)^2, \ldots

4, -8, 16, \ldots

There are two possible G.P.s that satisfy the conditions.

Final Answer: The G.P. can be

-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \ldots

or

4, -8, 16, \ldots

.

Q17EXERCISE 8.2

If the $4^{\text {th }}, 10^{\text {th }}$ and $16^{\text {th }}$ terms of a G.P. are $x, y$ and $z$ , respectively. Prove that $x, y, z$ are in G.P.

Solution

Given: In a G.P.,

a_4 = x

,

a_{10} = y

, and

a_{16} = z

.

To Prove:

x, y, z

are in G.P.

Proof: To prove that

x, y, z

are in G.P., we need to show that the ratio of consecutive terms is constant, i.e.,

\frac{y}{x} = \frac{z}{y}

, which is equivalent to showing

y^2 = xz

.

Let the first term of the original G.P. be

a

and the common ratio be

r

.

Express

x, y, z

in terms of

a

and

r

:

x = a_4 = ar^{4-1} = ar^3 \quad ...(1)

y = a_{10} = ar^{10-1} = ar^9 \quad ...(2)

z = a_{16} = ar^{16-1} = ar^{15} \quad ...(3)

Now, let's evaluate the LHS and RHS of the condition

y^2 = xz

.

LHS =

y^2

From equation (2),

y = ar^9

. So, LHS =

(ar^9)^2 = a^2 r^{18}

RHS =

xz

From equations (1) and (3),

x = ar^3

and

z = ar^{15}

. So, RHS =

(ar^3)(ar^{15}) = a^{1+1} r^{3+15} = a^2 r^{18}

Since LHS =

a^2 r^{18}

and RHS =

a^2 r^{18}

, we have LHS = RHS. Thus,

y^2 = xz

.

This shows that the numbers

x, y, z

form a geometric progression.

Hence Proved.

Q18EXERCISE 8.2

Find the sum to $n$ terms of the sequence, $8,88,888,8888 \ldots$ .

Solution

Given: The sequence

8, 88, 888, 8888, \ldots

To Find: The sum of the first

n

terms,

S_n

.

Solution: This sequence is not a G.P. directly, but we can express it in terms of a G.P. Let

S_n

be the sum of the first

n

terms:

S_n = 8 + 88 + 888 + 8888 + \ldots

(to

n

terms)

Step 1: Factor out the common digit.

S_n = 8(1 + 11 + 111 + 1111 + \ldots

to

n

terms)

Step 2: Multiply and divide by 9 to create terms of the form

(10^k - 1)

.

S_n = \frac{8}{9}(9 + 99 + 999 + 9999 + \ldots

to

n

terms)

Step 3: Express each term as a power of 10 minus 1.

S_n = \frac{8}{9}((10-1) + (10^2-1) + (10^3-1) + (10^4-1) + \ldots

to

n

terms)

Step 4: Group the powers of 10 and the -1s separately.

S_n = \frac{8}{9} [ (10 + 10^2 + 10^3 + \ldots + 10^n) - (1 + 1 + 1 + \ldots \text{to } n \text{ terms}) ]

Step 5: The first group is a G.P. and the second is a simple sum. The series

10, 10^2, 10^3, \ldots, 10^n

is a G.P. with first term

a=10

, common ratio

r=10

, and

n

terms. Its sum is

\frac{a(r^n-1)}{r-1} = \frac{10(10^n-1)}{10-1} = \frac{10(10^n-1)}{9}

.

The sum of

(1 + 1 + 1 + \ldots \text{to } n \text{ terms})

is simply

n

.

Step 6: Substitute these sums back into the expression for

S_n

.

S_n = \frac{8}{9} \left[ \frac{10(10^n-1)}{9} - n \right]

This can be simplified further:

S_n = \frac{80}{81}(10^n-1) - \frac{8n}{9}

Final Answer: The sum to

n

terms of the sequence is

\frac{8}{9} \left[ \frac{10(10^n-1)}{9} - n \right]

.

Q19EXERCISE 8.2

Find the sum of the products of the corresponding terms of the sequences $2,4,8, 16,32$ and $128,32,8,2, \frac{1}{2}$ .

Solution

Given: Sequence 1:

2, 4, 8, 16, 32

Sequence 2:

128, 32, 8, 2, \frac{1}{2}

To Find: The sum of the products of the corresponding terms.

Solution: First, let's find the products of the corresponding terms. 1st term product:

2 \times 128 = 256

2nd term product:

4 \times 32 = 128

3rd term product:

8 \times 8 = 64

4th term product:

16 \times 2 = 32

5th term product:

32 \times \frac{1}{2} = 16

The new sequence formed by the products is:

256, 128, 64, 32, 16

.

We need to find the sum of this new sequence. Let's check if it is a G.P. First term

a = 256

. Common ratio

r = \frac{128}{256} = \frac{1}{2}

. The ratio

\frac{64}{128}

is also

\frac{1}{2}

, so this is a G.P. Number of terms

n = 5

.

We use the formula for the sum of a G.P. with

|r|<1

:

S_n = \frac{a(1-r^n)}{1-r}

Substitute the values

a=256

,

r=\frac{1}{2}

, and

n=5

:

S_5 = \frac{256\left(1 - \left(\frac{1}{2}\right)^5\right)}{1 - \frac{1}{2}}

S_5 = \frac{256\left(1 - \frac{1}{32}\right)}{\frac{1}{2}}

S_5 = 2 \times 256\left(\frac{32-1}{32}\right)

S_5 = 512\left(\frac{31}{32}\right)

S_5 = \frac{512}{32} \times 31

S_5 = 16 \times 31

S_5 = 496

Final Answer: The sum of the products of the corresponding terms is 496.

Q20EXERCISE 8.2

Show that the products of the corresponding terms of the sequences $a, a r, a r^{2}, \ldots a r^{n-1}$ and $\mathrm{A}, \mathrm{AR}, \mathrm{AR}^{2}, \ldots \mathrm{AR}^{n-1}$ form a G.P, and find the common ratio.

Solution

Given: Sequence 1:

a, ar, ar^2, \ldots, ar^{n-1}

(a G.P. with first term

a

and common ratio

r

) Sequence 2:

A, AR, AR^2, \ldots, AR^{n-1}

(a G.P. with first term

A

and common ratio

R

)

To Show: The sequence of products of corresponding terms forms a G.P. and to find its common ratio.

Proof: Let's form the new sequence by multiplying the corresponding terms of the two given sequences. Let the new sequence be denoted by

c_1, c_2, c_3, \ldots, c_n

.

1st term:

c_1 = a \times A = aA

2nd term:

c_2 = (ar) \times (AR) = aA(rR)

3rd term:

c_3 = (ar^2) \times (AR^2) = aA(r^2R^2) = aA(rR)^2

...

k^{\text{th}}

term:

c_k = (ar^{k-1}) \times (AR^{k-1}) = aA(r^{k-1}R^{k-1}) = aA(rR)^{k-1}

So, the new sequence is:

aA, aA(rR), aA(rR)^2, \ldots, aA(rR)^{n-1}

.

To check if this sequence is a G.P., we need to see if the ratio of any term to its preceding term is constant. Let's find the ratio of the

(k+1)^{\text{th}}

term to the

k^{\text{th}}

term:

\frac{c_{k+1}}{c_k} = \frac{aA(rR)^{(k+1)-1}}{aA(rR)^{k-1}} = \frac{aA(rR)^k}{aA(rR)^{k-1}} = (rR)^{k - (k-1)} = rR

Since the ratio

\frac{c_{k+1}}{c_k}

is equal to the constant value

rR

for all

k \geq 1

, the resulting sequence is a G.P.

The first term of this new G.P. is

aA

. The common ratio of this new G.P. is

rR

.

Hence Proved.

Final Answer: The sequence of products forms a G.P. with the common ratio

rR

.

Q21EXERCISE 8.2

Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the $4^{\text {th }}$ by 18.

Solution

Given: Four numbers are in a G.P. Third term is greater than the first term by 9. Second term is greater than the fourth term by 18.

To Find: The four numbers.

Solution: Let the four numbers in G.P. be

a, ar, ar^2, ar^3

.

From the given conditions, we can form two equations:

Condition 1:

a_3 = a_1 + 9

ar^2 = a + 9

a(r^2 - 1) = 9 \quad ...(1)

Condition 2:

a_2 = a_4 + 18

ar = ar^3 + 18

ar - ar^3 = 18

ar(1 - r^2) = 18 \quad ...(2)

Notice that

(1 - r^2) = -(r^2 - 1)

. We can rewrite equation (2) as:

-ar(r^2 - 1) = 18 \quad ...(3)

Now, we can substitute the value of

a(r^2 - 1)

from equation (1) into equation (3):

-r [a(r^2 - 1)] = 18

-r(9) = 18

-9r = 18

r = -2

Now substitute the value of

r = -2

back into equation (1) to find

a

:

a((-2)^2 - 1) = 9

a(4 - 1) = 9

3a = 9

a = 3

So, the first term is

a=3

and the common ratio is

r=-2

.

The four numbers are:

a_1 = a = 3

a_2 = ar = 3(-2) = -6

a_3 = ar^2 = 3(-2)^2 = 3(4) = 12

a_4 = ar^3 = 3(-2)^3 = 3(-8) = -24

Let's check the conditions:

a_3 = 12

,

a_1 = 3

.

12 = 3 + 9

. (Correct)

a_2 = -6

,

a_4 = -24

.

-6 = -24 + 18

. (Correct)

Final Answer: The four numbers are 3, -6, 12, -24.

Q22EXERCISE 8.2

If the $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of a G.P. are $a, b$ and $c$ , respectively. Prove that $a^{q-r} b^{r-p} c^{p-q}=1 .$

Solution

Given: In a G.P., the

p^{\text{th}}

term is

a

, the

q^{\text{th}}

term is

b

, and the

r^{\text{th}}

term is

c

.

To Prove:

a^{q-r} b^{r-p} c^{p-q} = 1

.

Proof: Let the first term of the G.P. be

A

and the common ratio be

R

. (We use

A

and

R

to avoid confusion with the given variables

a,b,c,r

).

The

n^{\text{th}}

term is given by

A_n = AR^{n-1}

.

According to the given information:

a = A_p = AR^{p-1} \quad ...(1)

b = A_q = AR^{q-1} \quad ...(2)

c = A_r = AR^{r-1} \quad ...(3)

Now, let's evaluate the Left Hand Side (LHS) of the equation to be proved: LHS =

a^{q-r} b^{r-p} c^{p-q}

Substitute the expressions for

a, b, c

from (1), (2), and (3): LHS =

(AR^{p-1})^{q-r} (AR^{q-1})^{r-p} (AR^{r-1})^{p-q}

Using the rule

(xy)^m = x^m y^m

and

(x^m)^n = x^{mn}

: LHS =

(A^{q-r} R^{(p-1)(q-r)}) (A^{r-p} R^{(q-1)(r-p)}) (A^{p-q} R^{(r-1)(p-q)})

Group the terms with base

A

and base

R

together: LHS =

A^{(q-r) + (r-p) + (p-q)} \times R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)}

Now, let's simplify the exponents.

Exponent of A:

(q-r) + (r-p) + (p-q) = q - r + r - p + p - q = 0

So,

A^0 = 1

.

Exponent of R:

(p-1)(q-r) = pq - pr - q + r

(q-1)(r-p) = qr - qp - r + p

(r-1)(p-q) = rp - rq - p + q

Sum of these three expressions:

(pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q)

= (pq - qp) + (-pr + rp) + (-q + q) + (r - r) + (qr - rq) + (p - p)

= 0 + 0 + 0 + 0 + 0 + 0 = 0

So,

R^0 = 1

.

Therefore, the LHS becomes: LHS =

A^0 \times R^0 = 1 \times 1 = 1

LHS = RHS.

Hence Proved.

Q23EXERCISE 8.2

If the first and the $n^{\text {th }}$ term of a G.P. are $a$ and $b$ , respectively, and if P is the product of $n$ terms, prove that $\mathrm{P}^{2}=(a b)^{n}$ .

Solution

Given: First term of a G.P. is

a_1 = a

.

n^{\text{th}}

term of the G.P. is

a_n = b

. P is the product of the first

n

terms.

To Prove:

P^2 = (ab)^n

.

Proof: Let the common ratio of the G.P. be

r

. We are given

a_n = b

. Using the formula

a_n = a_1 r^{n-1}

, we have:

b = ar^{n-1} \quad ...(1)

The product P of the first

n

terms is:

P = a_1 \times a_2 \times a_3 \times \ldots \times a_n

P = (a) \times (ar) \times (ar^2) \times \ldots \times (ar^{n-1})

Group the 'a' terms and 'r' terms:

P = (a \times a \times \ldots \times a \text{ (n times)}) \times (r^0 \times r^1 \times r^2 \times \ldots \times r^{n-1})

P = a^n \times r^{0+1+2+\ldots+(n-1)}

The sum of the exponents of

r

is an arithmetic series

0+1+2+\ldots+(n-1)

. The sum of the first

k

natural numbers is

\frac{k(k+1)}{2}

. Here we have the sum up to

n-1

, so the sum is

\frac{(n-1)(n-1+1)}{2} = \frac{n(n-1)}{2}

.

So, the product P is:

P = a^n r^{\frac{n(n-1)}{2}}

Now, let's evaluate the LHS and RHS of the equation to be proved,

P^2 = (ab)^n

.

LHS =

P^2

P^2 = \left(a^n r^{\frac{n(n-1)}{2}}\right)^2

P^2 = (a^n)^2 \times (r^{\frac{n(n-1)}{2}})^2

P^2 = a^{2n} r^{n(n-1)} \quad ...(2)

RHS =

(ab)^n

Substitute the expression for

b

from equation (1):

RHS = (a \times (ar^{n-1}))^n

RHS = (a^2 r^{n-1})^n

RHS = (a^2)^n (r^{n-1})^n

RHS = a^{2n} r^{n(n-1)} \quad ...(3)

From (2) and (3), we see that LHS = RHS.

Hence Proved.

Q24EXERCISE 8.2

Show that the ratio of the sum of first $n$ terms of a G.P. to the sum of terms from $(n+1)^{\text {th }}$ to $(2 n)^{\text {th }}$ term is $\frac{1}{r^{n}}$ .

Solution

To Show: The ratio of the sum of first

n

terms of a G.P. to the sum of terms from

(n+1)^{\text{th}}

to

(2n)^{\text{th}}

term is

\frac{1}{r^n}

.

Proof: Let the G.P. have the first term

a

and common ratio

r

(assuming

r \neq 1

).

Part 1: Sum of the first $n$ terms ( $S_n$ ) The sum of the first

n

terms is given by the formula:

S_n = \frac{a(r^n - 1)}{r-1} \quad ...(1)

Part 2: Sum of terms from $(n+1)^{\text{th}}$ to $(2n)^{\text{th}}$ term Let this sum be denoted by

S'

.

S' = a_{n+1} + a_{n+2} + \ldots + a_{2n}

This is a G.P. with: First term:

a_{n+1} = ar^{(n+1)-1} = ar^n

Number of terms:

(2n) - (n+1) + 1 = 2n - n - 1 + 1 = n

terms. Common ratio:

r

.

Using the sum formula for a G.P. with first term

ar^n

and

n

terms:

S' = \frac{(ar^n)(r^n - 1)}{r-1} \quad ...(2)

Part 3: The ratio We need to find the ratio

\frac{S_n}{S'}

.

\frac{S_n}{S'} = \frac{\frac{a(r^n - 1)}{r-1}}{\frac{ar^n(r^n - 1)}{r-1}}

Cancel the common terms

\frac{a(r^n-1)}{r-1}

from the numerator and denominator:

\frac{S_n}{S'} = \frac{1}{r^n}

This is the required ratio.

Alternative method for S': The sum of terms from

(n+1)^{\text{th}}

to

(2n)^{\text{th}}

can also be calculated as the sum of the first

2n

terms minus the sum of the first

n

terms.

S' = S_{2n} - S_n

S' = \frac{a(r^{2n} - 1)}{r-1} - \frac{a(r^n - 1)}{r-1}

S' = \frac{a}{r-1} [ (r^{2n} - 1) - (r^n - 1) ]

S' = \frac{a}{r-1} [ r^{2n} - r^n ]

S' = \frac{a}{r-1} [ r^n(r^n - 1) ] = \frac{ar^n(r^n - 1)}{r-1}

This is the same result as equation (2), and the rest of the proof follows.

Hence Proved.

Q25EXERCISE 8.2

If $a, b, c$ and $d$ are in G.P. show that $\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$ .

Solution

Given:

a, b, c, d

are in G.P.

To Show:

(a^2+b^2+c^2)(b^2+c^2+d^2) = (ab+bc+cd)^2

.

Proof: Let the common ratio of the G.P. be

r

. Since

a, b, c, d

are in G.P., we can express

b, c, d

in terms of

a

and

r

:

b = ar

c = ar^2

d = ar^3

Now, we will evaluate the Left Hand Side (LHS) and Right Hand Side (RHS) separately.

LHS = $(a^2+b^2+c^2)(b^2+c^2+d^2)$ Substitute the expressions for

b, c, d

: LHS =

(a^2 + (ar)^2 + (ar^2)^2)((ar)^2 + (ar^2)^2 + (ar^3)^2)

LHS =

(a^2 + a^2r^2 + a^2r^4)(a^2r^2 + a^2r^4 + a^2r^6)

Factor out common terms from each bracket: LHS =

[a^2(1 + r^2 + r^4)][a^2r^2(1 + r^2 + r^4)]

LHS =

a^2 \cdot a^2r^2 (1 + r^2 + r^4)^2

LHS =

a^4r^2(1 + r^2 + r^4)^2 \quad ...(1)

RHS = $(ab+bc+cd)^2$ Substitute the expressions for

b, c, d

: RHS =

(a(ar) + (ar)(ar^2) + (ar^2)(ar^3))^2

RHS =

(a^2r + a^2r^3 + a^2r^5)^2

Factor out the common term

a^2r

from the bracket: RHS =

[a^2r(1 + r^2 + r^4)]^2

Apply the exponent to each factor inside the bracket: RHS =

(a^2r)^2 (1 + r^2 + r^4)^2

RHS =

a^4r^2 (1 + r^2 + r^4)^2 \quad ...(2)

Comparing equations (1) and (2), we see that LHS = RHS.

Hence Proved.

Q26EXERCISE 8.2

Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Solution

Given: We need to insert two numbers between 3 and 81 to form a G.P.

To Find: The two numbers.

Solution: Let the two numbers be

G_1

and

G_2

. The resulting G.P. will be

3, G_1, G_2, 81

. This G.P. has 4 terms. The first term is

a = 3

. The fourth term is

a_4 = 81

.

Let the common ratio be

r

. Using the formula for the

n^{\text{th}}

term,

a_n = ar^{n-1}

:

a_4 = ar^{4-1} = ar^3

81 = 3 \times r^3

\frac{81}{3} = r^3

27 = r^3

3^3 = r^3

r = 3

Now we can find the two numbers we inserted,

G_1

and

G_2

.

G_1

is the second term of the G.P.:

G_1 = a_2 = ar = 3 \times 3 = 9

G_2

is the third term of the G.P.:

G_2 = a_3 = ar^2 = 3 \times 3^2 = 3 \times 9 = 27

The resulting sequence is 3, 9, 27, 81, which is a G.P. with common ratio 3.

Final Answer: The two numbers to be inserted between 3 and 81 are 9 and 27.

Q27EXERCISE 8.2

Find the value of $n$ so that $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$ may be the geometric mean between $a$ and $b$ .

Solution

Given: The expression

\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}

is the geometric mean (G.M.) between

a

and

b

.

To Find: The value of

n

.

Concept: The geometric mean between two positive numbers

a

and

b

is

\sqrt{ab}

.

Solution: According to the problem statement:

\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}} = \sqrt{ab} = (ab)^{1/2} = a^{1/2}b^{1/2}

Now, we solve this equation for

n

. Cross-multiply:

a^{n+1} + b^{n+1} = (a^n + b^n)(a^{1/2}b^{1/2})

a^{n+1} + b^{n+1} = a^n \cdot a^{1/2}b^{1/2} + b^n \cdot a^{1/2}b^{1/2}

a^{n+1} + b^{n+1} = a^{n+1/2}b^{1/2} + a^{1/2}b^{n+1/2}

Rearrange the terms, bringing all terms with

a

to one side and terms with

b

to the other:

a^{n+1} - a^{n+1/2}b^{1/2} = a^{1/2}b^{n+1/2} - b^{n+1}

Factor out the lowest power of

a

from the LHS and the lowest power of

b

from the RHS. LHS:

a^{n+1/2}(a^{1/2} - b^{1/2})

RHS:

b^{n+1/2}(a^{1/2} - b^{1/2})

So the equation becomes:

a^{n+1/2}(a^{1/2} - b^{1/2}) = b^{n+1/2}(a^{1/2} - b^{1/2})

Assuming

a \neq b

, we have

a^{1/2} - b^{1/2} \neq 0

, so we can divide both sides by

(a^{1/2} - b^{1/2})

:

a^{n+1/2} = b^{n+1/2}

Divide both sides by

b^{n+1/2}

:

\frac{a^{n+1/2}}{b^{n+1/2}} = 1

\left(\frac{a}{b}\right)^{n+1/2} = 1

Since

a \neq b

,

\frac{a}{b} \neq 1

. Any non-one number raised to the power of 0 is 1. So, the exponent must be zero.

n + \frac{1}{2} = 0

n = -\frac{1}{2}

Final Answer: The value of

n

is

-\frac{1}{2}

.

Q28EXERCISE 8.2

The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio $(3+2 \sqrt{2}):(3-2 \sqrt{2})$ .

Solution

Given: Let the two numbers be

a

and

b

. Their sum is 6 times their geometric mean (G.M.).

To Show: The ratio of the numbers,

a:b

, is

(3+2\sqrt{2}) : (3-2\sqrt{2})

.

Proof: The geometric mean of

a

and

b

is

\sqrt{ab}

. According to the given condition:

a+b = 6\sqrt{ab} \quad ...(1)

Divide both sides by

\sqrt{ab}

(assuming

a,b > 0

):

\frac{a+b}{\sqrt{ab}} = 6

\frac{a}{\sqrt{ab}} + \frac{b}{\sqrt{ab}} = 6

\frac{\sqrt{a}}{\sqrt{b}} + \frac{\sqrt{b}}{\sqrt{a}} = 6

Let

x = \frac{\sqrt{a}}{\sqrt{b}}

. Then

\frac{\sqrt{b}}{\sqrt{a}} = \frac{1}{x}

. The equation becomes:

x + \frac{1}{x} = 6

Multiply by

x

to get a quadratic equation:

x^2 + 1 = 6x

x^2 - 6x + 1 = 0

We solve for

x

using the quadratic formula

x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}

:

x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(1)}}{2(1)}

x = \frac{6 \pm \sqrt{36 - 4}}{2}

x = \frac{6 \pm \sqrt{32}}{2}

x = \frac{6 \pm \sqrt{16 \times 2}}{2}

x = \frac{6 \pm 4\sqrt{2}}{2}

x = 3 \pm 2\sqrt{2}

So, we have two possible values for

x = \frac{\sqrt{a}}{\sqrt{b}}

.

Case 1:

\frac{\sqrt{a}}{\sqrt{b}} = 3 + 2\sqrt{2}

Squaring both sides:

\frac{a}{b} = (3 + 2\sqrt{2})^2

\frac{a}{b} = 3^2 + (2\sqrt{2})^2 + 2(3)(2\sqrt{2})

\frac{a}{b} = 9 + 8 + 12\sqrt{2} = 17 + 12\sqrt{2}

(This is not the required ratio)

Let's re-examine the problem. The question asks for the ratio

a:b

. We have found

\frac{\sqrt{a}}{\sqrt{b}}

. Let's use Componendo and Dividendo on equation (1) after rearranging it. From

a+b = 6\sqrt{ab}

, we can write

\frac{a+b}{2\sqrt{ab}} = 3

. Or

\frac{a+b}{2\sqrt{ab}} = \frac{3}{1}

.

Applying Componendo and Dividendo, which states if

\frac{p}{q} = \frac{r}{s}

then

\frac{p+q}{p-q} = \frac{r+s}{r-s}

:

\frac{(a+b) + 2\sqrt{ab}}{(a+b) - 2\sqrt{ab}} = \frac{3+1}{3-1}

The numerator is

(\sqrt{a} + \sqrt{b})^2

and the denominator is

(\sqrt{a} - \sqrt{b})^2

.

\frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = \frac{4}{2} = 2

Taking the square root of both sides:

\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \pm\sqrt{2}

Let's take the positive root first:

\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \frac{\sqrt{2}}{1}

Apply Componendo and Dividendo again:

\frac{(\sqrt{a} + \sqrt{b}) + (\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b}) - (\sqrt{a} - \sqrt{b})} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}

\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}

\frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}

Rationalize the RHS:

\frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{(\sqrt{2}+1)^2}{2-1} = (\sqrt{2}+1)^2

Squaring both sides to find

\frac{a}{b}

:

\frac{a}{b} = ((\sqrt{2}+1)^2)^2 = (\sqrt{2}+1)^4

This is also not the required form.

Let's go back to

x = 3 \pm 2\sqrt{2}

.

x = \frac{\sqrt{a}}{\sqrt{b}}

. So

\frac{a}{b} = x^2

. If

x = 3+2\sqrt{2}

, then

\frac{a}{b} = (3+2\sqrt{2})^2 = 9 + 8 + 12\sqrt{2} = 17+12\sqrt{2}

. If

x = 3-2\sqrt{2}

, then

\frac{a}{b} = (3-2\sqrt{2})^2 = 9 + 8 - 12\sqrt{2} = 17-12\sqrt{2}

. The ratio of the two numbers is

a:b

. The question asks to show the numbers are in the ratio... this means one number is

(3+2\sqrt{2})k

and the other is

(3-2\sqrt{2})k

for some constant

k

.

Let the numbers be

a = (3+2\sqrt{2})k

and

b = (3-2\sqrt{2})k

. Sum:

a+b = (3+2\sqrt{2})k + (3-2\sqrt{2})k = 6k

. Product:

ab = (3+2\sqrt{2})k \times (3-2\sqrt{2})k = (3^2 - (2\sqrt{2})^2)k^2 = (9-8)k^2 = k^2

. Geometric Mean:

\sqrt{ab} = \sqrt{k^2} = k

. Check the condition: Is

a+b = 6\sqrt{ab}

?

6k = 6(k)

. Yes, it is.

So, the numbers are in the required ratio. The derivation using Componendo-Dividendo was the standard way to show this. Let's re-check the C&D method.

rac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{2}+1}{\sqrt{2}-1}

. Squaring both sides:

\frac{a}{b} = \frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2} = \frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}} = \frac{3+2\sqrt{2}}{3-2\sqrt{2}}

. This works perfectly.

Let's write the proof cleanly.

Proof: Let the two numbers be

a

and

b

. Their geometric mean is

\sqrt{ab}

. Given:

a+b = 6\sqrt{ab}

.

\frac{a+b}{2\sqrt{ab}} = \frac{3}{1}

Apply Componendo and Dividendo

(\frac{x+y}{x-y} = \frac{m+n}{m-n}

if

\frac{x}{y} = \frac{m}{n})

:

\frac{(a+b) + 2\sqrt{ab}}{(a+b) - 2\sqrt{ab}} = \frac{3+1}{3-1}

\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \frac{4}{2} = 2

The numerator is

(\sqrt{a}+\sqrt{b})^2

and the denominator is

(\sqrt{a}-\sqrt{b})^2

.

\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2} = 2

Taking the square root of both sides:

\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \sqrt{2} = \frac{\sqrt{2}}{1}

Apply Componendo and Dividendo again:

\frac{(\sqrt{a}+\sqrt{b}) + (\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b}) - (\sqrt{a}-\sqrt{b})} = \frac{\sqrt{2}+1}{\sqrt{2}-1}

\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{2}+1}{\sqrt{2}-1}

\frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{2}+1}{\sqrt{2}-1}

Squaring both sides:

\frac{a}{b} = \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)^2 = \frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2}

\frac{a}{b} = \frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}} = \frac{3+2\sqrt{2}}{3-2\sqrt{2}}

Thus, the ratio

a:b

is

(3+2\sqrt{2}) : (3-2\sqrt{2})

.

Hence Proved.

Q29EXERCISE 8.2

If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are $\mathrm{A} \pm \sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}$ .

Solution

Given: Let the two positive numbers be

a

and

b

. A is the Arithmetic Mean (A.M.) of

a

and

b

. G is the Geometric Mean (G.M.) of

a

and

b

.

To Prove: The numbers

a

and

b

are

A \pm \sqrt{(A+G)(A-G)}

.

Proof: By definition of A.M. and G.M.:

A = \frac{a+b}{2} \implies a+b = 2A \quad ...(1)

G = \sqrt{ab} \implies ab = G^2 \quad ...(2)

Consider a quadratic equation with roots

a

and

b

. The equation can be written as:

x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0

x^2 - (a+b)x + ab = 0

Substitute the values from equations (1) and (2):

x^2 - 2Ax + G^2 = 0

The solutions to this quadratic equation will give the numbers

a

and

b

. We use the quadratic formula

x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A_{coeff}}

:

x = \frac{-(-2A) \pm \sqrt{(-2A)^2 - 4(1)(G^2)}}{2(1)}

x = \frac{2A \pm \sqrt{4A^2 - 4G^2}}{2}

x = \frac{2A \pm \sqrt{4(A^2 - G^2)}}{2}

x = \frac{2A \pm 2\sqrt{A^2 - G^2}}{2}

x = A \pm \sqrt{A^2 - G^2}

We can write the term under the square root using the difference of squares formula,

A^2 - G^2 = (A+G)(A-G)

. So,

x = A \pm \sqrt{(A+G)(A-G)}

.

The two roots of the equation are

A + \sqrt{(A+G)(A-G)}

and

A - \sqrt{(A+G)(A-G)}

. These roots are the numbers

a

and

b

.

Therefore, the two numbers are

A \pm \sqrt{(A+G)(A-G)}

.

Hence Proved.

Q30EXERCISE 8.2

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of $2^{\text {nd }}$ hour, $4^{\text {th }}$ hour and $n^{\text {th }}$ hour?

Solution

Given: Initial number of bacteria = 30. The number of bacteria doubles every hour.

To Find: Number of bacteria at the end of the 2nd hour. Number of bacteria at the end of the 4th hour. Number of bacteria at the end of the nth hour.

Solution: The number of bacteria forms a geometric progression (G.P.). Let's list the number of bacteria at the end of each hour:

Originally (at time t=0): 30 bacteria. End of 1st hour:

30 \times 2 = 60

bacteria. End of 2nd hour:

60 \times 2 = 120

bacteria. End of 3rd hour:

120 \times 2 = 240

bacteria.

The sequence is

30, 60, 120, 240, \ldots

This is a G.P. where: The first term (number after 1 hour) is

a = 60

. Or we can think of the initial amount as

a_0 = 30

. The amount after n hours would be

a_0 \times r^n

. Let's use the standard G.P. notation. Let

a_n

be the number of bacteria at the end of the

n^{\text{th}}

hour.

a_0 = 30

(initial amount)

a_1 = 30 \times 2^1 = 60

a_2 = 30 \times 2^2 = 120

a_n = 30 \times 2^n

Let's verify this. The sequence of population at the end of the hour is

a_1, a_2, ...

. The first term of this sequence is

a_1=60

. The common ratio is 2. The

k

-th term of this sequence (population at the end of hour

k

) is

a_k = a_1 r^{k-1} = 60 \times 2^{k-1} = (30 \times 2) \times 2^{k-1} = 30 \times 2^k

. This matches the formula derived above. So let's use

P(n) = 30 \times 2^n

where

P(n)

is the population at the end of the

n^{th}

hour.

Number of bacteria at the end of the 2nd hour: Substitute

n=2

into the formula

P(n) = 30 \times 2^n

.

P(2) = 30 \times 2^2 = 30 \times 4 = 120

Number of bacteria at the end of the 4th hour: Substitute

n=4

into the formula.

P(4) = 30 \times 2^4 = 30 \times 16 = 480

Number of bacteria at the end of the nth hour: The formula is already derived for the

n^{\text{th}}

hour.

P(n) = 30 \times 2^n

Final Answer: At the end of the 2nd hour, there will be 120 bacteria. At the end of the 4th hour, there will be 480 bacteria. At the end of the nth hour, there will be

30 \times 2^n

bacteria.

Q31EXERCISE 8.2

What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of $10\%$ compounded annually?

Solution

Given: Principal amount (P) = Rs 500. Annual interest rate (R) = 10% = 0.10. Time period (n) = 10 years. The interest is compounded annually.

To Find: The total amount (A) after 10 years.

Formula: The formula for the amount A after n years with compound interest is:

A = P(1 + R)^n

Solution: Substitute the given values into the formula:

P = 500

R = 0.10

n = 10

A = 500(1 + 0.10)^{10}

A = 500(1.1)^{10}

This calculation can be left in this form, or a calculator can be used for the final value. The sequence of amounts at the end of each year forms a G.P: Year 1:

500(1.1)

Year 2:

500(1.1)^2

... Year 10:

500(1.1)^{10}

This is the 10th term of a G.P. with first term

a = 500(1.1)

and common ratio

r=1.1

.

Calculating

(1.1)^{10}

:

(1.1)^2 = 1.21

(1.1)^4 = (1.21)^2 = 1.4641

(1.1)^8 = (1.4641)^2 \approx 2.14358881

(1.1)^{10} = (1.1)^8 \times (1.1)^2 \approx 2.14358881 \times 1.21 \approx 2.59374246

Now, calculate the final amount:

A = 500 \times 2.59374246 \approx 1296.87

Final Answer: Rs 500 will amount to Rs

500(1.1)^{10}

(approximately Rs 1296.87) in 10 years.

Q32EXERCISE 8.2

If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

Solution

Given: Let the roots of a quadratic equation be

\alpha

and

\beta

. Arithmetic Mean (A.M.) of the roots is 8. Geometric Mean (G.M.) of the roots is 5.

To Find: The quadratic equation.

Solution: From the definitions of A.M. and G.M.:

A.M. =

\frac{\alpha + \beta}{2} = 8

This implies the sum of the roots is

\alpha + \beta = 2 \times 8 = 16

.

G.M. =

\sqrt{\alpha \beta} = 5

This implies the product of the roots is

\alpha \beta = 5^2 = 25

.

Formula for a quadratic equation: A quadratic equation with roots

\alpha

and

\beta

can be written as:

x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0

x^2 - (\alpha + \beta)x + \alpha\beta = 0

Substitute the calculated sum and product of the roots:

x^2 - (16)x + (25) = 0

x^2 - 16x + 25 = 0

Final Answer: The required quadratic equation is

x^2 - 16x + 25 = 0

.

Q1Miscellaneous Exercise On Chapter 8

If $f$ is a function satisfying $f(x+y)=f(x) f(y)$ for all $x, y \in \mathbf{N}$ such that $f(1)=3$ and $\sum_{x=1}^{n} f(x)=120$ , find the value of $n$ .

Solution

Given: A function

f

satisfies

f(x+y) = f(x)f(y)

for all

x, y \in \mathbf{N}

.

f(1) = 3

.

\sum_{x=1}^{n} f(x) = 120

.

To Find: The value of

n

.

Solution: First, let's find the form of the function

f(x)

.

f(2) = f(1+1) = f(1)f(1) = 3 \times 3 = 3^2 = 9

.

f(3) = f(2+1) = f(2)f(1) = 3^2 \times 3 = 3^3 = 27

. By induction, we can see that

f(x) = 3^x

.

The sequence of values

f(1), f(2), f(3), \ldots, f(n)

is

3, 3^2, 3^3, \ldots, 3^n

. This is a geometric progression (G.P.).

Now, we are given the sum of this sequence:

\sum_{x=1}^{n} f(x) = f(1) + f(2) + \ldots + f(n) = 120

3 + 3^2 + 3^3 + \ldots + 3^n = 120

This is the sum of a G.P. with: First term

a = 3

. Common ratio

r = 3

. Number of terms =

n

. Sum

S_n = 120

.

We use the formula for the sum of a G.P.,

S_n = \frac{a(r^n-1)}{r-1}

:

120 = \frac{3(3^n - 1)}{3 - 1}

120 = \frac{3(3^n - 1)}{2}

Now, we solve for

n

:

120 \times 2 = 3(3^n - 1)

240 = 3(3^n - 1)

\frac{240}{3} = 3^n - 1

80 = 3^n - 1

81 = 3^n

Since

81 = 3^4

, we have:

3^4 = 3^n

Equating the exponents:

n = 4

Final Answer: The value of

n

is 4.

Q2Miscellaneous Exercise On Chapter 8

The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

Solution

Given: A G.P. with: First term

a = 5

. Common ratio

r = 2

. Sum of

n

terms

S_n = 315

.

To Find: The last term (

a_n

) and the number of terms (

n

).

Solution: First, we find the number of terms,

n

, using the sum formula for a G.P. Since

r=2 > 1

, we use

S_n = \frac{a(r^n-1)}{r-1}

.

315 = \frac{5(2^n - 1)}{2 - 1}

315 = \frac{5(2^n - 1)}{1}

315 = 5(2^n - 1)

Divide by 5:

\frac{315}{5} = 2^n - 1

63 = 2^n - 1

64 = 2^n

Since

64 = 2^6

, we have:

2^6 = 2^n

So,

n = 6

. The number of terms is 6.

Now, we find the last term, which is the

6^{\text{th}}

term (

a_6

). The formula for the

n^{\text{th}}

term is

a_n = ar^{n-1}

.

a_6 = 5 \times (2)^{6-1}

a_6 = 5 \times 2^5

a_6 = 5 \times 32

a_6 = 160

The last term is 160.

Final Answer: The last term is 160 and the number of terms is 6.

Q3Miscellaneous Exercise On Chapter 8

The first term of a G.P. is 1 . The sum of the third term and fifth term is 90 . Find the common ratio of G.P.

Solution

Given: A G.P. with: First term

a = 1

. Sum of the third and fifth terms is 90, i.e.,

a_3 + a_5 = 90

.

To Find: The common ratio,

r

.

Solution: The formula for the

n^{\text{th}}

term of a G.P. is

a_n = ar^{n-1}

.

Third term:

a_3 = ar^{3-1} = ar^2

. Since

a=1

,

a_3 = r^2

.

Fifth term:

a_5 = ar^{5-1} = ar^4

. Since

a=1

,

a_5 = r^4

.

Now, use the given condition:

a_3 + a_5 = 90

r^2 + r^4 = 90

r^4 + r^2 - 90 = 0

This is a quadratic equation in terms of

r^2

. Let

x = r^2

. The equation becomes:

x^2 + x - 90 = 0

We can solve this by factorization. We need two numbers that multiply to -90 and add to 1. These are 10 and -9.

x^2 + 10x - 9x - 90 = 0

x(x+10) - 9(x+10) = 0

(x-9)(x+10) = 0

So,

x = 9

or

x = -10

. Substitute back

x = r^2

:

r^2 = 9

or

r^2 = -10

.

If the terms of the G.P. are real numbers, then

r^2

cannot be negative. So we discard

r^2 = -10

. We have

r^2 = 9

. Taking the square root, we get:

r = \pm 3

Both values are possible for the common ratio.

Final Answer: The common ratio of the G.P. is

3

or

-3

.

Q4Miscellaneous Exercise On Chapter 8

The sum of three numbers in G.P. is 56 . If we subtract $1,7,21$ from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Solution

Given: Three numbers in G.P. have a sum of 56. Subtracting 1, 7, 21 from these numbers respectively results in an A.P.

To Find: The three numbers.

Solution: Let the three numbers in G.P. be

a, ar, ar^2

.

From the first condition (sum):

a + ar + ar^2 = 56

a(1 + r + r^2) = 56 \quad ...(1)

After subtracting 1, 7, 21, the new numbers are

a-1, ar-7, ar^2-21

. These new numbers are in an Arithmetic Progression (A.P.).

For an A.P., the middle term is the average of the first and third terms. So:

2(ar - 7) = (a - 1) + (ar^2 - 21)

2ar - 14 = a + ar^2 - 22

2ar = a + ar^2 - 8

a + ar^2 - 2ar = 8

a(1 + r^2 - 2r) = 8

a(r-1)^2 = 8 \quad ...(2)

Now we have a system of two equations with two variables,

a

and

r

. From (2),

a = \frac{8}{(r-1)^2}

. Substitute this expression for

a

into equation (1):

\frac{8}{(r-1)^2} (1 + r + r^2) = 56

Divide both sides by 8:

\frac{1 + r + r^2}{(r-1)^2} = 7

1 + r + r^2 = 7(r-1)^2

1 + r + r^2 = 7(r^2 - 2r + 1)

1 + r + r^2 = 7r^2 - 14r + 7

Rearrange into a standard quadratic form:

7r^2 - r^2 - 14r - r + 7 - 1 = 0

6r^2 - 15r + 6 = 0

Divide by 3:

2r^2 - 5r + 2 = 0

Factorize the quadratic equation:

2r^2 - 4r - r + 2 = 0

2r(r - 2) - 1(r - 2) = 0

(2r - 1)(r - 2) = 0

This gives two possible values for

r

:

r = 2

or

r = \frac{1}{2}

.

Case 1: $r = 2$ Substitute

r=2

into equation (2) to find

a

:

a(2-1)^2 = 8

a(1)^2 = 8 \implies a = 8

. The numbers are

a, ar, ar^2

, which are

8, 8(2), 8(2)^2 \implies 8, 16, 32

. Let's check: Sum =

8+16+32 = 56

. (Correct) Subtracting 1, 7, 21 gives

7, 9, 11

, which is an A.P. with common difference 2. (Correct)

Case 2: $r = \frac{1}{2}$ Substitute

r=1/2

into equation (2) to find

a

:

a(\frac{1}{2}-1)^2 = 8

a(-\frac{1}{2})^2 = 8

a(\frac{1}{4}) = 8 \implies a = 32

. The numbers are

a, ar, ar^2

, which are

32, 32(\frac{1}{2}), 32(\frac{1}{2})^2 \implies 32, 16, 8

. Let's check: Sum =

32+16+8 = 56

. (Correct) Subtracting 1, 7, 21 gives

31, 9, -13

, which is an A.P. with common difference -22. (Correct)

Both sets of numbers are valid solutions.

Final Answer: The numbers are 8, 16, 32 or 32, 16, 8.

Q5Miscellaneous Exercise On Chapter 8

A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

Solution

Given: A G.P. has an even number of terms. Let the number of terms be

2n

. The sum of all terms is 5 times the sum of terms occupying odd places.

To Find: The common ratio,

r

.

Solution: Let the G.P. be

a, ar, ar^2, \ldots, ar^{2n-1}

.

Sum of all terms ( $S_{2n}$ ):

S_{2n} = a + ar + ar^2 + \ldots + ar^{2n-1}

Using the sum formula

S_N = \frac{a(R^N-1)}{R-1}

:

S_{2n} = \frac{a(r^{2n}-1)}{r-1} \quad ...(1)

Sum of terms occupying odd places ( $S_{odd}$ ): The terms at odd places are the 1st, 3rd, 5th, ..., (2n-1)th terms. These terms are

a, ar^2, ar^4, \ldots, ar^{2n-2}

. This is a new G.P. with: First term:

A = a

Common ratio:

R = \frac{ar^2}{a} = r^2

Number of terms: There are

n

odd-placed terms in a sequence of

2n

terms.

Sum of this new G.P. is:

S_{odd} = \frac{A(R^n-1)}{R-1} = \frac{a((r^2)^n-1)}{r^2-1} = \frac{a(r^{2n}-1)}{r^2-1} \quad ...(2)

Given condition:

S_{2n} = 5 \times S_{odd}

Substitute the expressions from (1) and (2):

\frac{a(r^{2n}-1)}{r-1} = 5 \times \frac{a(r^{2n}-1)}{r^2-1}

Assuming the sum is not zero,

a(r^{2n}-1) \neq 0

. We can cancel this term from both sides.

\frac{1}{r-1} = \frac{5}{r^2-1}

Factor the denominator on the right side:

r^2 - 1 = (r-1)(r+1)

.

\frac{1}{r-1} = \frac{5}{(r-1)(r+1)}

Since

r \neq 1

(otherwise the sum formula is not valid), we can multiply both sides by

(r-1)

:

1 = \frac{5}{r+1}

Cross-multiply:

r+1 = 5

r = 4

Final Answer: The common ratio is 4.

Q6Miscellaneous Exercise On Chapter 8

If $\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}(x \neq 0)$ , then show that $a, b, c$ and $d$ are in G.P.

Solution

Given:

\frac{a+bx}{a-bx} = \frac{b+cx}{b-cx} = \frac{c+dx}{c-dx}

where

x \neq 0

.

To Show:

a, b, c, d

are in G.P.

Proof: To show that

a, b, c, d

are in G.P., we need to prove that their common ratio is constant, i.e.,

\frac{b}{a} = \frac{c}{b} = \frac{d}{c}

.

We will use the property of Componendo and Dividendo, which states that if

\frac{p}{q} = \frac{r}{s}

, then

\frac{p+q}{p-q} = \frac{r+s}{r-s}

.

Let's consider the first equality:

\frac{a+bx}{a-bx} = \frac{b+cx}{b-cx}

Applying Componendo and Dividendo:

\frac{(a+bx) + (a-bx)}{(a+bx) - (a-bx)} = \frac{(b+cx) + (b-cx)}{(b+cx) - (b-cx)}

\frac{2a}{2bx} = \frac{2b}{2cx}

Since

x \neq 0

, we can simplify:

\frac{a}{bx} = \frac{b}{cx}

\frac{a}{b} = \frac{b}{c}

This implies

b^2 = ac

. This shows that

a, b, c

are in G.P.

\quad ...(1)

Now, let's consider the second equality:

\frac{b+cx}{b-cx} = \frac{c+dx}{c-dx}

Applying Componendo and Dividendo again:

\frac{(b+cx) + (b-cx)}{(b+cx) - (b-cx)} = \frac{(c+dx) + (c-dx)}{(c+dx) - (c-dx)}

\frac{2b}{2cx} = \frac{2c}{2dx}

Since

x \neq 0

, we can simplify:

\frac{b}{cx} = \frac{c}{dx}

\frac{b}{c} = \frac{c}{d}

This implies

c^2 = bd

. This shows that

b, c, d

are in G.P.

\quad ...(2)

From (1), we have

\frac{b}{a} = \frac{c}{b}

. From (2), we have

\frac{c}{b} = \frac{d}{c}

.

Combining these results:

\frac{b}{a} = \frac{c}{b} = \frac{d}{c}

This is the condition for the numbers

a, b, c, d

to be in a geometric progression.

Hence Proved.

Q7Miscellaneous Exercise On Chapter 8

Let S be the sum, P the product and R the sum of reciprocals of $n$ terms in a G.P. Prove that $\mathrm{P}^{2} \mathrm{R}^{n}=\mathrm{S}^{n}$ .

Solution

Given: Let a G.P. have

n

terms:

a, ar, ar^2, \ldots, ar^{n-1}

. S = Sum of the

n

terms. P = Product of the

n

terms. R = Sum of the reciprocals of the

n

terms.

To Prove:

P^2 R^n = S^n

.

Proof: Let's express S, P, and R in terms of

a

and

r

.

Sum (S):

S = a + ar + ar^2 + \ldots + ar^{n-1}

S = \frac{a(r^n - 1)}{r-1} \quad ...(1)

Product (P):

P = (a)(ar)(ar^2)\ldots(ar^{n-1})

P = a^n r^{0+1+2+\ldots+(n-1)}

The sum of the exponents is

\frac{n(n-1)}{2}

.

P = a^n r^{\frac{n(n-1)}{2}} \quad ...(2)

Sum of Reciprocals (R): The sequence of reciprocals is

\frac{1}{a}, \frac{1}{ar}, \frac{1}{ar^2}, \ldots, \frac{1}{ar^{n-1}}

. This is also a G.P. with: First term:

A = \frac{1}{a}

Common ratio:

R_{ratio} = \frac{1/ar}{1/a} = \frac{1}{r}

Number of terms:

n

.

R = \frac{A(1 - R_{ratio}^n)}{1 - R_{ratio}} = \frac{\frac{1}{a}(1 - (\frac{1}{r})^n)}{1 - \frac{1}{r}}

R = \frac{\frac{1}{a}(1 - \frac{1}{r^n})}{\frac{r-1}{r}} = \frac{1}{a} \frac{\frac{r^n-1}{r^n}}{\frac{r-1}{r}}

R = \frac{1}{a} \frac{r^n-1}{r^n} \frac{r}{r-1} = \frac{1}{a r^{n-1}} \frac{r^n-1}{r-1} \quad ...(3)

Now, let's evaluate the LHS of the equation to be proved:

P^2 R^n

.

From (2),

P^2 = (a^n r^{\frac{n(n-1)}{2}})^2 = a^{2n} r^{n(n-1)}

.

From (3),

R^n = \left( \frac{1}{a r^{n-1}} \frac{r^n-1}{r-1} \right)^n = \frac{1}{a^n (r^{n-1})^n} \left( \frac{r^n-1}{r-1} \right)^n = \frac{1}{a^n r^{n(n-1)}} \left( \frac{r^n-1}{r-1} \right)^n

.

Now, multiply

P^2

and

R^n

:

P^2 R^n = (a^{2n} r^{n(n-1)}) \times \left( \frac{1}{a^n r^{n(n-1)}} \left( \frac{r^n-1}{r-1} \right)^n \right)

P^2 R^n = \frac{a^{2n} r^{n(n-1)}}{a^n r^{n(n-1)}} \left( \frac{r^n-1}{r-1} \right)^n

P^2 R^n = a^{2n-n} \times r^{n(n-1)-n(n-1)} \times \left( \frac{r^n-1}{r-1} \right)^n

P^2 R^n = a^n r^0 \left( \frac{r^n-1}{r-1} \right)^n = a^n \left( \frac{r^n-1}{r-1} \right)^n

P^2 R^n = \left( a \frac{r^n-1}{r-1} \right)^n

From equation (1), we know

S = \frac{a(r^n-1)}{r-1}

. So,

P^2 R^n = (S)^n = S^n

.

LHS = RHS.

Hence Proved.

Q8Miscellaneous Exercise On Chapter 8

If $a, b, c, d$ are in G.P, prove that $\left(a^{n}+b^{n}\right),\left(b^{n}+c^{n}\right),\left(c^{n}+d^{n}\right)$ are in G.P.

Solution

Given:

a, b, c, d

are in G.P.

To Prove:

(a^n+b^n), (b^n+c^n), (c^n+d^n)

are in G.P.

Proof: To prove that the three terms are in G.P., we need to show that the square of the middle term is equal to the product of the other two terms. That is:

(b^n+c^n)^2 = (a^n+b^n)(c^n+d^n)

Since

a, b, c, d

are in G.P., let the common ratio be

r

. Then:

b = ar

c = ar^2

d = ar^3

Now, let's evaluate the LHS and RHS of the equation to be proved.

LHS = $(b^n+c^n)^2$ Substitute the expressions for

b

and

c

: LHS =

((ar)^n + (ar^2)^n)^2

LHS =

(a^n r^n + a^n r^{2n})^2

Factor out the common term

a^n r^n

: LHS =

[a^n r^n (1 + r^n)]^2

LHS =

(a^n r^n)^2 (1 + r^n)^2

LHS =

a^{2n} r^{2n} (1 + r^n)^2 \quad ...(1)

RHS = $(a^n+b^n)(c^n+d^n)$ Substitute the expressions for

b, c, d

: RHS =

(a^n + (ar)^n)((ar^2)^n + (ar^3)^n)

RHS =

(a^n + a^n r^n)(a^n r^{2n} + a^n r^{3n})

Factor out common terms from each bracket: RHS =

[a^n(1 + r^n)][a^n r^{2n}(1 + r^n)]

RHS =

a^n \cdot a^n r^{2n} (1 + r^n)(1 + r^n)

RHS =

a^{2n} r^{2n} (1 + r^n)^2 \quad ...(2)

Comparing equations (1) and (2), we see that LHS = RHS.

Since

(b^n+c^n)^2 = (a^n+b^n)(c^n+d^n)

, the terms

(a^n+b^n), (b^n+c^n), (c^n+d^n)

are in G.P.

Hence Proved.

Q9Miscellaneous Exercise On Chapter 8

If $a$ and $b$ are the roots of $x^{2}-3 x+p=0$ and $c, d$ are roots of $x^{2}-12 x+q=0$ , where $a, b, c, d$ form a G.P. Prove that $(q+p):(q-p)=17: 15$ .

Solution

Given:

a, b

are roots of

x^2 - 3x + p = 0

.

c, d

are roots of

x^2 - 12x + q = 0

.

a, b, c, d

form a G.P.

To Prove:

(q+p):(q-p) = 17:15

.

Proof: From the properties of quadratic equations (sum and product of roots): For

x^2 - 3x + p = 0

: Sum of roots:

a+b = 3 \quad ...(1)

Product of roots:

ab = p \quad ...(2)

For

x^2 - 12x + q = 0

: Sum of roots:

c+d = 12 \quad ...(3)

Product of roots:

cd = q \quad ...(4)

Since

a, b, c, d

form a G.P., let the common ratio be

r

. We can write:

b = ar

c = ar^2

d = ar^3

Now substitute these into the equations for sum and product of roots. From (1):

a + ar = 3 \implies a(1+r) = 3 \quad ...(5)

From (3):

ar^2 + ar^3 = 12 \implies ar^2(1+r) = 12 \quad ...(6)

Divide equation (6) by equation (5):

\frac{ar^2(1+r)}{a(1+r)} = \frac{12}{3}

r^2 = 4

r = \pm 2

Let's assume

r=2

. (The result will be the same for

r=-2

as we will see). Substitute

r=2

into equation (5):

a(1+2) = 3 \implies 3a = 3 \implies a = 1

.

Now we have the terms of the G.P.:

a = 1

b = ar = 1(2) = 2

c = ar^2 = 1(2)^2 = 4

d = ar^3 = 1(2)^3 = 8

Now we can find the values of

p

and

q

. From (2):

p = ab = 1 \times 2 = 2

. From (4):

q = cd = 4 \times 8 = 32

.

Now, let's find the required ratio:

\frac{q+p}{q-p} = \frac{32+2}{32-2} = \frac{34}{30} = \frac{17}{15}

This gives the ratio

17:15

.

(If we had chosen

r=-2

, from (5)

a(1-2)=3 \implies -a=3 \implies a=-3

. Then

b=(-3)(-2)=6

,

c=(-3)(-2)^2=-12

,

d=(-3)(-2)^3=24

. Then

p=ab=(-3)(6)=-18

and

q=cd=(-12)(24)=-288

. The ratio would be

\frac{-288-18}{-288-(-18)} = \frac{-306}{-270} = \frac{306}{270} = \frac{17}{15}

. The result is the same.)

Hence Proved.

Q10Miscellaneous Exercise On Chapter 8

The ratio of the A.M. and G.M. of two positive numbers $a$ and $b$ , is $m: n$ . Show that $a: b=\left(m+\sqrt{m^{2}-n^{2}}\right):\left(m-\sqrt{m^{2}-n^{2}}\right)$ .

Solution

Given: Let the two positive numbers be

a

and

b

. A.M. =

\frac{a+b}{2}

G.M. =

\sqrt{ab}

The ratio A.M. : G.M. is

m:n

.

To Show:

a:b = (m+\sqrt{m^2-n^2}) : (m-\sqrt{m^2-n^2})

.

Proof: From the given ratio:

\frac{\text{A.M.}}{\text{G.M.}} = \frac{m}{n}

\frac{(a+b)/2}{\sqrt{ab}} = \frac{m}{n}

\frac{a+b}{2\sqrt{ab}} = \frac{m}{n}

Apply Componendo and Dividendo, which states that if

\frac{x}{y} = \frac{p}{q}

, then

\frac{x+y}{x-y} = \frac{p+q}{p-q}

.

\frac{(a+b) + 2\sqrt{ab}}{(a+b) - 2\sqrt{ab}} = \frac{m+n}{m-n}

The numerator is

(\sqrt{a} + \sqrt{b})^2

and the denominator is

(\sqrt{a} - \sqrt{b})^2

.

\frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = \frac{m+n}{m-n}

Take the square root of both sides:

\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \frac{\sqrt{m+n}}{\sqrt{m-n}}

Apply Componendo and Dividendo again:

\frac{(\sqrt{a} + \sqrt{b}) + (\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b}) - (\sqrt{a} - \sqrt{b})} = \frac{\sqrt{m+n} + \sqrt{m-n}}{\sqrt{m+n} - \sqrt{m-n}}

\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{m+n} + \sqrt{m-n}}{\sqrt{m+n} - \sqrt{m-n}}

\frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{m+n} + \sqrt{m-n}}{\sqrt{m+n} - \sqrt{m-n}}

Square both sides to find the ratio

\frac{a}{b}

:

\frac{a}{b} = \frac{(\sqrt{m+n} + \sqrt{m-n})^2}{(\sqrt{m+n} - \sqrt{m-n})^2}

Expand the numerator and denominator: Numerator:

(\sqrt{m+n})^2 + (\sqrt{m-n})^2 + 2\sqrt{m+n}\sqrt{m-n}

= (m+n) + (m-n) + 2\sqrt{(m+n)(m-n)}

= 2m + 2\sqrt{m^2-n^2}

Denominator:

(\sqrt{m+n})^2 + (\sqrt{m-n})^2 - 2\sqrt{m+n}\sqrt{m-n}

= (m+n) + (m-n) - 2\sqrt{(m+n)(m-n)}

= 2m - 2\sqrt{m^2-n^2}

So the ratio becomes:

\frac{a}{b} = \frac{2m + 2\sqrt{m^2-n^2}}{2m - 2\sqrt{m^2-n^2}}

Factor out 2 from the numerator and denominator:

\frac{a}{b} = \frac{2(m + \sqrt{m^2-n^2})}{2(m - \sqrt{m^2-n^2})} = \frac{m + \sqrt{m^2-n^2}}{m - \sqrt{m^2-n^2}}

Therefore,

a:b = (m+\sqrt{m^2-n^2}) : (m-\sqrt{m^2-n^2})

.

Hence Proved.

Q11Miscellaneous Exercise On Chapter 8

Find the sum of the following series up to $n$ terms:

(i)

$5+55+555+\ldots$

(ii)

$.6+.66+.666+\ldots$

Solution

Solution:

(i) $5+55+555+\ldots$

Let

S_n

be the sum of the first

n

terms.

S_n = 5 + 55 + 555 + \ldots

to

n

terms.

Step 1: Factor out the common digit 5.

S_n = 5(1 + 11 + 111 + \ldots

to

n

terms)

Step 2: Multiply and divide by 9.

S_n = \frac{5}{9}(9 + 99 + 999 + \ldots

to

n

terms)

Step 3: Express each term as a power of 10 minus 1.

S_n = \frac{5}{9}((10-1) + (10^2-1) + (10^3-1) + \ldots

to

n

terms)

Step 4: Group the powers of 10 and the -1s.

S_n = \frac{5}{9} [ (10 + 10^2 + 10^3 + \ldots + 10^n) - (1 + 1 + \ldots \text{to } n \text{ terms}) ]

Step 5: The first part is a G.P. with

a=10, r=10

. Its sum is

\frac{10(10^n-1)}{10-1} = \frac{10}{9}(10^n-1)

. The second part is simply

n

.

Step 6: Substitute back.

S_n = \frac{5}{9} \left[ \frac{10(10^n-1)}{9} - n \right]

Final Answer (i): The sum is

\frac{5}{9} \left[ \frac{10(10^n-1)}{9} - n \right]

.

(ii) $.6+.66+.666+\ldots$

Let

S_n

be the sum of the first

n

terms.

S_n = 0.6 + 0.66 + 0.666 + \ldots

to

n

terms.

Step 1: Factor out 6.

S_n = 6(0.1 + 0.11 + 0.111 + \ldots

to

n

terms)

Step 2: Multiply and divide by 9.

S_n = \frac{6}{9}(0.9 + 0.99 + 0.999 + \ldots

to

n

terms)

S_n = \frac{2}{3}(0.9 + 0.99 + 0.999 + \ldots

to

n

terms)

Step 3: Express each term as 1 minus a power of 10.

S_n = \frac{2}{3}((1-0.1) + (1-0.01) + (1-0.001) + \ldots

to

n

terms)

S_n = \frac{2}{3}((1 - \frac{1}{10}) + (1 - \frac{1}{100}) + (1 - \frac{1}{1000}) + \ldots

to

n

terms)

S_n = \frac{2}{3}((1 - 10^{-1}) + (1 - 10^{-2}) + (1 - 10^{-3}) + \ldots

to

n

terms)

Step 4: Group the 1s and the powers of 10.

S_n = \frac{2}{3} [ (1 + 1 + \ldots \text{to } n \text{ terms}) - (10^{-1} + 10^{-2} + 10^{-3} + \ldots + 10^{-n}) ]

Step 5: The first part is

n

. The second part is a G.P. with

a=\frac{1}{10}, r=\frac{1}{10}

. Its sum is

\frac{a(1-r^n)}{1-r} = \frac{\frac{1}{10}(1 - (\frac{1}{10})^n)}{1 - \frac{1}{10}} = \frac{\frac{1}{10}(1 - 10^{-n})}{\frac{9}{10}} = \frac{1}{9}(1-10^{-n})

.

Step 6: Substitute back.

S_n = \frac{2}{3} \left[ n - \frac{1}{9}(1-10^{-n}) \right]

Final Answer (ii): The sum is

\frac{2}{3} \left[ n - \frac{1}{9}(1-10^{-n}) \right]

.

Q12Miscellaneous Exercise On Chapter 8

Find the $20^{\text {th }}$ term of the series $2 \times 4+4 \times 6+6 \times 8+\ldots+n$ terms.

Solution

Given: The series

2 \times 4 + 4 \times 6 + 6 \times 8 + \ldots

To Find: The

20^{\text{th}}

term of the series.

Solution: First, let's find the general formula for the

n^{\text{th}}

term,

a_n

.

The series is a sum of products. Let's look at the factors separately. First factors:

2, 4, 6, \ldots

. This is an A.P. with first term 2 and common difference 2. The

n^{\text{th}}

term is

2 + (n-1)2 = 2 + 2n - 2 = 2n

.

Second factors:

4, 6, 8, \ldots

. This is an A.P. with first term 4 and common difference 2. The

n^{\text{th}}

term is

4 + (n-1)2 = 4 + 2n - 2 = 2n+2

.

So, the

n^{\text{th}}

term of the series is the product of the

n^{\text{th}}

terms of these two sequences:

a_n = (2n)(2n+2)

Now, we need to find the

20^{\text{th}}

term. We substitute

n=20

into the formula for

a_n

.

a_{20} = (2 \times 20)(2 \times 20 + 2)

a_{20} = (40)(40 + 2)

a_{20} = 40 \times 42

a_{20} = 1680

Final Answer: The

20^{\text{th}}

term of the series is 1680.

Q13Miscellaneous Exercise On Chapter 8

A farmer buys a used tractor for Rs 12000 . He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus $12\%$ interest on the unpaid amount. How much will the tractor cost him?

Solution

Given: Cost of tractor = Rs 12000. Down payment = Rs 6000. Balance amount = Rs 12000 - Rs 6000 = Rs 6000. Annual instalment of principal = Rs 500. Interest rate = 12% on the unpaid amount.

To Find: The total cost of the tractor for the farmer.

Solution: The total cost will be the down payment plus all the instalments (principal + interest). Total cost = Down Payment + Total Principal Paid + Total Interest Paid. Total Principal Paid in instalments is the balance amount, which is Rs 6000.

We need to calculate the total interest paid. The number of instalments will be

\frac{\text{Balance Amount}}{\text{Instalment Principal}} = \frac{6000}{500} = 12

instalments.

Interest Calculation: Interest for the 1st year (paid with the 1st instalment) is on the unpaid amount of Rs 6000. Interest 1 =

12\%

of 6000 =

0.12 \times 6000 = 720

.

After the 1st instalment, the unpaid amount is

6000 - 500 = 5500

. Interest for the 2nd year =

12\%

of 5500 =

0.12 \times 5500 = 660

.

After the 2nd instalment, the unpaid amount is

5500 - 500 = 5000

. Interest for the 3rd year =

12\%

of 5000 =

0.12 \times 5000 = 600

.

This pattern continues for 12 years. The interest payments form an Arithmetic Progression (A.P.):

720, 660, 600, \ldots

First term

a = 720

. Common difference

d = 660 - 720 = -60

. Number of terms

n = 12

.

Total interest paid is the sum of this A.P.,

S_{12}

. Formula for sum of A.P.:

S_n = \frac{n}{2}[2a + (n-1)d]

S_{12} = \frac{12}{2}[2(720) + (12-1)(-60)]

S_{12} = 6[1440 + 11(-60)]

S_{12} = 6[1440 - 660]

S_{12} = 6[780]

S_{12} = 4680

.

Total interest paid is Rs 4680.

Total Cost of the Tractor: Total Cost = Down Payment + Balance Principal + Total Interest Total Cost =

6000 + 6000 + 4680

Total Cost =

16680

.

Final Answer: The tractor will cost the farmer Rs 16680.

Q14Miscellaneous Exercise On Chapter 8

Shamshad Ali buys a scooter for Rs 22000 . He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus $10\%$ interest on the unpaid amount. How much will the scooter cost him?

Solution

Given: Cost of scooter = Rs 22000. Down payment = Rs 4000. Balance amount = Rs 22000 - Rs 4000 = Rs 18000. Annual instalment of principal = Rs 1000. Interest rate = 10% on the unpaid amount.

To Find: The total cost of the scooter.

Solution: Total cost = Down Payment + Total Principal Paid in instalments + Total Interest Paid. Total Principal Paid in instalments is the balance amount, Rs 18000.

We need to calculate the total interest paid. The number of instalments will be

\frac{\text{Balance Amount}}{\text{Instalment Principal}} = \frac{18000}{1000} = 18

instalments.

Interest Calculation: Interest for the 1st year (on Rs 18000) =

10\%

of 18000 =

0.10 \times 18000 = 1800

.

Unpaid amount for 2nd year =

18000 - 1000 = 17000

. Interest for the 2nd year =

10\%

of 17000 =

0.10 \times 17000 = 1700

.

Unpaid amount for 3rd year =

17000 - 1000 = 16000

. Interest for the 3rd year =

10\%

of 16000 =

0.10 \times 16000 = 1600

.

The interest payments form an Arithmetic Progression (A.P.):

1800, 1700, 1600, \ldots

First term

a = 1800

. Common difference

d = 1700 - 1800 = -100

. Number of terms

n = 18

.

Total interest paid is the sum of this A.P.,

S_{18}

. Formula for sum of A.P.:

S_n = \frac{n}{2}[2a + (n-1)d]

S_{18} = \frac{18}{2}[2(1800) + (18-1)(-100)]

S_{18} = 9[3600 + 17(-100)]

S_{18} = 9[3600 - 1700]

S_{18} = 9[1900]

S_{18} = 17100

.

Total interest paid is Rs 17100.

Total Cost of the Scooter: Total Cost = Down Payment + Balance Principal + Total Interest Total Cost =

4000 + 18000 + 17100

Total Cost =

39100

.

Final Answer: The scooter will cost him Rs 39100.

Q15Miscellaneous Exercise On Chapter 8

A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when $8^{\text {th }}$ set of letter is mailed.

Solution

Given: One person sends letters to 4 friends. Each recipient sends letters to 4 new people. The cost to mail one letter is 50 paise = Rs 0.50.

To Find: The amount spent on postage when the 8th set of letters is mailed.

Solution: Let's analyze the number of letters sent in each set.

1st set: The initial person sends 4 letters. Number of letters = 4.

2nd set: Each of the 4 friends sends 4 letters. Number of letters =

4 \times 4 = 16 = 4^2

.

3rd set: Each of the 16 recipients sends 4 letters. Number of letters =

16 \times 4 = 64 = 4^3

.

The number of letters sent in each set forms a geometric progression (G.P.):

4, 16, 64, \ldots

or

4^1, 4^2, 4^3, \ldots

The number of letters in the

n^{\text{th}}

set is given by

a_n = 4^n

.

We need to find the number of letters in the 8th set, which is

a_8

.

a_8 = 4^8

4^2 = 16

4^4 = 16^2 = 256

4^8 = (4^4)^2 = 256^2 = 65536

.

So, 65536 letters are mailed in the 8th set.

Now, we calculate the total cost for this set. Cost per letter = Rs 0.50. Total cost = (Number of letters)

\times

(Cost per letter) Total cost =

65536 \times 0.50

Total cost =

32768

.

Final Answer: The amount spent on postage when the 8th set of letters is mailed is Rs 32768.

Q16Miscellaneous Exercise On Chapter 8

A man deposited Rs 10000 in a bank at the rate of $5\%$ simple interest annually. Find the amount in $15^{\text {th }}$ year since he deposited the amount and also calculate the total amount after 20 years.

Solution

Given: Principal (P) = Rs 10000. Rate of simple interest (R) = 5% per annum.

To Find:

The amount in the 15th year.
The total amount after 20 years.

Solution: Part 1: Amount in the 15th year

First, calculate the annual simple interest (I).

I = P \times R \times T = 10000 \times \frac{5}{100} \times 1 = 500

. So, the interest earned each year is Rs 500.

The amounts at the end of each year will form an Arithmetic Progression (A.P.). Amount at end of year 1 (

A_1

) = Principal + Interest for 1 year =

10000 + 500 = 10500

. Amount at end of year 2 (

A_2

) = Principal + Interest for 2 years =

10000 + 2 \times 500 = 11000

. Amount at end of year n (

A_n

) = Principal + Interest for n years =

10000 + n \times 500

.

The question asks for the amount in the 15th year. This phrasing can be ambiguous. It could mean the amount at the end of the 14th year (which exists throughout the 15th year before interest is added) or the amount at the end of the 15th year. Standard interpretation is the amount at the end of the 15th year.

Amount at the end of 15 years (

A_{15}

) =

10000 + 15 \times 500 = 10000 + 7500 = 17500

.

Part 2: Total amount after 20 years

Using the same formula for the amount at the end of year n: Amount at the end of 20 years (

A_{20}

) =

10000 + 20 \times 500

A_{20} = 10000 + 10000

A_{20} = 20000

.

Final Answer: The amount at the end of the 15th year is Rs 17500. The total amount after 20 years is Rs 20000.

Q17Miscellaneous Exercise On Chapter 8

A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by $20\%$ . Find the estimated value at the end of 5 years.

Solution

Given: Initial cost of the machine (P) = Rs 15625. Rate of depreciation (R) = 20% per year. Time period (n) = 5 years.

To Find: The estimated value of the machine at the end of 5 years.

Solution: Depreciation means the value decreases by a certain percentage of its value at the beginning of that year. This is similar to compound interest, but with a negative rate.

Value after 1 year = Initial Value - 20% of Initial Value Value after 1 year =

P - 0.20P = P(1 - 0.20) = P(0.80)

.

Value after 2 years = (Value after 1 year)

\times (0.80) = P(0.80)(0.80) = P(0.80)^2

.

This forms a geometric progression. The value of the machine at the end of

n

years is given by the formula:

V_n = P(1 - R)^n

Here, P = 15625, R = 20% = 0.20, and n = 5.

V_5 = 15625 (1 - 0.20)^5

V_5 = 15625 (0.8)^5

We can write

0.8

as

\frac{4}{5}

.

V_5 = 15625 \left(\frac{4}{5}\right)^5

V_5 = 15625 \times \frac{4^5}{5^5}

We know that

15625 = 5^6

and

4^5 = 1024

.

V_5 = 5^6 \times \frac{4^5}{5^5}

V_5 = 5^{6-5} \times 4^5

V_5 = 5^1 \times 1024

V_5 = 5 \times 1024

V_5 = 5120

.

Final Answer: The estimated value of the machine at the end of 5 years is Rs 5120.

Q18Miscellaneous Exercise On Chapter 8

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.

Solution

Given: Initial number of workers = 150. Number of workers dropping out each day (from the second day) = 4. The work took 8 more days than planned.

To Find: The number of days in which the work was completed.

Solution: Let the original number of days planned to finish the work be

n

.

Total work (in man-days): If 150 workers worked for

n

days, the total work would be: Total Work =

150 \times n

man-days.

Actual work done: The number of workers on each day is as follows: Day 1: 150 workers Day 2:

150 - 4 = 146

workers Day 3:

146 - 4 = 142

workers and so on. This forms an Arithmetic Progression (A.P.) with first term

a=150

and common difference

d=-4

.

The work was completed in

n+8

days. So, the total work done is the sum of the number of workers over these

n+8

days. Total Work = Sum of the A.P. for

(n+8)

terms.

Using the sum formula for an A.P.,

S_k = \frac{k}{2}[2a + (k-1)d]

: Total Work =

S_{n+8} = \frac{n+8}{2}[2(150) + ((n+8)-1)(-4)]

Total Work =

\frac{n+8}{2}[300 + (n+7)(-4)]

Total Work =

\frac{n+8}{2}[300 - 4n - 28]

Total Work =

\frac{n+8}{2}[272 - 4n]

Total Work =

(n+8)(136 - 2n)

Equating the planned work and actual work:

150n = (n+8)(136 - 2n)

150n = 136n - 2n^2 + 8 \times 136 - 16n

150n = 120n - 2n^2 + 1088

Rearrange into a standard quadratic equation:

2n^2 + 150n - 120n - 1088 = 0

2n^2 + 30n - 1088 = 0

Divide by 2:

n^2 + 15n - 544 = 0

We need to find two numbers that multiply to -544 and add to 15. Let's factorize 544.

544 = 2 \times 272 = 2^2 \times 136 = 2^3 \times 68 = 2^4 \times 34 = 2^5 \times 17 = 32 \times 17

. The difference between 32 and 17 is 15. So the numbers are 32 and -17.

n^2 + 32n - 17n - 544 = 0

n(n + 32) - 17(n + 32) = 0

(n - 17)(n + 32) = 0

This gives two possible values for

n

:

n = 17

or

n = -32

. Since the number of days cannot be negative, we have

n = 17

.

This is the original planned number of days. The question asks for the number of days in which the work was completed. Actual number of days =

n + 8 = 17 + 8 = 25

.

Final Answer: The work was completed in 25 days.

Sequences and SeriesClass 11 Mathematics NCERT Solutions

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth n^{\text {th }}nth terms are: an=n(n+2)a_{n}=n(n+2)an​=n(n+2)

Solution

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth n^{\text {th }}nth terms are: 2. an=nn+1a_{n}=\frac{n}{n+1}an​=n+1n​

Solution

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth n^{\text {th }}nth terms are: 3. an=2na_{n}=2^{n}an​=2n

Solution

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth n^{\text {th }}nth terms are: 4. an=2n−36a_{n}=\frac{2 n-3}{6}an​=62n−3​

Solution

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth n^{\text {th }}nth terms are: 5. an=(−1)n−15n+1a_{n}=(-1)^{n-1} 5^{n+1}an​=(−1)n−15n+1

Solution

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth n^{\text {th }}nth terms are: 6. an=nn2+54a_{n}=n \frac{n^{2}+5}{4}an​=n4n2+5​

Solution

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth n^{\text {th }}nth terms are: 7. an=4n−3;a17,a24a_{n}=4 n-3 ; a_{17}, a_{24}an​=4n−3;a17​,a24​

Solution

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth n^{\text {th }}nth terms are: 8. an=n22n;a7a_{n}=\frac{n^{2}}{2^{n}} ; a_{7}an​=2nn2​;a7​

Solution

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth n^{\text {th }}nth terms are: 9. an=(−1)n−1n3;a9a_{n}=(-1)^{n-1} n^{3} ; a_{9}an​=(−1)n−1n3;a9​

Solution

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth n^{\text {th }}nth terms are: 10. an=n(n−2)n+3;a20a_{n}=\frac{n(n-2)}{n+3} ; a_{20}an​=n+3n(n−2)​;a20​

Solution

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: 11. a1=3,an=3an−1+2a_{1}=3, a_{n}=3 a_{n-1}+2a1​=3,an​=3an−1​+2 for all n>1n>1n>1

Solution

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: 12. a1=−1,an=an−1n,n≥2a_{1}=-1, a_{n}=\frac{a_{n-1}}{n}, n \geq 2a1​=−1,an​=nan−1​​,n≥2

Solution

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: 13. a1=a2=2,an=an−1−1,n>2a_{1}=a_{2}=2, a_{n}=a_{n-1}-1, n>2a1​=a2​=2,an​=an−1​−1,n>2

Solution

The Fibonacci sequence is defined by 1=a1=a21=a_{1}=a_{2}1=a1​=a2​ and an=an−1+an−2,n>2a_{n}=a_{n-1}+a_{n-2}, n>2an​=an−1​+an−2​,n>2 Find an+1an\frac{a_{n+1}}{a_{n}}an​an+1​​, for n=1,2,3,4,5n=1,2,3,4,5n=1,2,3,4,5

Solution

Find the 20th 20^{\text {th }}20th and nth n^{\text {th }}nth terms of the G.P. 52,54,58,…\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots25​,45​,85​,…

Solution

Find the 12th 12^{\text {th }}12th term of a G.P. whose 8th 8^{\text {th }}8th term is 192 and the common ratio is 2.

Solution

The 5th ,8th 5^{\text {th }}, 8^{\text {th }}5th ,8th and 11th 11^{\text {th }}11th terms of a G.P. are p,qp, qp,q and sss, respectively. Show that q2=psq^{2}=p sq2=ps.

Solution

The 4th 4^{\text {th }}4th term of a G.P. is square of its second term, and the first term is -3 . Determine its 7th 7^{\text {th }}7th term.

Solution

Which term of the following sequences:

Solution

For what values of xxx, the numbers −27,x,−72-\frac{2}{7}, x,-\frac{7}{2}−72​,x,−27​ are in G.P.?

Solution

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 7. 0.15,0.015,0.0015,…200.15,0.015,0.0015, \ldots 200.15,0.015,0.0015,…20 terms.

Solution

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 8. 7,21,37,…n\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots n7​,21​,37​,…n terms.

Solution

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 9. 1,−a,a2,−a3,…n1,-a, a^{2},-a^{3}, \ldots n1,−a,a2,−a3,…n terms (if a≠−1a \neq-1a=−1 ).

Solution

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 10. x3,x5,x7,…nx^{3}, x^{5}, x^{7}, \ldots nx3,x5,x7,…n terms (if x≠±1x \neq \pm 1x=±1 ).

Solution

Evaluate ∑k=111(2+3k)\sum_{k=1}^{11}\left(2+3^{k}\right)∑k=111​(2+3k).

Solution

The sum of first three terms of a G.P. is 3910\frac{39}{10}1039​ and their product is 1 . Find the common ratio and the terms.

Solution

How many terms of G.P. 3,32,33,…3,3^{2}, 3^{3}, \ldots3,32,33,… are needed to give the sum 120 ?

Solution

The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to nnn terms of the G.P.

Solution

Given a G.P. with a=729a=729a=729 and 7th 7^{\text {th }}7th term 64, determine S7\mathrm{S}_{7}S7​.

Solution

Find a G.P. for which sum of the first two terms is -4 and the fifth term is 4 times the third term.

Solution

If the 4th ,10th 4^{\text {th }}, 10^{\text {th }}4th ,10th and 16th 16^{\text {th }}16th terms of a G.P. are x,yx, yx,y and zzz, respectively. Prove that x,y,zx, y, zx,y,z are in G.P.

Solution

Find the sum to nnn terms of the sequence, 8,88,888,8888…8,88,888,8888 \ldots8,88,888,8888….

Solution

Find the sum of the products of the corresponding terms of the sequences 2,4,8,16,322,4,8, 16,322,4,8,16,32 and 128,32,8,2,12128,32,8,2, \frac{1}{2}128,32,8,2,21​.

Solution

Show that the products of the corresponding terms of the sequences a,ar,ar2,…arn−1a, a r, a r^{2}, \ldots a r^{n-1}a,ar,ar2,…arn−1 and A,AR,AR2,…ARn−1\mathrm{A}, \mathrm{AR}, \mathrm{AR}^{2}, \ldots \mathrm{AR}^{n-1}A,AR,AR2,…ARn−1 form a G.P, and find the common ratio.

Solution

Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th 4^{\text {th }}4th by 18.

Solution

If the pth ,qth p^{\text {th }}, q^{\text {th }}pth ,qth and rth r^{\text {th }}rth terms of a G.P. are a,ba, ba,b and ccc, respectively. Prove that aq−rbr−pcp−q=1.a^{q-r} b^{r-p} c^{p-q}=1 .aq−rbr−pcp−q=1.

Solution

If the first and the nth n^{\text {th }}nth term of a G.P. are aaa and bbb, respectively, and if P is the product of nnn terms, prove that P2=(ab)n\mathrm{P}^{2}=(a b)^{n}P2=(ab)n.

Solution

Show that the ratio of the sum of first nnn terms of a G.P. to the sum of terms from (n+1)th (n+1)^{\text {th }}(n+1)th to (2n)th (2 n)^{\text {th }}(2n)th term is 1rn\frac{1}{r^{n}}rn1​.

Solution

If a,b,ca, b, ca,b,c and ddd are in G.P. show that (a2+b2+c2)(b2+c2+d2)=(ab+bc+cd)2\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}(a2+b2+c2)(b2+c2+d2)=(ab+bc+cd)2.

Solution

Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Write the first five terms of each of the sequences in Exercises 1 to 6 whose $n^{\text {th }}$ terms are: $a_{n}=n(n+2)$

Write the first five terms of each of the sequences in Exercises 1 to 6 whose $n^{\text {th }}$ terms are: 2. $a_{n}=\frac{n}{n+1}$

Write the first five terms of each of the sequences in Exercises 1 to 6 whose $n^{\text {th }}$ terms are: 3. $a_{n}=2^{n}$

Write the first five terms of each of the sequences in Exercises 1 to 6 whose $n^{\text {th }}$ terms are: 4. $a_{n}=\frac{2 n-3}{6}$

Write the first five terms of each of the sequences in Exercises 1 to 6 whose $n^{\text {th }}$ terms are: 5. $a_{n}=(-1)^{n-1} 5^{n+1}$

Write the first five terms of each of the sequences in Exercises 1 to 6 whose $n^{\text {th }}$ terms are: 6. $a_{n}=n \frac{n^{2}+5}{4}$

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose $n^{\text {th }}$ terms are: 7. $a_{n}=4 n-3 ; a_{17}, a_{24}$

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose $n^{\text {th }}$ terms are: 8. $a_{n}=\frac{n^{2}}{2^{n}} ; a_{7}$

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose $n^{\text {th }}$ terms are: 9. $a_{n}=(-1)^{n-1} n^{3} ; a_{9}$

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose $n^{\text {th }}$ terms are: 10. $a_{n}=\frac{n(n-2)}{n+3} ; a_{20}$

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: 11. $a_{1}=3, a_{n}=3 a_{n-1}+2$ for all $n>1$

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: 12. $a_{1}=-1, a_{n}=\frac{a_{n-1}}{n}, n \geq 2$

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: 13. $a_{1}=a_{2}=2, a_{n}=a_{n-1}-1, n>2$

The Fibonacci sequence is defined by $1=a_{1}=a_{2}$ and $a_{n}=a_{n-1}+a_{n-2}, n>2$ Find $\frac{a_{n+1}}{a_{n}}$ , for $n=1,2,3,4,5$

Find the $20^{\text {th }}$ and $n^{\text {th }}$ terms of the G.P. $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots$

Find the $12^{\text {th }}$ term of a G.P. whose $8^{\text {th }}$ term is 192 and the common ratio is 2.

The $5^{\text {th }}, 8^{\text {th }}$ and $11^{\text {th }}$ terms of a G.P. are $p, q$ and $s$ , respectively. Show that $q^{2}=p s$ .

The $4^{\text {th }}$ term of a G.P. is square of its second term, and the first term is -3 . Determine its $7^{\text {th }}$ term.

For what values of $x$ , the numbers $-\frac{2}{7}, x,-\frac{7}{2}$ are in G.P.?

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 7. $0.15,0.015,0.0015, \ldots 20$ terms.

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 8. $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots n$ terms.

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 9. $1,-a, a^{2},-a^{3}, \ldots n$ terms (if $a \neq-1$ ).

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 10. $x^{3}, x^{5}, x^{7}, \ldots n$ terms (if $x \neq \pm 1$ ).

Evaluate $\sum_{k=1}^{11}\left(2+3^{k}\right)$ .

The sum of first three terms of a G.P. is $\frac{39}{10}$ and their product is 1 . Find the common ratio and the terms.

How many terms of G.P. $3,3^{2}, 3^{3}, \ldots$ are needed to give the sum 120 ?

The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to $n$ terms of the G.P.

Given a G.P. with $a=729$ and $7^{\text {th }}$ term 64, determine $\mathrm{S}_{7}$ .

If the $4^{\text {th }}, 10^{\text {th }}$ and $16^{\text {th }}$ terms of a G.P. are $x, y$ and $z$ , respectively. Prove that $x, y, z$ are in G.P.

Find the sum to $n$ terms of the sequence, $8,88,888,8888 \ldots$ .

Find the sum of the products of the corresponding terms of the sequences $2,4,8, 16,32$ and $128,32,8,2, \frac{1}{2}$ .

Show that the products of the corresponding terms of the sequences $a, a r, a r^{2}, \ldots a r^{n-1}$ and $\mathrm{A}, \mathrm{AR}, \mathrm{AR}^{2}, \ldots \mathrm{AR}^{n-1}$ form a G.P, and find the common ratio.

Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the $4^{\text {th }}$ by 18.

If the $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of a G.P. are $a, b$ and $c$ , respectively. Prove that $a^{q-r} b^{r-p} c^{p-q}=1 .$

If the first and the $n^{\text {th }}$ term of a G.P. are $a$ and $b$ , respectively, and if P is the product of $n$ terms, prove that $\mathrm{P}^{2}=(a b)^{n}$ .

Show that the ratio of the sum of first $n$ terms of a G.P. to the sum of terms from $(n+1)^{\text {th }}$ to $(2 n)^{\text {th }}$ term is $\frac{1}{r^{n}}$ .

If $a, b, c$ and $d$ are in G.P. show that $\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$ .

Find the value of $n$ so that $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$ may be the geometric mean between $a$ and $b$ .

The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio $(3+2 \sqrt{2}):(3-2 \sqrt{2})$ .

If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are $\mathrm{A} \pm \sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}$ .

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of $2^{\text {nd }}$ hour, $4^{\text {th }}$ hour and $n^{\text {th }}$ hour?

What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of $10\%$ compounded annually?

If $f$ is a function satisfying $f(x+y)=f(x) f(y)$ for all $x, y \in \mathbf{N}$ such that $f(1)=3$ and $\sum_{x=1}^{n} f(x)=120$ , find the value of $n$ .

The sum of three numbers in G.P. is 56 . If we subtract $1,7,21$ from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

If $\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}(x \neq 0)$ , then show that $a, b, c$ and $d$ are in G.P.

Let S be the sum, P the product and R the sum of reciprocals of $n$ terms in a G.P. Prove that $\mathrm{P}^{2} \mathrm{R}^{n}=\mathrm{S}^{n}$ .

If $a, b, c, d$ are in G.P, prove that $\left(a^{n}+b^{n}\right),\left(b^{n}+c^{n}\right),\left(c^{n}+d^{n}\right)$ are in G.P.

If $a$ and $b$ are the roots of $x^{2}-3 x+p=0$ and $c, d$ are roots of $x^{2}-12 x+q=0$ , where $a, b, c, d$ form a G.P. Prove that $(q+p):(q-p)=17: 15$ .

The ratio of the A.M. and G.M. of two positive numbers $a$ and $b$ , is $m: n$ . Show that $a: b=\left(m+\sqrt{m^{2}-n^{2}}\right):\left(m-\sqrt{m^{2}-n^{2}}\right)$ .

Find the sum of the following series up to $n$ terms:

$5+55+555+\ldots$

$.6+.66+.666+\ldots$

Find the $20^{\text {th }}$ term of the series $2 \times 4+4 \times 6+6 \times 8+\ldots+n$ terms.

A farmer buys a used tractor for Rs 12000 . He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus $12\%$ interest on the unpaid amount. How much will the tractor cost him?

Shamshad Ali buys a scooter for Rs 22000 . He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus $10\%$ interest on the unpaid amount. How much will the scooter cost him?

A man deposited Rs 10000 in a bank at the rate of $5\%$ simple interest annually. Find the amount in $15^{\text {th }}$ year since he deposited the amount and also calculate the total amount after 20 years.

A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by $20\%$ . Find the estimated value at the end of 5 years.