Arithmetic Progressions Class 10 Mathematics NCERT Solutions

Q1EXERCISE 5.1

In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(i)

The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km .

(ii)

The amount of air present in a cylinder when a vacuum pump removes $\frac{1}{4}$ of the air remaining in the cylinder at a time.

(iii)

The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.

(iv)

The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at $8\%$ per annum.

Solution

(i)

The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.

Given: Fare for the first km = ₹ 15 Fare for each additional km = ₹ 8

Solution: Let's find the fare for the first few kilometers. Fare for 1 km = ₹ 15 Fare for 2 km = Fare for 1 km + Fare for additional km = ₹ 15 + ₹ 8 = ₹ 23 Fare for 3 km = Fare for 2 km + Fare for additional km = ₹ 23 + ₹ 8 = ₹ 31 Fare for 4 km = Fare for 3 km + Fare for additional km = ₹ 31 + ₹ 8 = ₹ 39

The list of fares is: 15, 23, 31, 39, ... To check if this forms an Arithmetic Progression (AP), we check the difference between consecutive terms.

a_2 - a_1 = 23 - 15 = 8

a_3 - a_2 = 31 - 23 = 8

a_4 - a_3 = 39 - 31 = 8

Since the difference between consecutive terms is constant (8), the list of numbers forms an AP.

Final Answer: Yes, this situation forms an AP because the fare increases by a fixed amount (₹ 8) for each additional kilometer.

(ii)

The amount of air present in a cylinder when a vacuum pump removes

\frac{1}{4}

of the air remaining in the cylinder at a time.

Given: A vacuum pump removes

\frac{1}{4}

of the remaining air each time.

Solution: Let the initial volume of air in the cylinder be

V

. Amount of air after the 1st removal =

V - \frac{1}{4}V = \frac{3}{4}V

Amount of air after the 2nd removal =

\frac{3}{4}V - \frac{1}{4} \left( \frac{3}{4}V \right) = \frac{3}{4}V - \frac{3}{16}V = \left( \frac{12-3}{16} \right)V = \frac{9}{16}V = \left( \frac{3}{4} \right)^2 V

Amount of air after the 3rd removal =

\frac{9}{16}V - \frac{1}{4} \left( \frac{9}{16}V \right) = \frac{9}{16}V - \frac{9}{64}V = \left( \frac{36-9}{64} \right)V = \frac{27}{64}V = \left( \frac{3}{4} \right)^3 V

The list of amounts of air is:

V, \frac{3}{4}V, \frac{9}{16}V, \frac{27}{64}V, ...

Let's check the difference between consecutive terms.

a_2 - a_1 = \frac{3}{4}V - V = -\frac{1}{4}V

a_3 - a_2 = \frac{9}{16}V - \frac{3}{4}V = \frac{9V - 12V}{16} = -\frac{3}{16}V

Since the difference between consecutive terms is not constant, the list of numbers does not form an AP.

Final Answer: No, this situation does not form an AP because the amount of air removed is not a fixed quantity but a fraction of the remaining air, so the difference between consecutive terms is not constant.

(iii)

The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.

Given: Cost for the first metre = ₹ 150 Cost increase for each subsequent metre = ₹ 50

Solution: Cost of digging for 1 metre = ₹ 150 Cost of digging for 2 metres = ₹ 150 + ₹ 50 = ₹ 200 Cost of digging for 3 metres = ₹ 200 + ₹ 50 = ₹ 250 Cost of digging for 4 metres = ₹ 250 + ₹ 50 = ₹ 300

The list of costs is: 150, 200, 250, 300, ... To check if this forms an AP, we check the difference between consecutive terms.

a_2 - a_1 = 200 - 150 = 50

a_3 - a_2 = 250 - 200 = 50

a_4 - a_3 = 300 - 250 = 50

Since the difference between consecutive terms is constant (50), the list of numbers forms an AP.

Final Answer: Yes, this situation forms an AP because the cost increases by a fixed amount (₹ 50) for each subsequent metre.

(iv)

The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at

8\%

per annum.

Given: Principal (P) = ₹ 10000 Rate of interest (R) =

8\%

per annum, compounded annually.

Formula: Amount after

n

years,

A_n = P \left( 1 + \frac{R}{100} \right)^n

Solution: Amount after 1 year (

A_1

) =

10000 \left( 1 + \frac{8}{100} \right)^1 = 10000(1.08) = 10800

Amount after 2 years (

A_2

) =

10000 \left( 1 + \frac{8}{100} \right)^2 = 10000(1.08)^2 = 10000(1.1664) = 11664

Amount after 3 years (

A_3

) =

10000 \left( 1 + \frac{8}{100} \right)^3 = 10000(1.08)^3 = 10000(1.259712) = 12597.12

The list of amounts is: 10800, 11664, 12597.12, ... Let's check the difference between consecutive terms.

A_2 - A_1 = 11664 - 10800 = 864

A_3 - A_2 = 12597.12 - 11664 = 933.12

Since the difference between consecutive terms is not constant, the list of numbers does not form an AP.

Final Answer: No, this situation does not form an AP because the amount of interest earned each year is not constant, so the difference between consecutive year-end amounts is not constant.

Q2EXERCISE 5.1

Write first four terms of the AP, when the first term $a$ and the common difference $d$ are given as follows:

(i)

$a=10, \quad d=10$

(ii)

$a=-2, \quad d=0$

(iii)

$a=4, \quad d=-3$

(iv)

$a=-1, d=\frac{1}{2}$

(v)

$a=-1.25, d=-0.25$

Solution

An Arithmetic Progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. The terms of an AP can be written as

a, a+d, a+2d, a+3d, \ldots

, where

a

is the first term and

d

is the common difference.

(i)

a=10, \quad d=10

Given: First term

a = 10

, common difference

d = 10

.

Solution: First term,

a_1 = a = 10

Second term,

a_2 = a + d = 10 + 10 = 20

Third term,

a_3 = a + 2d = 10 + 2(10) = 10 + 20 = 30

Fourth term,

a_4 = a + 3d = 10 + 3(10) = 10 + 30 = 40

Final Answer: The first four terms of the AP are 10, 20, 30, 40.

(ii)

a=-2, \quad d=0

Given: First term

a = -2

, common difference

d = 0

.

Solution: First term,

a_1 = a = -2

Second term,

a_2 = a + d = -2 + 0 = -2

Third term,

a_3 = a + 2d = -2 + 2(0) = -2

Fourth term,

a_4 = a + 3d = -2 + 3(0) = -2

Final Answer: The first four terms of the AP are -2, -2, -2, -2.

(iii)

a=4, \quad d=-3

Given: First term

a = 4

, common difference

d = -3

.

Solution: First term,

a_1 = a = 4

Second term,

a_2 = a + d = 4 + (-3) = 1

Third term,

a_3 = a + 2d = 4 + 2(-3) = 4 - 6 = -2

Fourth term,

a_4 = a + 3d = 4 + 3(-3) = 4 - 9 = -5

Final Answer: The first four terms of the AP are 4, 1, -2, -5.

(iv)

a=-1, d=\frac{1}{2}

Given: First term

a = -1

, common difference

d = \frac{1}{2}

.

Solution: First term,

a_1 = a = -1

Second term,

a_2 = a + d = -1 + \frac{1}{2} = -\frac{1}{2}

Third term,

a_3 = a + 2d = -1 + 2\left(\frac{1}{2}\right) = -1 + 1 = 0

Fourth term,

a_4 = a + 3d = -1 + 3\left(\frac{1}{2}\right) = -1 + \frac{3}{2} = \frac{1}{2}

Final Answer: The first four terms of the AP are -1,

-\frac{1}{2}

, 0,

\frac{1}{2}

.

(v)

a=-1.25, d=-0.25

Given: First term

a = -1.25

, common difference

d = -0.25

.

Solution: First term,

a_1 = a = -1.25

Second term,

a_2 = a + d = -1.25 + (-0.25) = -1.50

Third term,

a_3 = a + 2d = -1.25 + 2(-0.25) = -1.25 - 0.50 = -1.75

Fourth term,

a_4 = a + 3d = -1.25 + 3(-0.25) = -1.25 - 0.75 = -2.00

Final Answer: The first four terms of the AP are -1.25, -1.50, -1.75, -2.00.

Q3EXERCISE 5.1

For the following APs, write the first term and the common difference:

(i)

$3,1,-1,-3, \ldots$

(ii)

$-5,-1,3,7, \ldots$

(iii)

$\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots$

(iv)

$0.6,1.7,2.8,3.9, \ldots$

Solution

For an Arithmetic Progression (AP), the first term is the first number in the sequence. The common difference (

d

) is the difference between any term and its preceding term, i.e.,

d = a_n - a_{n-1}

.

(i)

3, 1, -1, -3, \ldots

Given AP:

3, 1, -1, -3, \ldots

Solution: The first term is the first number in the sequence. First term,

a = 3

.

The common difference is the difference between the second term and the first term. Common difference,

d = a_2 - a_1 = 1 - 3 = -2

.

Final Answer: The first term is

a = 3

and the common difference is

d = -2

.

(ii)

-5, -1, 3, 7, \ldots

Given AP:

-5, -1, 3, 7, \ldots

Solution: The first term is the first number in the sequence. First term,

a = -5

.

The common difference is the difference between the second term and the first term. Common difference,

d = a_2 - a_1 = -1 - (-5) = -1 + 5 = 4

.

Final Answer: The first term is

a = -5

and the common difference is

d = 4

.

(iii)

\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots

Given AP:

\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots

Solution: The first term is the first number in the sequence. First term,

a = \frac{1}{3}

.

The common difference is the difference between the second term and the first term. Common difference,

d = a_2 - a_1 = \frac{5}{3} - \frac{1}{3} = \frac{5-1}{3} = \frac{4}{3}

.

Final Answer: The first term is

a = \frac{1}{3}

and the common difference is

d = \frac{4}{3}

.

(iv)

0.6, 1.7, 2.8, 3.9, \ldots

Given AP:

0.6, 1.7, 2.8, 3.9, \ldots

Solution: The first term is the first number in the sequence. First term,

a = 0.6

.

The common difference is the difference between the second term and the first term. Common difference,

d = a_2 - a_1 = 1.7 - 0.6 = 1.1

.

Final Answer: The first term is

a = 0.6

and the common difference is

d = 1.1

.

Q4EXERCISE 5.1

Which of the following are APs ? If they form an AP, find the common difference $d$ and write three more terms.

(i)

$2,4,8,16, \ldots$

(ii)

$2, \frac{5}{2}, 3, \frac{7}{2}, \ldots$

(iii)

$-1.2,-3.2,-5.2,-7.2, \ldots$

(iv)

$-10,-6,-2,2, \ldots$

(v)

$3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \cdots$

(vi)

$0.2,0.22,0.222,0.2222, \ldots$

(vii)

$0,-4,-8,-12, \ldots$

(viii)

$-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots$

(ix)

$1,3,9,27, \ldots$

(x)

$a, 2 a, 3 a, 4 a, \ldots$ (xi) $a, a^{2}, a^{3}, a^{4}, \ldots$ (xii) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \cdots$ (xiii) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \cdots$ (xiv) $1^{2}, 3^{2}, 5^{2}, 7^{2}, \ldots$ (xv) $1^{2}, 5^{2}, 7^{2}, 73, \ldots$

Solution

To check if a sequence is an Arithmetic Progression (AP), we must verify that the difference between consecutive terms is constant. This constant value is the common difference,

d

. If it is an AP, the next terms are found by adding

d

to the last known term.

(i)

2, 4, 8, 16, \ldots

a_2 - a_1 = 4 - 2 = 2

a_3 - a_2 = 8 - 4 = 4

Since

a_2 - a_1 \neq a_3 - a_2

, the sequence is not an AP.

(ii)

2, \frac{5}{2}, 3, \frac{7}{2}, \ldots

a_2 - a_1 = \frac{5}{2} - 2 = \frac{5-4}{2} = \frac{1}{2}

a_3 - a_2 = 3 - \frac{5}{2} = \frac{6-5}{2} = \frac{1}{2}

a_4 - a_3 = \frac{7}{2} - 3 = \frac{7-6}{2} = \frac{1}{2}

This is an AP with

d = \frac{1}{2}

. Next three terms:

a_5 = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4

,

a_6 = 4 + \frac{1}{2} = \frac{9}{2}

,

a_7 = \frac{9}{2} + \frac{1}{2} = \frac{10}{2} = 5

.

(iii)

-1.2, -3.2, -5.2, -7.2, \ldots

a_2 - a_1 = -3.2 - (-1.2) = -2

a_3 - a_2 = -5.2 - (-3.2) = -2

This is an AP with

d = -2

. Next three terms:

a_5 = -7.2 + (-2) = -9.2

,

a_6 = -9.2 + (-2) = -11.2

,

a_7 = -11.2 + (-2) = -13.2

.

(iv)

-10, -6, -2, 2, \ldots

a_2 - a_1 = -6 - (-10) = 4

a_3 - a_2 = -2 - (-6) = 4

This is an AP with

d = 4

. Next three terms:

a_5 = 2 + 4 = 6

,

a_6 = 6 + 4 = 10

,

a_7 = 10 + 4 = 14

.

(v)

3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \ldots

a_2 - a_1 = (3+\sqrt{2}) - 3 = \sqrt{2}

a_3 - a_2 = (3+2\sqrt{2}) - (3+\sqrt{2}) = \sqrt{2}

This is an AP with

d = \sqrt{2}

. Next three terms:

a_5 = (3+3\sqrt{2}) + \sqrt{2} = 3+4\sqrt{2}

,

a_6 = (3+4\sqrt{2}) + \sqrt{2} = 3+5\sqrt{2}

,

a_7 = (3+5\sqrt{2}) + \sqrt{2} = 3+6\sqrt{2}

.

(vi)

0.2, 0.22, 0.222, 0.2222, \ldots

a_2 - a_1 = 0.22 - 0.2 = 0.02

a_3 - a_2 = 0.222 - 0.22 = 0.002

Since the differences are not equal, this is not an AP.

(vii)

0, -4, -8, -12, \ldots

a_2 - a_1 = -4 - 0 = -4

a_3 - a_2 = -8 - (-4) = -4

This is an AP with

d = -4

. Next three terms:

a_5 = -12 + (-4) = -16

,

a_6 = -16 + (-4) = -20

,

a_7 = -20 + (-4) = -24

.

(viii)

-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \ldots

a_2 - a_1 = -\frac{1}{2} - (-\frac{1}{2}) = 0

This is an AP with

d = 0

. Next three terms:

a_5 = -\frac{1}{2} + 0 = -\frac{1}{2}

,

a_6 = -\frac{1}{2}

,

a_7 = -\frac{1}{2}

.

(ix)

1, 3, 9, 27, \ldots

a_2 - a_1 = 3 - 1 = 2

a_3 - a_2 = 9 - 3 = 6

Since the differences are not equal, this is not an AP.

(x)

a, 2a, 3a, 4a, \ldots

a_2 - a_1 = 2a - a = a

a_3 - a_2 = 3a - 2a = a

This is an AP with

d = a

. Next three terms:

a_5 = 4a + a = 5a

,

a_6 = 5a + a = 6a

,

a_7 = 6a + a = 7a

.

(xi)

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

a_2 - a_1 = a^2 - a = a(a-1)

a_3 - a_2 = a^3 - a^2 = a^2(a-1)

For the differences to be equal,

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

. This holds if

a=0

,

a=1

, or

a=a^2

. In general, for an arbitrary

a

, this is not an AP.

(xii)

\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots

Simplifying the terms:

\sqrt{2}, \sqrt{4 \times 2}, \sqrt{9 \times 2}, \sqrt{16 \times 2}, \ldots

which is

\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \ldots

a_2 - a_1 = 2\sqrt{2} - \sqrt{2} = \sqrt{2}

a_3 - a_2 = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2}

This is an AP with

d = \sqrt{2}

. Next three terms:

a_5 = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{50}

,

a_6 = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{72}

,

a_7 = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{98}

.

(xiii)

\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots

Simplifying the terms:

\sqrt{3}, \sqrt{6}, 3, 2\sqrt{3}, \ldots

a_2 - a_1 = \sqrt{6} - \sqrt{3}

a_3 - a_2 = 3 - \sqrt{6}

Since the differences are not equal, this is not an AP.

(xiv)

1^2, 3^2, 5^2, 7^2, \ldots

The sequence is

1, 9, 25, 49, \ldots

a_2 - a_1 = 9 - 1 = 8

a_3 - a_2 = 25 - 9 = 16

Since the differences are not equal, this is not an AP.

(xv)

1^2, 5^2, 7^2, 73, \ldots

The sequence is

1, 25, 49, 73, \ldots

a_2 - a_1 = 25 - 1 = 24

a_3 - a_2 = 49 - 25 = 24

a_4 - a_3 = 73 - 49 = 24

This is an AP with

d = 24

. Next three terms:

a_5 = 73 + 24 = 97

,

a_6 = 97 + 24 = 121 = 11^2

,

a_7 = 121 + 24 = 145

.

Q1EXERCISE 5.2

Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_{n}$ the $n$ th term of the AP: \begin{tabular}{|r|r|r|r|} \hline $a$ & $d$ & $n$ & $a_{n}$
\hline 7 & 3 & 8 & $\ldots$
-18 & $\ldots$ & 10 & 0
$\ldots$ & -3 & 18 & -5
-18.9 & 2.5 & $\ldots$ & 3.6
3.5 & 0 & 105 & $\ldots$
\hline \end{tabular}

Solution

The formula for the

n

th term of an Arithmetic Progression (AP) is:

a_n = a + (n-1)d

where

a

is the first term,

d

is the common difference, and

n

is the term number.

(i)

Given:

a=7, d=3, n=8

To Find:

a_n

Solution: Using the formula

a_n = a + (n-1)d

:

a_8 = 7 + (8-1)3

a_8 = 7 + (7)3

a_8 = 7 + 21

a_8 = 28

Final Answer:

a_n = 28

(ii)

Given:

a=-18, n=10, a_n=0

To Find:

d

Solution: Using the formula

a_n = a + (n-1)d

:

0 = -18 + (10-1)d

0 = -18 + 9d

18 = 9d

d = \frac{18}{9}

d = 2

Final Answer:

d = 2

(iii)

Given:

d=-3, n=18, a_n=-5

To Find:

a

Solution: Using the formula

a_n = a + (n-1)d

:

-5 = a + (18-1)(-3)

-5 = a + (17)(-3)

-5 = a - 51

a = -5 + 51

a = 46

Final Answer:

a = 46

(iv)

Given:

a=-18.9, d=2.5, a_n=3.6

To Find:

n

Solution: Using the formula

a_n = a + (n-1)d

:

3.6 = -18.9 + (n-1)2.5

3.6 + 18.9 = (n-1)2.5

22.5 = (n-1)2.5

n-1 = \frac{22.5}{2.5}

n-1 = \frac{225}{25}

n-1 = 9

n = 9 + 1

n = 10

Final Answer:

n = 10

(v)

Given:

a=3.5, d=0, n=105

To Find:

a_n

Solution: Using the formula

a_n = a + (n-1)d

:

a_{105} = 3.5 + (105-1)0

a_{105} = 3.5 + (104)0

a_{105} = 3.5 + 0

a_{105} = 3.5

Final Answer:

a_n = 3.5

Q2EXERCISE 5.2

Choose the correct choice in the following and justify :

(i)

30 th term of the AP: $10,7,4, \ldots$ , is

(A)

97

(B)

77

(C)

-77

(D)

-87

(ii)

11th term of the AP: $-3,-\frac{1}{2}, 2, \ldots$ , is

(A)

28

(B)

22

(C)

-38

(D)

-48 \frac{1}{2}

Solution

(i) 30th term of the AP: $10, 7, 4, \ldots$ , is

Given: The Arithmetic Progression (AP) is

10, 7, 4, \ldots

.

To Find: The 30th term (

a_{30}

) and choose the correct option.

Solution: The first term of the AP is

a = 10

. The common difference is

d = 7 - 10 = -3

. We need to find the 30th term, so

n = 30

.

The formula for the

n

-th term of an AP is:

a_n = a + (n-1)d

Substituting the values:

a_{30} = 10 + (30-1)(-3)

a_{30} = 10 + (29)(-3)

a_{30} = 10 - 87

a_{30} = -77

Final Answer: The 30th term is -77. This corresponds to option (C).

(ii) 11th term of the AP: $-3, -\frac{1}{2}, 2, \ldots$ , is

Given: The AP is

-3, -\frac{1}{2}, 2, \ldots

.

To Find: The 11th term (

a_{11}

) and choose the correct option.

Solution: The first term of the AP is

a = -3

. The common difference is

d = a_2 - a_1 = -\frac{1}{2} - (-3) = -\frac{1}{2} + 3 = \frac{-1+6}{2} = \frac{5}{2}

. We need to find the 11th term, so

n = 11

.

The formula for the

n

-th term of an AP is:

a_n = a + (n-1)d

Substituting the values:

a_{11} = -3 + (11-1)\left(\frac{5}{2}\right)

a_{11} = -3 + (10)\left(\frac{5}{2}\right)

a_{11} = -3 + 5 \times 5

a_{11} = -3 + 25

a_{11} = 22

Final Answer: The 11th term is 22. This corresponds to option (B).

Q3EXERCISE 5.2

In the following APs, find the missing terms in the boxes :

(i)

$2, \square, 26$

(ii)

$\square, 13, \square, 3$

(iii)

$5, \square, \square, 9 \frac{1}{2}$

(iv)

$-4, \square, \square, \square, \square, 6$

(v)

$\square, 38, \square, \square, \square,-22$

Solution

To Find: The missing terms in the boxes for the following APs.

(i) $2, \square, 26$

Solution: Let the AP be

a_1, a_2, a_3

. We are given

a_1 = 2

and

a_3 = 26

. The missing term is

a_2

. In an AP, any term is the average of its preceding and succeeding terms.

a_2 = \frac{a_1 + a_3}{2}

a_2 = \frac{2 + 26}{2} = \frac{28}{2} = 14

Alternatively, using the formula

a_n = a + (n-1)d

:

a_1 = a = 2

a_3 = a + 2d = 26

2 + 2d = 26 \implies 2d = 24 \implies d = 12

.

a_2 = a + d = 2 + 12 = 14

. Final Answer: The missing term is 14.

(ii) $\square, 13, \square, 3$

Solution: Let the AP be

a_1, a_2, a_3, a_4

. We are given

a_2 = 13

and

a_4 = 3

. Using the formula

a_n = a + (n-1)d

:

a_2 = a + d = 13 \quad ...(1)

a_4 = a + 3d = 3 \quad ...(2)

Subtracting equation (1) from (2):

(a + 3d) - (a + d) = 3 - 13

2d = -10 \implies d = -5

. Substituting

d = -5

into equation (1):

a + (-5) = 13 \implies a = 18

. The missing terms are

a_1 = a = 18

and

a_3 = a + 2d = 18 + 2(-5) = 18 - 10 = 8

. Final Answer: The missing terms are 18 and 8.

(iii) $5, \square, \square, 9 \frac{1}{2}$

Solution: Let the AP be

a_1, a_2, a_3, a_4

. We are given

a_1 = 5

and

a_4 = 9 \frac{1}{2} = \frac{19}{2}

. Using the formula

a_n = a + (n-1)d

:

a_1 = a = 5

a_4 = a + 3d = \frac{19}{2}

5 + 3d = \frac{19}{2}

3d = \frac{19}{2} - 5 = \frac{19 - 10}{2} = \frac{9}{2}

d = \frac{9}{2} \times \frac{1}{3} = \frac{3}{2}

. The missing terms are:

a_2 = a + d = 5 + \frac{3}{2} = \frac{10+3}{2} = \frac{13}{2} = 6 \frac{1}{2}

.

a_3 = a + 2d = 5 + 2(\frac{3}{2}) = 5 + 3 = 8

. Final Answer: The missing terms are

6 \frac{1}{2}

and 8.

(iv) $-4, \square, \square, \square, \square, 6$

Solution: We are given

a_1 = -4

and

a_6 = 6

. Using the formula

a_n = a + (n-1)d

:

a_1 = a = -4

a_6 = a + 5d = 6

-4 + 5d = 6 \implies 5d = 10 \implies d = 2

. The missing terms are:

a_2 = a + d = -4 + 2 = -2

.

a_3 = a + 2d = -4 + 2(2) = 0

.

a_4 = a + 3d = -4 + 3(2) = 2

.

a_5 = a + 4d = -4 + 4(2) = 4

. Final Answer: The missing terms are -2, 0, 2, and 4.

(v) $\square, 38, \square, \square, \square, -22$

Solution: We are given

a_2 = 38

and

a_6 = -22

. Using the formula

a_n = a + (n-1)d

:

a_2 = a + d = 38 \quad ...(1)

a_6 = a + 5d = -22 \quad ...(2)

Subtracting equation (1) from (2):

(a + 5d) - (a + d) = -22 - 38

4d = -60 \implies d = -15

. Substituting

d = -15

into equation (1):

a + (-15) = 38 \implies a = 53

. The missing terms are:

a_1 = a = 53

.

a_3 = a + 2d = 53 + 2(-15) = 53 - 30 = 23

.

a_4 = a + 3d = 53 + 3(-15) = 53 - 45 = 8

.

a_5 = a + 4d = 53 + 4(-15) = 53 - 60 = -7

. Final Answer: The missing terms are 53, 23, 8, and -7.

Q4EXERCISE 5.2

Which term of the AP: $3,8,13,18, \ldots$ , is 78 ?

Solution

Given: The AP is

3, 8, 13, 18, \ldots

and a term of this AP is 78.

To Find: Which term of the AP is 78.

Solution: Let's identify the properties of the given AP. First term,

a = 3

. Common difference,

d = 8 - 3 = 5

.

Let the

n

-th term of the AP be 78. So,

a_n = 78

. We need to find the value of

n

.

The formula for the

n

-th term of an AP is:

a_n = a + (n-1)d

Substitute the known values into the formula:

78 = 3 + (n-1)5

Now, we solve for

n

:

78 - 3 = (n-1)5

75 = (n-1)5

\frac{75}{5} = n-1

15 = n-1

n = 15 + 1

n = 16

This means that the 16th term of the given AP is 78.

Final Answer: The 16th term of the AP is 78.

Q5EXERCISE 5.2

Find the number of terms in each of the following APs :

(i)

$7,13,19, \ldots, 205$

(ii)

$18,15 \frac{1}{2}, 13, \ldots,-47$

Solution

(i) $7, 13, 19, \ldots, 205$

Given: An AP with first term 7, last term 205, and a common difference.

To Find: The number of terms in the AP.

Solution: The given AP is

7, 13, 19, \ldots, 205

. First term,

a = 7

. Common difference,

d = 13 - 7 = 6

. The last term of the AP is

a_n = 205

.

We use the formula for the

n

-th term of an AP:

a_n = a + (n-1)d

Substitute the values:

205 = 7 + (n-1)6

205 - 7 = (n-1)6

198 = (n-1)6

n-1 = \frac{198}{6}

n-1 = 33

n = 33 + 1 = 34

Final Answer: There are 34 terms in this AP.

(ii) $18, 15 \frac{1}{2}, 13, \ldots, -47$

Given: An AP with first term 18, last term -47, and a common difference.

To Find: The number of terms in the AP.

Solution: The given AP is

18, 15 \frac{1}{2}, 13, \ldots, -47

. First term,

a = 18

. Common difference,

d = 15 \frac{1}{2} - 18 = \frac{31}{2} - 18 = \frac{31 - 36}{2} = -\frac{5}{2}

. The last term of the AP is

a_n = -47

.

We use the formula for the

n

-th term of an AP:

a_n = a + (n-1)d

Substitute the values:

-47 = 18 + (n-1)\left(-\frac{5}{2}\right)

-47 - 18 = (n-1)\left(-\frac{5}{2}\right)

-65 = (n-1)\left(-\frac{5}{2}\right)

n-1 = -65 \times \left(-\frac{2}{5}\right)

n-1 = 13 \times 2

n-1 = 26

n = 26 + 1 = 27

Final Answer: There are 27 terms in this AP.

Q6EXERCISE 5.2

Check whether -150 is a term of the AP : $11,8,5,2 \ldots$

Solution

Given: The AP is

11, 8, 5, 2, \ldots

To Find: Whether -150 is a term of this AP.

Solution: First, let's identify the properties of the given AP. First term,

a = 11

. Common difference,

d = 8 - 11 = -3

.

Let us assume that -150 is the

n

-th term of this AP. So,

a_n = -150

. We need to find the value of

n

. If

n

is a positive integer, then -150 is a term of the AP. Otherwise, it is not.

The formula for the

n

-th term of an AP is:

a_n = a + (n-1)d

Substitute the known values into the formula:

-150 = 11 + (n-1)(-3)

Now, we solve for

n

:

-150 - 11 = (n-1)(-3)

-161 = (n-1)(-3)

n-1 = \frac{-161}{-3}

n-1 = \frac{161}{3}

Since 161 is not divisible by 3, the value of

n-1

is not an integer. Let us calculate

n

anyway.

n = \frac{161}{3} + 1

n = \frac{161 + 3}{3}

n = \frac{164}{3} = 54.66...

Since

n

is not a positive integer, -150 is not a term of the given AP. The term number must be a natural number.

Final Answer: No, -150 is not a term of the AP

11, 8, 5, 2, \ldots

because the term number

n

is not a positive integer.

Q7EXERCISE 5.2

Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Solution

Given: An Arithmetic Progression (AP) where: The 11th term,

a_{11} = 38

. The 16th term,

a_{16} = 73

.

To Find: The 31st term,

a_{31}

.

Formula: The formula for the

n

-th term of an AP is:

a_n = a + (n-1)d

where

a

is the first term and

d

is the common difference.

Solution: Using the formula, we can write two equations based on the given information: For the 11th term:

a_{11} = a + (11-1)d = a + 10d = 38 \quad ...(1)

For the 16th term:

a_{16} = a + (16-1)d = a + 15d = 73 \quad ...(2)

Now, we solve these two linear equations for

a

and

d

. Subtracting equation (1) from equation (2):

(a + 15d) - (a + 10d) = 73 - 38

5d = 35

d = \frac{35}{5} = 7

Now, substitute the value of

d=7

into equation (1):

a + 10(7) = 38

a + 70 = 38

a = 38 - 70

a = -32

We have found the first term

a = -32

and the common difference

d = 7

. Now we can find the 31st term:

a_{31} = a + (31-1)d

a_{31} = a + 30d

a_{31} = -32 + 30(7)

a_{31} = -32 + 210

a_{31} = 178

Final Answer: The 31st term of the AP is 178.

Q8EXERCISE 5.2

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106 . Find the 29th term.

Solution

Given: An Arithmetic Progression (AP) with: Total number of terms,

n = 50

. The 3rd term,

a_3 = 12

. The last term (which is the 50th term),

a_{50} = 106

.

To Find: The 29th term,

a_{29}

.

Formula: The formula for the

n

-th term of an AP is:

a_n = a + (n-1)d

where

a

is the first term and

d

is the common difference.

Solution: Using the formula, we can write two equations based on the given information: For the 3rd term:

a_3 = a + (3-1)d = a + 2d = 12 \quad ...(1)

For the 50th term:

a_{50} = a + (50-1)d = a + 49d = 106 \quad ...(2)

Now, we solve these two linear equations for

a

and

d

. Subtracting equation (1) from equation (2):

(a + 49d) - (a + 2d) = 106 - 12

47d = 94

d = \frac{94}{47} = 2

Now, substitute the value of

d=2

into equation (1):

a + 2(2) = 12

a + 4 = 12

a = 12 - 4

a = 8

We have found the first term

a = 8

and the common difference

d = 2

. Now we can find the 29th term:

a_{29} = a + (29-1)d

a_{29} = a + 28d

a_{29} = 8 + 28(2)

a_{29} = 8 + 56

a_{29} = 64

Final Answer: The 29th term of the AP is 64.

Q9EXERCISE 5.2

If the 3 rd and the 9 th terms of an AP are 4 and -8 respectively, which term of this AP is zero?

Solution

Given: An Arithmetic Progression (AP) where: The 3rd term,

a_3 = 4

. The 9th term,

a_9 = -8

.

To Find: Which term of this AP is zero. Let this be the

n

-th term, so we need to find

n

such that

a_n = 0

.

Formula: The formula for the

n

-th term of an AP is:

a_n = a + (n-1)d

where

a

is the first term and

d

is the common difference.

Solution: Using the formula, we can write two equations based on the given information: For the 3rd term:

a_3 = a + (3-1)d = a + 2d = 4 \quad ...(1)

For the 9th term:

a_9 = a + (9-1)d = a + 8d = -8 \quad ...(2)

Now, we solve these two linear equations for

a

and

d

. Subtracting equation (1) from equation (2):

(a + 8d) - (a + 2d) = -8 - 4

6d = -12

d = \frac{-12}{6} = -2

Now, substitute the value of

d=-2

into equation (1):

a + 2(-2) = 4

a - 4 = 4

a = 4 + 4

a = 8

We have found the first term

a = 8

and the common difference

d = -2

. Now, we need to find which term is zero. Let the

n

-th term,

a_n

, be 0.

a_n = a + (n-1)d = 0

8 + (n-1)(-2) = 0

-2(n-1) = -8

n-1 = \frac{-8}{-2}

n-1 = 4

n = 5

Final Answer: The 5th term of this AP is zero.

Q10EXERCISE 5.2

The 17 th term of an AP exceeds its 10 th term by 7 . Find the common difference.

Solution

Given: In an Arithmetic Progression (AP), the 17th term exceeds its 10th term by 7. This can be written as:

a_{17} = a_{10} + 7

.

To Find: The common difference,

d

.

Formula: The formula for the

n

-th term of an AP is:

a_n = a + (n-1)d

where

a

is the first term and

d

is the common difference.

Solution: First, let's express the 17th and 10th terms using the formula:

a_{17} = a + (17-1)d = a + 16d

a_{10} = a + (10-1)d = a + 9d

Now, substitute these expressions into the given condition

a_{17} = a_{10} + 7

:

a + 16d = (a + 9d) + 7

Now, we solve this equation for

d

.

a + 16d = a + 9d + 7

Subtract

a

from both sides:

16d = 9d + 7

Subtract

9d

from both sides:

16d - 9d = 7

7d = 7

d = \frac{7}{7}

d = 1

Final Answer: The common difference is 1.

Q11EXERCISE 5.2

Which term of the AP: $3,15,27,39, \ldots$ will be 132 more than its 54 th term?

Solution

Given: The Arithmetic Progression (AP) is

3, 15, 27, 39, \ldots

.

To Find: Which term of the AP is 132 more than its 54th term. Let the required term be the

n

-th term,

a_n

. So, we need to find

n

such that

a_n = a_{54} + 132

.

Formula: The formula for the

n

-th term of an AP is:

a_n = a + (n-1)d

where

a

is the first term and

d

is the common difference.

Solution: First, let's identify the first term and common difference from the given AP. First term,

a = 3

. Common difference,

d = a_2 - a_1 = 15 - 3 = 12

.

Let the required term be the

n

-th term,

a_n

. According to the question:

a_n = a_{54} + 132

Now, we express

a_n

and

a_{54}

using the formula:

a + (n-1)d = [a + (54-1)d] + 132

a + (n-1)d = a + 53d + 132

Subtract

a

from both sides:

(n-1)d = 53d + 132

Now, substitute the value of the common difference,

d=12

:

(n-1)12 = 53(12) + 132

12(n-1) = 636 + 132

12(n-1) = 768

Divide both sides by 12:

n-1 = \frac{768}{12}

n-1 = 64

n = 64 + 1

n = 65

So, the 65th term of the AP is 132 more than its 54th term.

Verification (Optional):

a_{54} = a + 53d = 3 + 53(12) = 3 + 636 = 639

.

a_{65} = a + 64d = 3 + 64(12) = 3 + 768 = 771

. Difference =

a_{65} - a_{54} = 771 - 639 = 132

. This matches the condition.

Final Answer: The 65th term of the AP will be 132 more than its 54th term.

Q12EXERCISE 5.2

Two APs have the same common difference. The difference between their 100 th terms is 100, what is the difference between their 1000th terms?

Solution

Given: Two APs have the same common difference, let it be

d

. The difference between their 100th terms is 100.

To Find: The difference between their 1000th terms.

Let: Let the first term of the first AP be

a_1

and its

n

th term be

A_n

. Let the first term of the second AP be

a_2

and its

n

th term be

B_n

.

Formula: The

n

th term of an AP is given by

a_n = a + (n-1)d

, where

a

is the first term and

d

is the common difference.

Solution: For the first AP, the 100th term is:

A_{100} = a_1 + (100-1)d = a_1 + 99d

For the second AP, the 100th term is:

B_{100} = a_2 + (100-1)d = a_2 + 99d

It is given that the difference between their 100th terms is 100.

A_{100} - B_{100} = 100

(a_1 + 99d) - (a_2 + 99d) = 100

a_1 + 99d - a_2 - 99d = 100

a_1 - a_2 = 100 \quad ... (1)

Now, we need to find the difference between their 1000th terms, which is

A_{1000} - B_{1000}

.

For the first AP, the 1000th term is:

A_{1000} = a_1 + (1000-1)d = a_1 + 999d

For the second AP, the 1000th term is:

B_{1000} = a_2 + (1000-1)d = a_2 + 999d

The difference between their 1000th terms is:

A_{1000} - B_{1000} = (a_1 + 999d) - (a_2 + 999d)

= a_1 + 999d - a_2 - 999d

= a_1 - a_2

From equation (1), we know that

a_1 - a_2 = 100

. Therefore, the difference between their 1000th terms is also 100.

Final Answer: The difference between their 1000th terms is 100.

Q13EXERCISE 5.2

How many three-digit numbers are divisible by 7 ?

Solution

To Find: How many three-digit numbers are divisible by 7.

Solution: First, we need to identify the first and the last three-digit numbers that are divisible by 7. These numbers will form an Arithmetic Progression (AP).

The smallest three-digit number is 100. To find the first three-digit number divisible by 7, we divide 100 by 7.

100 = 7 \times 14 + 2

The remainder is 2. So, 100 is not divisible by 7. The next number divisible by 7 will be

100 - 2 + 7 = 105

. So, the first three-digit number divisible by 7 is

a = 105

.

The largest three-digit number is 999. To find the last three-digit number divisible by 7, we divide 999 by 7.

999 = 7 \times 142 + 5

The remainder is 5. To get a number divisible by 7, we subtract the remainder from 999.

999 - 5 = 994

So, the last three-digit number divisible by 7 is

a_n = 994

.

The list of three-digit numbers divisible by 7 forms an AP:

105, 112, 119, \ldots, 994

. Here, the first term is

a = 105

. The common difference is

d = 112 - 105 = 7

. The last term is

a_n = 994

.

We need to find the number of terms,

n

, in this AP.

Formula: The

n

th term of an AP is given by

a_n = a + (n-1)d

.

Calculation: Substituting the values into the formula:

994 = 105 + (n-1) \times 7

994 - 105 = (n-1) \times 7

889 = (n-1) \times 7

n-1 = \frac{889}{7}

n-1 = 127

n = 127 + 1

n = 128

So, there are 128 three-digit numbers that are divisible by 7.

Final Answer: There are 128 three-digit numbers divisible by 7.

Q14EXERCISE 5.2

How many multiples of 4 lie between 10 and 250 ?

Solution

To Find: How many multiples of 4 lie between 10 and 250.

Solution: First, we need to identify the first and the last multiples of 4 that lie between 10 and 250. These numbers will form an Arithmetic Progression (AP).

The first number greater than 10 that is a multiple of 4 is 12. So, the first term of the AP is

a = 12

.

The numbers are multiples of 4, so the common difference is

d = 4

.

Now, we need to find the last multiple of 4 before 250. We divide 250 by 4.

250 = 4 \times 62 + 2

The remainder is 2. To get a number divisible by 4, we subtract the remainder from 250.

250 - 2 = 248

So, the last multiple of 4 before 250 is

a_n = 248

.

The AP is

12, 16, 20, \ldots, 248

. Here, the first term is

a = 12

. The common difference is

d = 4

. The last term is

a_n = 248

.

We need to find the number of terms,

n

, in this AP.

Formula: The

n

th term of an AP is given by

a_n = a + (n-1)d

.

Calculation: Substituting the values into the formula:

248 = 12 + (n-1) \times 4

248 - 12 = (n-1) \times 4

236 = (n-1) \times 4

n-1 = \frac{236}{4}

n-1 = 59

n = 59 + 1

n = 60

So, there are 60 multiples of 4 that lie between 10 and 250.

Final Answer: There are 60 multiples of 4 between 10 and 250.

Q15EXERCISE 5.2

For what value of $n$ , are the $n$ th terms of two APs: $63,65,67, \ldots$ and $3,10,17, \ldots$ equal?

Solution

Given: Two APs: First AP:

63, 65, 67, \ldots

Second AP:

3, 10, 17, \ldots

To Find: The value of

n

for which the

n

th terms of the two APs are equal.

Formula: The

n

th term of an AP is given by

a_n = a + (n-1)d

, where

a

is the first term and

d

is the common difference.

Solution: For the first AP:

63, 65, 67, \ldots

First term,

a_1 = 63

. Common difference,

d_1 = 65 - 63 = 2

. The

n

th term of this AP is:

A_n = a_1 + (n-1)d_1

A_n = 63 + (n-1)2

A_n = 63 + 2n - 2

A_n = 61 + 2n \quad ... (1)

For the second AP:

3, 10, 17, \ldots

First term,

a_2 = 3

. Common difference,

d_2 = 10 - 3 = 7

. The

n

th term of this AP is:

B_n = a_2 + (n-1)d_2

B_n = 3 + (n-1)7

B_n = 3 + 7n - 7

B_n = 7n - 4 \quad ... (2)

We are given that the

n

th terms of the two APs are equal, i.e.,

A_n = B_n

. Equating equations (1) and (2):

61 + 2n = 7n - 4

61 + 4 = 7n - 2n

65 = 5n

n = \frac{65}{5}

n = 13

Thus, the 13th terms of both APs are equal.

Verification: 13th term of the first AP:

A_{13} = 61 + 2(13) = 61 + 26 = 87

. 13th term of the second AP:

B_{13} = 7(13) - 4 = 91 - 4 = 87

. Since

A_{13} = B_{13}

, our answer is correct.

Final Answer: For

n=13

, the

n

th terms of the two given APs are equal.

Q16EXERCISE 5.2

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12 .

Solution

Given: For an Arithmetic Progression (AP): The third term is 16. The 7th term exceeds the 5th term by 12.

To Find: Determine the AP.

Let: Let the first term of the AP be

a

and the common difference be

d

.

Formula: The

n

th term of an AP is given by

a_n = a + (n-1)d

.

Solution: From the given information, we can form two equations.

The third term is 16. $a_3 = 16$ $a + (3-1)d = 16$ $a + 2d = 16 \quad ... (1)$
The 7th term exceeds the 5th term by 12. $a_7 = a_5 + 12$ $a + (7-1)d = [a + (5-1)d] + 12$ $a + 6d = a + 4d + 12$ $a - a + 6d - 4d = 12$ $2d = 12$ $d = \frac{12}{2}$ $d = 6$

Now, substitute the value of

d

in equation (1) to find

a

.

a + 2(6) = 16

a + 12 = 16

a = 16 - 12

a = 4

So, the first term is

a = 4

and the common difference is

d = 6

.

The AP is formed by the terms

a, a+d, a+2d, a+3d, \ldots

Substituting the values of

a

and

d

: First term = 4 Second term =

4 + 6 = 10

Third term =

4 + 2(6) = 4 + 12 = 16

Fourth term =

4 + 3(6) = 4 + 18 = 22

and so on.

The required AP is

4, 10, 16, 22, \ldots

Final Answer: The AP is

4, 10, 16, 22, \ldots

Q17EXERCISE 5.2

Find the 20 th term from the last term of the AP : $3,8,13, \ldots, 253$ .

Solution

Given: The Arithmetic Progression (AP) is

3, 8, 13, \ldots, 253

.

To Find: The 20th term from the last term of the given AP.

Method: To find the 20th term from the last term, we can reverse the AP and find the 20th term from the beginning of the new AP.

The given AP is

3, 8, 13, \ldots, 253

. First term,

a = 3

. Common difference,

d = 8 - 3 = 5

. Last term,

l = 253

.

Reversing the AP, we get a new AP:

253, 248, \ldots, 13, 8, 3

. For this new AP: First term,

a' = 253

. Common difference,

d' = 8 - 13 = -5

.

We need to find the 20th term of this new AP.

Formula: The

n

-th term of an AP is given by

a_n = a + (n-1)d

.

Solution: For the reversed AP, we want to find the 20th term (

n=20

).

a'_{20} = a' + (20-1)d'

a'_{20} = 253 + (19)(-5)

a'_{20} = 253 - 95

a'_{20} = 158

Thus, the 20th term from the last term of the original AP is 158.

Final Answer: The 20th term from the last term of the AP is 158.

Q18EXERCISE 5.2

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution

Given: Let the first term of the AP be

a

and the common difference be

d

. The sum of the 4th and 8th terms is 24:

a_4 + a_8 = 24

. The sum of the 6th and 10th terms is 44:

a_6 + a_{10} = 44

.

To Find: The first three terms of the AP.

Formula: The

n

-th term of an AP is given by

a_n = a + (n-1)d

.

Solution: Using the formula for the

n

-th term:

a_4 = a + 3d

a_8 = a + 7d

a_6 = a + 5d

a_{10} = a + 9d

From the first condition:

a_4 + a_8 = 24

(a + 3d) + (a + 7d) = 24

2a + 10d = 24

Dividing by 2, we get:

a + 5d = 12 \quad \ldots(1)

From the second condition:

a_6 + a_{10} = 44

(a + 5d) + (a + 9d) = 44

2a + 14d = 44

Dividing by 2, we get:

a + 7d = 22 \quad \ldots(2)

Now, we solve the two linear equations. Subtracting equation (1) from equation (2):

(a + 7d) - (a + 5d) = 22 - 12

2d = 10

d = 5

Substitute the value of

d

in equation (1):

a + 5(5) = 12

a + 25 = 12

a = 12 - 25

a = -13

So, the first term is

a = -13

and the common difference is

d = 5

.

The first three terms of the AP are: First term:

a_1 = a = -13

Second term:

a_2 = a + d = -13 + 5 = -8

Third term:

a_3 = a + 2d = -13 + 2(5) = -13 + 10 = -3

Final Answer: The first three terms of the AP are

-13, -8, -3

.

Q19EXERCISE 5.2

Subba Rao started work in 1995 at an annual salary of $₹ 5000$ and received an increment of $₹ 200$ each year. In which year did his income reach $₹ 7000$ ?

Solution

Given: Subba Rao's starting annual salary in 1995 =

₹ 5000

. Annual increment =

₹ 200

. Final annual salary =

₹ 7000

.

To Find: The year in which his income reached

₹ 7000

.

Let: The annual salaries form an Arithmetic Progression (AP). First term,

a = 5000

. Common difference,

d = 200

. Let the

n

-th term be the year his salary reaches

₹ 7000

. So,

a_n = 7000

.

Formula: The

n

-th term of an AP is given by

a_n = a + (n-1)d

.

Solution: Substituting the values into the formula:

7000 = 5000 + (n-1)200

7000 - 5000 = (n-1)200

2000 = (n-1)200

n-1 = \frac{2000}{200}

n-1 = 10

n = 11

This means it took 11 years for his salary to reach

₹ 7000

. The starting year was 1995 (which is the 1st year). The 11th year will be: Year = Starting Year +

(n-1)

Year =

1995 + (11-1)

Year =

1995 + 10

Year =

2005

Final Answer: In the year 2005, Subba Rao's income reached

₹ 7000

.

Q20EXERCISE 5.2

Ramkali saved $₹ 5$ in the first week of a year and then increased her weekly savings by $₹ 1.75$ . If in the $n$ th week, her weekly savings become $₹ 20.75$ , find $n$ .

Solution

Given: Ramkali's savings in the first week =

₹ 5

. Increase in weekly savings =

₹ 1.75

. Her weekly savings in the

n

-th week =

₹ 20.75

.

To Find: The value of

n

.

Let: The weekly savings form an Arithmetic Progression (AP). First term (savings in 1st week),

a = 5

. Common difference (weekly increase),

d = 1.75

. The savings in the

n

-th week,

a_n = 20.75

.

Formula: The

n

-th term of an AP is given by

a_n = a + (n-1)d

.

Solution: Substituting the values into the formula:

20.75 = 5 + (n-1)1.75

20.75 - 5 = (n-1)1.75

15.75 = (n-1)1.75

n-1 = \frac{15.75}{1.75}

To simplify the fraction, we can multiply the numerator and denominator by 100:

n-1 = \frac{1575}{175}

Dividing both by 25:

n-1 = \frac{63}{7}

n-1 = 9

n = 10

Final Answer: The value of

n

is 10.

Q1EXERCISE 5.3

Find the sum of the following APs:

(i)

$2,7,12, \ldots$ , to 10 terms.

(ii)

$-37,-33,-29, \ldots$ , to 12 terms.

(iii)

$0.6,1.7,2.8, \ldots$ , to 100 terms.

(iv)

$\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots$ , to 11 terms.

Solution

To Find: The sum of the given Arithmetic Progressions (APs).

Formula: The sum of the first

n

terms of an AP is given by

S_n = \frac{n}{2}[2a + (n-1)d]

, where

a

is the first term,

d

is the common difference, and

n

is the number of terms.

(i) $2, 7, 12, \ldots$ , to 10 terms. Given: First term,

a = 2

. Common difference,

d = 7 - 2 = 5

. Number of terms,

n = 10

.

Solution:

S_{10} = \frac{10}{2}[2(2) + (10-1)5]

S_{10} = 5[4 + (9)(5)]

S_{10} = 5[4 + 45]

S_{10} = 5[49]

S_{10} = 245

(ii) $-37, -33, -29, \ldots$ , to 12 terms. Given: First term,

a = -37

. Common difference,

d = -33 - (-37) = -33 + 37 = 4

. Number of terms,

n = 12

.

Solution:

S_{12} = \frac{12}{2}[2(-37) + (12-1)4]

S_{12} = 6[-74 + (11)(4)]

S_{12} = 6[-74 + 44]

S_{12} = 6[-30]

S_{12} = -180

(iii) $0.6, 1.7, 2.8, \ldots$ , to 100 terms. Given: First term,

a = 0.6

. Common difference,

d = 1.7 - 0.6 = 1.1

. Number of terms,

n = 100

.

Solution:

S_{100} = \frac{100}{2}[2(0.6) + (100-1)1.1]

S_{100} = 50[1.2 + (99)(1.1)]

S_{100} = 50[1.2 + 108.9]

S_{100} = 50[110.1]

S_{100} = 5505

(iv) $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots$ , to 11 terms. Given: First term,

a = \frac{1}{15}

. Common difference,

d = \frac{1}{12} - \frac{1}{15} = \frac{5-4}{60} = \frac{1}{60}

. Number of terms,

n = 11

.

Solution:

S_{11} = \frac{11}{2}[2(\frac{1}{15}) + (11-1)\frac{1}{60}]

S_{11} = \frac{11}{2}[\frac{2}{15} + (10)\frac{1}{60}]

S_{11} = \frac{11}{2}[\frac{2}{15} + \frac{10}{60}]

S_{11} = \frac{11}{2}[\frac{2}{15} + \frac{1}{6}]

To add the fractions, find a common denominator (30):

S_{11} = \frac{11}{2}[\frac{2 \times 2}{15 \times 2} + \frac{1 \times 5}{6 imes 5}]

S_{11} = \frac{11}{2}[\frac{4}{30} + \frac{5}{30}]

S_{11} = \frac{11}{2}[\frac{9}{30}]

S_{11} = \frac{11}{2} \times \frac{3}{10}

S_{11} = \frac{33}{20}

Final Answer:

(i)

The sum is 245.

(ii)

The sum is -180.

(iii)

The sum is 5505.

(iv)

The sum is

\frac{33}{20}

.

Q2EXERCISE 5.3

Find the sums given below :

(i)

$7+10 \frac{1}{2}+14+\ldots+84$

(ii)

$34+32+30+\ldots+10$

(iii)

$-5+(-8)+(-11)+\ldots+(-230)$

Solution

(i) $7+10 \frac{1}{2}+14+\ldots+84$

Given: The arithmetic progression is

7, 10.5, 14, \ldots, 84

. First term,

a = 7

. Last term,

a_n = 84

. Common difference,

d = 10.5 - 7 = 3.5 = \frac{7}{2}

.

To Find: The sum of the series,

S_n

.

Solution: First, we need to find the number of terms,

n

. Using the formula for the nth term,

a_n = a + (n-1)d

:

84 = 7 + (n-1)\frac{7}{2}

84 - 7 = (n-1)\frac{7}{2}

77 = (n-1)\frac{7}{2}

77 \times \frac{2}{7} = n-1

11 \times 2 = n-1

22 = n-1

n = 23

Now, we can find the sum of the 23 terms using the formula

S_n = \frac{n}{2}(a + a_n)

.

S_{23} = \frac{23}{2}(7 + 84)

S_{23} = \frac{23}{2}(91)

S_{23} = \frac{2093}{2} = 1046.5

Final Answer: The sum of the series is

1046.5

.

(ii) $34+32+30+\ldots+10$

Given: The arithmetic progression is

34, 32, 30, \ldots, 10

. First term,

a = 34

. Last term,

a_n = 10

. Common difference,

d = 32 - 34 = -2

.

To Find: The sum of the series,

S_n

.

Solution: First, we find the number of terms,

n

, using

a_n = a + (n-1)d

.

10 = 34 + (n-1)(-2)

10 - 34 = -2(n-1)

-24 = -2(n-1)

12 = n-1

n = 13

Now, we find the sum of the 13 terms using the formula

S_n = \frac{n}{2}(a + a_n)

.

S_{13} = \frac{13}{2}(34 + 10)

S_{13} = \frac{13}{2}(44)

S_{13} = 13 \times 22 = 286

Final Answer: The sum of the series is

286

.

(iii) $-5+(-8)+(-11)+\ldots+(-230)$

Given: The arithmetic progression is

-5, -8, -11, \ldots, -230

. First term,

a = -5

. Last term,

a_n = -230

. Common difference,

d = -8 - (-5) = -3

.

To Find: The sum of the series,

S_n

.

Solution: First, we find the number of terms,

n

, using

a_n = a + (n-1)d

.

-230 = -5 + (n-1)(-3)

-230 + 5 = -3(n-1)

-225 = -3(n-1)

\frac{-225}{-3} = n-1

75 = n-1

n = 76

Now, we find the sum of the 76 terms using the formula

S_n = \frac{n}{2}(a + a_n)

.

S_{76} = \frac{76}{2}(-5 + (-230))

S_{76} = 38(-235)

S_{76} = -8930

Final Answer: The sum of the series is

-8930

.

Q3EXERCISE 5.3

In an $A P$ :

(i)

given $a=5, d=3, a_{n}=50$ , find $n$ and $\mathrm{S}_{n}$ .

(ii)

given $a=7, a_{13}=35$ , find $d$ and $\mathrm{S}_{13}$ .

(iii)

given $a_{12}=37, d=3$ , find $a$ and $\mathrm{S}_{12}$ .

(iv)

given $a_{3}=15, \mathrm{~S}_{10}=125$ , find $d$ and $a_{10}$ .

(v)

given $d=5, S_{9}=75$ , find $a$ and $a_{9}$ .

(vi)

given $a=2, d=8, \mathrm{~S}_{n}=90$ , find $n$ and $a_{n}$ .

(vii)

given $a=8, a_{n}=62, \mathrm{~S}_{n}=210$ , find $n$ and $d$ .

(viii)

given $a_{n}=4, d=2, \mathrm{~S}_{n}=-14$ , find $n$ and $a$ .

(ix)

given $a=3, n=8, \mathrm{~S}=192$ , find $d$ .

(x)

given $l=28, \mathrm{~S}=144$ , and there are total 9 terms. Find $a$ .

Solution

In this problem, we use the formulas for the nth term of an AP,

a_n = a + (n-1)d

, and the sum of the first n terms,

S_n = \frac{n}{2}(2a + (n-1)d) = \frac{n}{2}(a + a_n)

.

(i) given $a=5, d=3, a_{n}=50$ , find $n$ and $\mathrm{S}_{n}$ Given:

a=5, d=3, a_{n}=50

. Solution: Using

a_n = a + (n-1)d

:

50 = 5 + (n-1)3

45 = 3(n-1)

15 = n-1

n = 16

Now, using

S_n = \frac{n}{2}(a + a_n)

:

S_{16} = \frac{16}{2}(5 + 50) = 8(55) = 440

. Final Answer:

n=16

,

S_{16}=440

.

(ii) given $a=7, a_{13}=35$ , find $d$ and $\mathrm{S}_{13}$ Given:

a=7, a_{13}=35

. Solution: Using

a_{13} = a + (13-1)d

:

35 = 7 + 12d

28 = 12d

d = \frac{28}{12} = \frac{7}{3}

Now, using

S_{13} = \frac{13}{2}(a + a_{13})

:

S_{13} = \frac{13}{2}(7 + 35) = \frac{13}{2}(42) = 13 \times 21 = 273

. Final Answer:

d=\frac{7}{3}

,

S_{13}=273

.

(iii) given $a_{12}=37, d=3$ , find $a$ and $\mathrm{S}_{12}$ Given:

a_{12}=37, d=3

. Solution: Using

a_{12} = a + (12-1)d

:

37 = a + 11(3)

37 = a + 33

a = 4

Now, using

S_{12} = \frac{12}{2}(a + a_{12})

:

S_{12} = 6(4 + 37) = 6(41) = 246

. Final Answer:

a=4

,

S_{12}=246

.

(iv) given $a_{3}=15, \mathrm{S}_{10}=125$ , find $d$ and $a_{10}$ Given:

a_3=15, S_{10}=125

. Solution:

a_3 = a + 2d = 15 \quad ...(1)

S_{10} = \frac{10}{2}(2a + 9d) = 5(2a+9d) = 125 \Rightarrow 2a+9d=25 \quad ...(2)

From (1),

a = 15 - 2d

. Substitute this into (2):

2(15 - 2d) + 9d = 25

30 - 4d + 9d = 25

5d = -5 d = -1

Substitute

d=-1

into (1):

a + 2(-1) = 15 \Rightarrow a=17

. We need to find

a_{10} = a + 9d = 17 + 9(-1) = 17 - 9 = 8

. Final Answer:

d=-1

,

a_{10}=8

.

(v) given $d=5, S_{9}=75$ , find $a$ and $a_{9}$ Given:

d=5, S_9=75

. Solution:

S_9 = \frac{9}{2}(2a + (9-1)5) = 75

\frac{9}{2}(2a + 40) = 75

9(a + 20) = 75

3(a+20) = 25

3a + 60 = 25

3a = -35 \Rightarrow a = -\frac{35}{3}

Now,

a_9 = a + 8d = -\frac{35}{3} + 8(5) = -\frac{35}{3} + 40 = \frac{-35+120}{3} = \frac{85}{3}

. Final Answer:

a=-\frac{35}{3}

,

a_9=\frac{85}{3}

.

(vi) given $a=2, d=8, \mathrm{S}_{n}=90$ , find $n$ and $a_{n}$ Given:

a=2, d=8, S_n=90

. Solution:

S_n = \frac{n}{2}(2a + (n-1)d) = 90

\frac{n}{2}(2(2) + (n-1)8) = 90

\frac{n}{2}(4 + 8n - 8) = 90

\frac{n}{2}(8n - 4) = 90

n(4n - 2) = 90

4n^2 - 2n - 90 = 0

2n^2 - n - 45 = 0

(2n+9)(n-5) = 0

Since

n

must be a positive integer,

n=5

. (The other solution

n=-9/2

is not valid). Now,

a_n = a_5 = a + (5-1)d = 2 + 4(8) = 2 + 32 = 34

. Final Answer:

n=5

,

a_5=34

.

(vii) given $a=8, a_{n}=62, \mathrm{S}_{n}=210$ , find $n$ and $d$ Given:

a=8, a_n=62, S_n=210

. Solution: Using

S_n = \frac{n}{2}(a + a_n)

:

210 = \frac{n}{2}(8 + 62) = \frac{n}{2}(70) = 35n

n = \frac{210}{35} = 6

Now using

a_n = a + (n-1)d

with

n=6

:

a_6 = 62

8 + (6-1)d = 62

5d = 54 \Rightarrow d = \frac{54}{5}

. Final Answer:

n=6

,

d=\frac{54}{5}

.

(viii) given $a_{n}=4, d=2, \mathrm{S}_{n}=-14$ , find $n$ and $a$ Given:

a_n=4, d=2, S_n=-14

. Solution:

a_n = a + (n-1)d \Rightarrow 4 = a + (n-1)2 \Rightarrow a = 4 - 2(n-1) = 6 - 2n \quad ...(1)

S_n = \frac{n}{2}(a + a_n) \Rightarrow -14 = \frac{n}{2}(a + 4) \quad ...(2)

Substitute (1) into (2):

-14 = \frac{n}{2}((6-2n) + 4) = \frac{n}{2}(10 - 2n) = n(5-n)

-14 = 5n - n^2

n^2 - 5n - 14 = 0

(n-7)(n+2)=0

Since

n

must be a positive integer,

n=7

. (The other solution

n=-2

is not valid). Substitute

n=7

into (1):

a = 6 - 2(7) = 6 - 14 = -8

. Final Answer:

n=7

,

a=-8

.

(ix) given $a=3, n=8, \mathrm{S}=192$ , find $d$ Given:

a=3, n=8, S_8=192

. Solution:

S_8 = \frac{8}{2}(2a + (8-1)d) = 192

4(2(3) + 7d) = 192

6 + 7d = \frac{192}{4} = 48

7d = 42 \Rightarrow d=6

. Final Answer:

d=6

.

(x) given $l=28, \mathrm{S}=144$ , and there are total 9 terms. Find $a$ Given: Last term

l=a_9=28

, Sum

S_9=144

, number of terms

n=9

. Solution: Using

S_n = \frac{n}{2}(a + l)

:

S_9 = \frac{9}{2}(a + 28) = 144

a + 28 = 144 \times \frac{2}{9}

a + 28 = 16 \times 2 = 32

a = 32 - 28 = 4

. Final Answer:

a=4

.

Q4EXERCISE 5.3

How many terms of the AP : $9,17,25, \ldots$ must be taken to give a sum of 636 ?

Solution

Given: The arithmetic progression (AP) is

9, 17, 25, \ldots

The sum of

n

terms,

S_n = 636

. First term,

a = 9

. Common difference,

d = 17 - 9 = 8

.

To Find: The number of terms,

n

, that must be taken to give a sum of 636.

Formula: The sum of the first

n

terms of an AP is given by

S_n = \frac{n}{2}[2a + (n-1)d]

.

Solution: Substitute the given values into the formula:

636 = \frac{n}{2}[2(9) + (n-1)8]

636 = \frac{n}{2}[18 + 8n - 8]

636 = \frac{n}{2}[10 + 8n]

636 = n(5 + 4n)

636 = 5n + 4n^2

Rearranging the terms, we get a quadratic equation:

4n^2 + 5n - 636 = 0

We can solve this quadratic equation for

n

using the quadratic formula

n = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

. Here,

a=4, b=5, c=-636

.

n = \frac{-5 \pm \sqrt{5^2 - 4(4)(-636)}}{2(4)}

n = \frac{-5 \pm \sqrt{25 + 16(636)}}{8}

n = \frac{-5 \pm \sqrt{25 + 10176}}{8}

n = \frac{-5 \pm \sqrt{10201}}{8}

The square root of 10201 is 101.

n = \frac{-5 \pm 101}{8}

We have two possible values for

n

:

n = \frac{-5 + 101}{8} = \frac{96}{8} = 12

n = \frac{-5 - 101}{8} = \frac{-106}{8} = -\frac{53}{4}

Since the number of terms (

n

) cannot be negative or a fraction, we discard

n = -\frac{53}{4}

. Therefore,

n = 12

.

Final Answer: 12 terms of the AP must be taken to give a sum of 636.

Q5EXERCISE 5.3

The first term of an AP is 5 , the last term is 45 and the sum is 400 . Find the number of terms and the common difference.

Solution

Given: First term of an AP,

a = 5

. Last term,

l = a_n = 45

. Sum of the terms,

S_n = 400

.

To Find: The number of terms,

n

, and the common difference,

d

.

Formulas:

Sum of $n$ terms: $S_n = \frac{n}{2}(a+l)$
Nth term: $a_n = a + (n-1)d$

Solution: First, we find the number of terms,

n

, using the sum formula.

S_n = \frac{n}{2}(a+l)

400 = \frac{n}{2}(5 + 45)

400 = \frac{n}{2}(50)

400 = 25n

n = \frac{400}{25}

n = 16

So, there are 16 terms in the AP.

Now, we find the common difference,

d

. We know that the 16th term is the last term, which is 45. So,

a_{16} = 45

. Using the formula for the nth term:

a_{16} = a + (16-1)d

45 = 5 + 15d

45 - 5 = 15d

40 = 15d

d = \frac{40}{15}

Simplifying the fraction:

d = \frac{8}{3}

Final Answer: The number of terms is 16 and the common difference is

\frac{8}{3}

.

Q6EXERCISE 5.3

The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9 , how many terms are there and what is their sum?

Solution

Given: The first term of an AP,

a = 17

. The last term of the AP,

l = a_n = 350

. The common difference,

d = 9

.

To Find: The number of terms,

n

, and the sum of the terms,

S_n

.

Formulas:

Nth term: $a_n = a + (n-1)d$
Sum of $n$ terms: $S_n = \frac{n}{2}(a+l)$

Solution: First, we find the number of terms,

n

, using the formula for the nth term.

a_n = a + (n-1)d

350 = 17 + (n-1)9

350 - 17 = 9(n-1)

333 = 9(n-1)

n-1 = \frac{333}{9}

n-1 = 37

n = 37 + 1 = 38

So, there are 38 terms in the AP.

Now, we find the sum of these 38 terms using the sum formula.

S_n = \frac{n}{2}(a+l)

S_{38} = \frac{38}{2}(17 + 350)

S_{38} = 19(367)

Calculation:

19 \times 367 = 19 \times (300 + 60 + 7)

= 19 \times 300 + 19 \times 60 + 19 \times 7

= 5700 + 1140 + 133

= 6840 + 133

= 6973

So,

S_{38} = 6973

.

Final Answer: There are 38 terms in the AP, and their sum is 6973.

Q7EXERCISE 5.3

Find the sum of first 22 terms of an AP in which $d=7$ and 22nd term is 149 .

Solution

Given: An Arithmetic Progression (AP) with: Common difference,

d = 7

22nd term,

a_{22} = 149

Number of terms,

n = 22

To Find: The sum of the first 22 terms,

S_{22}

.

Formulae:

The $n$ -th term of an AP: $a_n = a + (n-1)d$
The sum of the first $n$ terms of an AP: $S_n = \frac{n}{2}(a + a_n)$

Solution: First, we need to find the first term,

a

. We know that the 22nd term is 149. Using the formula for the

n

-th term:

a_{22} = a + (22-1)d

149 = a + 21 \times 7

149 = a + 147

a = 149 - 147

a = 2

Now that we have the first term (

a=2

) and the 22nd term (

a_{22}=149

), we can find the sum of the first 22 terms using the sum formula.

S_{n} = \frac{n}{2}(a + a_n)

S_{22} = \frac{22}{2}(a + a_{22})

S_{22} = 11(2 + 149)

S_{22} = 11(151)

S_{22} = 1661

Final Answer: The sum of the first 22 terms of the AP is 1661.

Q8EXERCISE 5.3

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution

Given: An Arithmetic Progression (AP) with: Second term,

a_2 = 14

Third term,

a_3 = 18

To Find: The sum of the first 51 terms,

S_{51}

.

Formulae:

Common difference: $d = a_{n} - a_{n-1}$
The sum of the first $n$ terms of an AP: $S_n = \frac{n}{2}[2a + (n-1)d]$

Solution: First, we find the common difference,

d

.

d = a_3 - a_2

d = 18 - 14

d = 4

Next, we find the first term,

a

. We know that

a_2 = a + d

.

14 = a + 4

a = 14 - 4

a = 10

Now we have the first term (

a=10

), the common difference (

d=4

), and the number of terms (

n=51

). We can calculate the sum of the first 51 terms.

S_n = \frac{n}{2}[2a + (n-1)d]

S_{51} = \frac{51}{2}[2(10) + (51-1)(4)]

S_{51} = \frac{51}{2}[20 + (50)(4)]

S_{51} = \frac{51}{2}[20 + 200]

S_{51} = \frac{51}{2}[220]

S_{51} = 51 \times 110

S_{51} = 5610

Final Answer: The sum of the first 51 terms of the AP is 5610.

Q9EXERCISE 5.3

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289 , find the sum of first $n$ terms.

Solution

Given: For an Arithmetic Progression (AP): The sum of the first 7 terms,

S_7 = 49

. The sum of the first 17 terms,

S_{17} = 289

.

To Find: The sum of the first

n

terms,

S_n

.

Formula: The sum of the first

n

terms of an AP is given by:

S_n = \frac{n}{2}[2a + (n-1)d]

where

a

is the first term and

d

is the common difference.

Solution: Using the given information, we can form two linear equations in terms of

a

and

d

.

For the sum of the first 7 terms:

S_7 = \frac{7}{2}[2a + (7-1)d] = 49

\frac{7}{2}[2a + 6d] = 49

7(a + 3d) = 49

a + 3d = 7 \quad \dots(1)

For the sum of the first 17 terms:

S_{17} = \frac{17}{2}[2a + (17-1)d] = 289

\frac{17}{2}[2a + 16d] = 289

17(a + 8d) = 289

a + 8d = 17 \quad \dots(2)

Now, we solve the two linear equations. Subtracting equation (1) from equation (2):

(a + 8d) - (a + 3d) = 17 - 7

5d = 10

d = 2

Substitute the value of

d=2

into equation (1):

a + 3(2) = 7

a + 6 = 7

a = 1

Now we have the first term

a=1

and the common difference

d=2

. We can find the sum of the first

n

terms,

S_n

.

S_n = \frac{n}{2}[2a + (n-1)d]

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

S_n = \frac{n}{2}[2 + 2n - 2]

S_n = \frac{n}{2}[2n]

S_n = n^2

Final Answer: The sum of the first

n

terms of the AP is

n^2

.

Q10EXERCISE 5.3

Show that $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ form an AP where $a_{n}$ is defined as below:

(i)

$a_{n}=3+4 n$

(ii)

$a_{n}=9-5 n$ Also find the sum of the first 15 terms in each case.

Solution

This question has two parts. We need to show that the sequence defined by

a_n

forms an AP and then find the sum of the first 15 terms.

A sequence is an AP if the difference between consecutive terms,

a_{k+1} - a_k

, is constant for all

k \geq 1

.

(i) $a_n = 3+4n$

To Show: The sequence forms an AP. Let's find the

(n+1)

-th term:

a_{n+1} = 3 + 4(n+1) = 3 + 4n + 4 = 7 + 4n

Now, find the difference between consecutive terms:

a_{n+1} - a_n = (7 + 4n) - (3 + 4n)

a_{n+1} - a_n = 4

Since the difference is a constant (4), the sequence

a_1, a_2, \ldots, a_n, \ldots

forms an AP with a common difference

d=4

.

To Find: The sum of the first 15 terms,

S_{15}

. First term,

a_1 = 3 + 4(1) = 7

. Common difference,

d = 4

. Number of terms,

n = 15

.

Using the formula for the sum of the first

n

terms:

S_n = \frac{n}{2}[2a + (n-1)d]

S_{15} = \frac{15}{2}[2(7) + (15-1)(4)]

S_{15} = \frac{15}{2}[14 + 14(4)]

S_{15} = \frac{15}{2}[14 + 56]

S_{15} = \frac{15}{2}[70]

S_{15} = 15 \times 35 = 525

Final Answer for (i): The sequence is an AP, and the sum of the first 15 terms is 525.

(ii) $a_n = 9-5n$

To Show: The sequence forms an AP. Let's find the

(n+1)

-th term:

a_{n+1} = 9 - 5(n+1) = 9 - 5n - 5 = 4 - 5n

Now, find the difference between consecutive terms:

a_{n+1} - a_n = (4 - 5n) - (9 - 5n)

a_{n+1} - a_n = -5

Since the difference is a constant (-5), the sequence

a_1, a_2, \ldots, a_n, \ldots

forms an AP with a common difference

d=-5

.

To Find: The sum of the first 15 terms,

S_{15}

. First term,

a_1 = 9 - 5(1) = 4

. Common difference,

d = -5

. Number of terms,

n = 15

.

Using the formula for the sum of the first

n

terms:

S_n = \frac{n}{2}[2a + (n-1)d]

S_{15} = \frac{15}{2}[2(4) + (15-1)(-5)]

S_{15} = \frac{15}{2}[8 + 14(-5)]

S_{15} = \frac{15}{2}[8 - 70]

S_{15} = \frac{15}{2}[-62]

S_{15} = 15 \times (-31) = -465

Final Answer for (ii): The sequence is an AP, and the sum of the first 15 terms is -465.

Q11EXERCISE 5.3

If the sum of the first $n$ terms of an AP is $4 n-n^{2}$ , what is the first term (that is $\mathrm{S}_{1}$ )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the $n$ th terms.

Solution

Given: The sum of the first

n

terms of an AP is

S_n = 4n - n^2

.

To Find:

The first term ( $a_1$ or $S_1$ ).
The sum of the first two terms ( $S_2$ ).
The second term ( $a_2$ ).
The 3rd term ( $a_3$ ), the 10th term ( $a_{10}$ ), and the $n$ th term ( $a_n$ ).

Formulae:

The first term $a_1$ is equal to the sum of the first term, $S_1$ .
The $n$ th term can be found using the relation $a_n = S_n - S_{n-1}$ .

Solution:

First term ( $a_1$ ) The first term is the sum of the first term itself. $a_1 = S_1 = 4(1) - (1)^2 = 4 - 1 = 3$ So, the first term is 3.
Sum of first two terms ( $S_2$ ) Substitute $n=2$ into the given formula for $S_n$ . $S_2 = 4(2) - (2)^2 = 8 - 4 = 4$ So, the sum of the first two terms is 4.
Second term ( $a_2$ ) The second term is the difference between the sum of the first two terms and the sum of the first term. $a_2 = S_2 - S_1 = 4 - 3 = 1$ So, the second term is 1.
3rd, 10th, and $n$ th terms To find these terms, we first find a general formula for the $n$ th term, $a_n$ . $a_n = S_n - S_{n-1}$ We have $S_n = 4n - n^2$ . Now, find $S_{n-1}$ by replacing $n$ with $(n-1)$ : $S_{n-1} = 4(n-1) - (n-1)^2$ $S_{n-1} = (4n - 4) - (n^2 - 2n + 1)$ $S_{n-1} = 4n - 4 - n^2 + 2n - 1$ $S_{n-1} = -n^2 + 6n - 5$

Now, calculate

a_n

:

a_n = (4n - n^2) - (-n^2 + 6n - 5)

a_n = 4n - n^2 + n^2 - 6n + 5

a_n = 5 - 2n

This is the formula for the

n

th term.

Now we can find the 3rd and 10th terms:

Third term ( $a_3$ )

a_3 = 5 - 2(3) = 5 - 6 = -1

Tenth term ( $a_{10}$ )

a_{10} = 5 - 2(10) = 5 - 20 = -15

Final Answer:

The first term ( $S_1$ ) is 3.
The sum of the first two terms is 4.
The second term is 1.
The 3rd term is -1.
The 10th term is -15.
The $n$ th term is $5 - 2n$ .

Q12EXERCISE 5.3

Find the sum of the first 40 positive integers divisible by 6 .

Solution

To Find: The sum of the first 40 positive integers divisible by 6.

Solution: The first 40 positive integers divisible by 6 are 6, 12, 18, ..., up to 40 terms. This sequence forms an Arithmetic Progression (AP).

First term,

a = 6

. Common difference,

d = 12 - 6 = 6

. Number of terms,

n = 40

.

We need to find the sum of these 40 terms,

S_{40}

.

Formula: The sum of the first

n

terms of an AP is given by:

S_n = \frac{n}{2} [2a + (n-1)d]

Calculation: Substituting the values

a=6

,

d=6

, and

n=40

into the formula:

S_{40} = \frac{40}{2} [2(6) + (40-1)6]

S_{40} = 20 [12 + (39)6]

S_{40} = 20 [12 + 234]

S_{40} = 20 [246]

S_{40} = 4920

Final Answer: The sum of the first 40 positive integers divisible by 6 is 4920.

Q13EXERCISE 5.3

Find the sum of the first 15 multiples of 8 .

Solution

To Find: The sum of the first 15 multiples of 8.

Solution: The first 15 multiples of 8 are 8, 16, 24, ..., up to 15 terms. This sequence forms an Arithmetic Progression (AP).

First term,

a = 8

. Common difference,

d = 16 - 8 = 8

. Number of terms,

n = 15

.

We need to find the sum of these 15 terms,

S_{15}

.

Formula: The sum of the first

n

terms of an AP is given by:

S_n = \frac{n}{2} [2a + (n-1)d]

Calculation: Substituting the values

a=8

,

d=8

, and

n=15

into the formula:

S_{15} = \frac{15}{2} [2(8) + (15-1)8]

S_{15} = \frac{15}{2} [16 + (14)8]

S_{15} = \frac{15}{2} [16 + 112]

S_{15} = \frac{15}{2} [128]

S_{15} = 15 \times 64

S_{15} = 960

Final Answer: The sum of the first 15 multiples of 8 is 960.

Q14EXERCISE 5.3

Find the sum of the odd numbers between 0 and 50 .

Solution

To Find: The sum of the odd numbers between 0 and 50.

Solution: The odd numbers between 0 and 50 are 1, 3, 5, ..., 49. This sequence forms an Arithmetic Progression (AP).

First term,

a = 1

. Common difference,

d = 3 - 1 = 2

. Last term,

a_n = l = 49

.

First, we need to find the number of terms,

n

.

Formula for the n-th term:

a_n = a + (n-1)d

Calculation of n:

49 = 1 + (n-1)2

49 - 1 = (n-1)2

48 = 2(n-1)

n-1 = \frac{48}{2}

n-1 = 24

n = 25

Now, we find the sum of these 25 terms.

Formula for the sum of n terms: We can use the formula

S_n = \frac{n}{2}(a+l)

since we know the first and last terms.

Calculation of Sum:

S_{25} = \frac{25}{2}(1 + 49)

S_{25} = \frac{25}{2}(50)

S_{25} = 25 \times 25

S_{25} = 625

Final Answer: The sum of the odd numbers between 0 and 50 is 625.

Q15EXERCISE 5.3

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

Solution

Given: A contract on a construction job has a penalty for delay. Penalty for the first day = ₹ 200. Penalty for the second day = ₹ 250. Penalty for the third day = ₹ 300. The penalty for each succeeding day is ₹ 50 more than the preceding day. The work is delayed by 30 days.

To Find: The total penalty the contractor has to pay for a delay of 30 days.

Solution: The penalties for each day form an Arithmetic Progression (AP). The sequence of penalties is ₹ 200, ₹ 250, ₹ 300, ...

First term,

a = 200

. Common difference,

d = 250 - 200 = 50

. Number of days of delay (number of terms),

n = 30

.

We need to find the sum of this AP for 30 terms,

S_{30}

.

Formula: The sum of the first

n

terms of an AP is:

S_n = \frac{n}{2} [2a + (n-1)d]

Calculation: Substituting the values

a=200

,

d=50

, and

n=30

:

S_{30} = \frac{30}{2} [2(200) + (30-1)50]

S_{30} = 15 [400 + (29)50]

S_{30} = 15 [400 + 1450]

S_{30} = 15 [1850]

S_{30} = 27750

Final Answer: The contractor has to pay ₹ 27,750 as penalty for delaying the work by 30 days.

Q16EXERCISE 5.3

A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.

Solution

Given: Total sum for prizes,

S_n = \text{₹ } 700

. Number of cash prizes,

n = 7

. Each prize is ₹ 20 less than its preceding prize.

To Find: The value of each of the seven prizes.

Solution: The values of the prizes form an Arithmetic Progression (AP). Let the value of the first prize be

a

. Since each prize is ₹ 20 less than the preceding one, the common difference is

d = -20

.

We are given the sum of the 7 prizes,

S_7 = 700

.

Formula: The sum of the first

n

terms of an AP is:

S_n = \frac{n}{2} [2a + (n-1)d]

Calculation of the first prize (a): Substitute the known values into the formula:

700 = \frac{7}{2} [2a + (7-1)(-20)]

700 = \frac{7}{2} [2a + (6)(-20)]

700 = \frac{7}{2} [2a - 120]

To solve for

a

, we can multiply both sides by

\frac{2}{7}

:

700 \times \frac{2}{7} = 2a - 120

100 \times 2 = 2a - 120

200 = 2a - 120

200 + 120 = 2a

320 = 2a

a = \frac{320}{2}

a = 160

So, the first prize is ₹ 160.

Calculating the values of all prizes: 1st prize:

a_1 = a = \text{₹ } 160

2nd prize:

a_2 = a + d = 160 - 20 = \text{₹ } 140

3rd prize:

a_3 = a + 2d = 160 - 40 = \text{₹ } 120

4th prize:

a_4 = a + 3d = 160 - 60 = \text{₹ } 100

5th prize:

a_5 = a + 4d = 160 - 80 = \text{₹ } 80

6th prize:

a_6 = a + 5d = 160 - 100 = \text{₹ } 60

7th prize:

a_7 = a + 6d = 160 - 120 = \text{₹ } 40

Final Answer: The values of the seven prizes are ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, and ₹ 40.

Q17EXERCISE 5.3

In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Solution

Given: Number of classes in the school = 12 (from Class I to Class XII). Number of sections in each class = 3. The number of trees planted by a section of any class is equal to the class number.

To Find: The total number of trees planted by the students.

Solution: Number of trees planted by a section of Class I = 1. Since there are 3 sections, total trees planted by Class I =

1 \times 3 = 3

.

Number of trees planted by a section of Class II = 2. Total trees planted by Class II =

2 \times 3 = 6

.

Number of trees planted by a section of Class III = 3. Total trees planted by Class III =

3 \times 3 = 9

.

This pattern continues up to Class XII. Number of trees planted by a section of Class XII = 12. Total trees planted by Class XII =

12 \times 3 = 36

.

The number of trees planted by each class forms a sequence:

3, 6, 9, \ldots, 36

. This sequence is an Arithmetic Progression (AP).

First term,

a = 3

. Common difference,

d = 6 - 3 = 3

. Number of terms,

n = 12

(since there are 12 classes). Last term,

l = 36

.

The total number of trees planted is the sum of this AP.

Formula: The sum of the first

n

terms of an AP is given by:

S_n = \frac{n}{2}(a + l)

Calculation: Substituting the values into the formula:

S_{12} = \frac{12}{2}(3 + 36)

S_{12} = 6(39)

S_{12} = 234

Final Answer: The total number of trees planted by the students is 234.

Q18EXERCISE 5.3

A spiral is made up of successive semicircles, with centres alternately at $A$ and $B$ , starting with centre at A, of radii $0.5 \mathrm{~cm}, 1.0 \mathrm{~cm}, 1.5 \mathrm{~cm}, 2.0 \mathrm{~cm}, \ldots$ as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? $\left(\text { Take } \pi=\frac{22}{7}\right)$ \begin{figure} \includegraphics[width=\textwidth]{https://cdn.mathpix.com/cropped/ef221bdc-fc6c-45ad-9119-3f04c686f63f-22.jpg?height=448&width=828&top_left_y=326&top_left_x=393} \captionsetup{labelformat=empty} \caption{Fig. 5.4} \end{figure} [Hint : Length of successive semicircles is $l_{1}, l_{2}, l_{3}, l_{4}, \ldots$ with centres at $\mathrm{A}, \mathrm{B}, \mathrm{A}, \mathrm{B}, \ldots$ , respectively.]

Solution

Given: A spiral is made of 13 consecutive semicircles with alternating centers A and B. The radii of the semicircles are

0.5 \text{ cm}, 1.0 \text{ cm}, 1.5 \text{ cm}, 2.0 \text{ cm}, \ldots

. We need to use

\pi = \frac{22}{7}

.

To Find: The total length of the spiral.

Formula: The length of a semicircle with radius

r

is given by the formula

L = \pi r

.

Solution: Let the radii of the consecutive semicircles be

r_1, r_2, r_3, \ldots, r_{13}

.

r_1 = 0.5 \text{ cm}

r_2 = 1.0 \text{ cm}

r_3 = 1.5 \text{ cm}

... and so on for 13 semicircles.

Let the lengths of the semicircles be

l_1, l_2, l_3, \ldots, l_{13}

. Length of the 1st semicircle,

l_1 = \pi r_1 = \pi(0.5) = 0.5\pi \text{ cm}

. Length of the 2nd semicircle,

l_2 = \pi r_2 = \pi(1.0) = 1.0\pi \text{ cm}

. Length of the 3rd semicircle,

l_3 = \pi r_3 = \pi(1.5) = 1.5\pi \text{ cm}

.

The sequence of lengths is

0.5\pi, 1.0\pi, 1.5\pi, \ldots

. This is an Arithmetic Progression (AP).

First term,

a = 0.5\pi

. Common difference,

d = l_2 - l_1 = 1.0\pi - 0.5\pi = 0.5\pi

. Number of terms,

n = 13

.

The total length of the spiral is the sum of the lengths of the 13 semicircles, which is the sum of this AP.

Formula for Sum of AP:

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

Calculation:

S_{13} = \frac{13}{2}[2(0.5\pi) + (13-1)(0.5\pi)]

S_{13} = \frac{13}{2}[\pi + 12(0.5\pi)]

S_{13} = \frac{13}{2}[\pi + 6\pi]

S_{13} = \frac{13}{2}(7\pi)

Now, substitute the value of

\pi = \frac{22}{7}

:

S_{13} = \frac{13}{2} \times 7 \times \frac{22}{7}

S_{13} = \frac{13}{2} \times 22

S_{13} = 13 \times 11

S_{13} = 143

Final Answer: The total length of the spiral made up of thirteen consecutive semicircles is 143 cm.

Q19EXERCISE 5.3

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Fig. 5.5). In how many rows are the 200 logs placed and how many logs are in the top row? \begin{figure} \includegraphics[width=\textwidth]{https://cdn.mathpix.com/cropped/ef221bdc-fc6c-45ad-9119-3f04c686f63f-22.jpg?height=209&width=918&top_left_y=1107&top_left_x=348} \captionsetup{labelformat=empty} \caption{Fig. 5.5} \end{figure}

Solution

Given: Total number of logs = 200. The logs are stacked in rows. The bottom row has 20 logs, the next row has 19, the next has 18, and so on.

To Find:

The number of rows in which the 200 logs are placed ( $n$ ).
The number of logs in the top row ( $a_n$ ).

Solution: The number of logs in the rows, starting from the bottom row, forms an Arithmetic Progression (AP). The sequence is

20, 19, 18, \ldots

.

First term,

a = 20

. Common difference,

d = 19 - 20 = -1

. The total number of logs is the sum of this AP, so

S_n = 200

.

Formula for Sum of AP:

S_n = \frac{n}{2}[2a + (n-1)d]

Calculation for number of rows (n):

200 = \frac{n}{2}[2(20) + (n-1)(-1)]

400 = n[40 - (n-1)]

400 = n[40 - n + 1]

400 = n(41 - n)

400 = 41n - n^2

Rearranging into a quadratic equation:

n^2 - 41n + 400 = 0

We need to find two numbers that multiply to 400 and add up to 41. These numbers are 16 and 25. Factorizing the quadratic equation:

n^2 - 16n - 25n + 400 = 0

n(n - 16) - 25(n - 16) = 0

(n - 16)(n - 25) = 0

This gives two possible values for

n

:

n = 16

or

n = 25

.

We need to check which value is valid by finding the number of logs in the top row for each case.

Formula for nth term:

a_n = a + (n-1)d

Case 1: $n = 16$ The number of logs in the top (16th) row is:

a_{16} = 20 + (16-1)(-1) = 20 + 15(-1) = 20 - 15 = 5

This is a valid scenario, as the number of logs is a positive integer.

Case 2: $n = 25$ The number of logs in the top (25th) row is:

a_{25} = 20 + (25-1)(-1) = 20 + 24(-1) = 20 - 24 = -4

The number of logs cannot be negative. Therefore,

n=25

is not a valid solution.

So, the only valid solution is

n = 16

.

Final Answer: The 200 logs are placed in 16 rows, and the number of logs in the top row is 5.

Q20EXERCISE 5.3

In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig. 5.6). \begin{figure} \includegraphics[width=\textwidth]{https://cdn.mathpix.com/cropped/ef221bdc-fc6c-45ad-9119-3f04c686f63f-22.jpg?height=212&width=1300&top_left_y=1536&top_left_x=166} \captionsetup{labelformat=empty} \caption{Fig. 5.6} \end{figure} A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run? [Hint : To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is $2 \times 5+2 \times(5+3)$ ]

Solution

Given: A bucket is at the starting point. The first potato is 5 m away. Other potatoes are placed 3 m apart in a straight line. There are 10 potatoes in total. A competitor starts from the bucket, picks up a potato, runs back to drop it in the bucket, and repeats for all potatoes.

To Find: The total distance the competitor has to run.

Solution: Let's calculate the distance the competitor runs to pick up each potato and return to the bucket.

Distance of the 1st potato from the bucket = 5 m. Distance run to pick up the 1st potato and return =

2 \times 5 = 10

m.

Distance of the 2nd potato from the bucket =

5 + 3 = 8

m. Distance run to pick up the 2nd potato and return =

2 \times 8 = 16

m.

Distance of the 3rd potato from the bucket =

5 + 3 + 3 = 11

m. Distance run to pick up the 3rd potato and return =

2 \times 11 = 22

m.

The distances run for each of the 10 potatoes form a sequence:

10, 16, 22, \ldots

. This sequence is an Arithmetic Progression (AP).

First term,

a = 10

. Common difference,

d = 16 - 10 = 6

. Number of terms (potatoes),

n = 10

.

The total distance run is the sum of this AP.

Formula for Sum of AP:

S_n = \frac{n}{2}[2a + (n-1)d]

Calculation:

S_{10} = \frac{10}{2}[2(10) + (10-1)(6)]

S_{10} = 5[20 + 9(6)]

S_{10} = 5[20 + 54]

S_{10} = 5[74]

S_{10} = 370

Alternative Method: The distances of the potatoes from the bucket are

5, 8, 11, \ldots

for 10 potatoes. This is an AP with first term

a' = 5

and common difference

d' = 3

. The sum of these distances is:

S'_{10} = \frac{10}{2}[2(5) + (10-1)3] = 5[10 + 27] = 5(37) = 185 \text{ m}

Since the competitor has to run to the potato and back to the bucket for each one, the total distance is twice this sum. Total distance =

2 \times 185 = 370 \text{ m}

.

Final Answer: The total distance the competitor has to run is 370 m.

Q1EXERCISE 5.4 (Optional)

Which term of the AP : $121, 117, 113, \ldots$ , is its first negative term? [Hint : Find $n$ for $a_{n}<0$ ]

Solution

Given: The Arithmetic Progression (AP) is

121, 117, 113, \ldots

.

To Find: Which term of the AP is its first negative term.

Solution: From the given AP, we can determine the first term and the common difference. First term,

a = 121

. Common difference,

d = 117 - 121 = -4

.

Let the

n

-th term of the AP be

a_n

. We want to find the smallest integer

n

for which

a_n

is negative. So, we need to find

n

such that

a_n < 0

.

Formula for the nth term of an AP:

a_n = a + (n-1)d

Calculation: Substituting the values of

a

and

d

into the inequality:

121 + (n-1)(-4) < 0

121 - 4(n-1) < 0

121 - 4n + 4 < 0

125 - 4n < 0

125 < 4n

\frac{125}{4} < n

31.25 < n

Since

n

must be an integer (as it represents the term number), the smallest integer value of

n

that satisfies the inequality

n > 31.25

is

n = 32

.

Therefore, the 32nd term will be the first negative term of the AP.

Verification: Let's calculate the 31st and 32nd terms. For

n=31

:

a_{31} = 121 + (31-1)(-4) = 121 + 30(-4) = 121 - 120 = 1

For

n=32

:

a_{32} = 121 + (32-1)(-4) = 121 + 31(-4) = 121 - 124 = -3

As we can see, the 31st term is positive (1) and the 32nd term is negative (-3). Thus, the 32nd term is indeed the first negative term.

Final Answer: The 32nd term of the given AP is its first negative term.

Q2EXERCISE 5.4 (Optional)

The sum of the third and the seventh terms of an AP is 6 and their product is 8 . Find the sum of first sixteen terms of the AP.

Solution

Given: The sum of the third and the seventh terms of an AP is 6. The product of the third and the seventh terms of an AP is 8.

To Find: The sum of the first sixteen terms of the AP.

Let: The first term of the AP be

a

and the common difference be

d

.

Formula: The

n

-th term of an AP is given by

a_n = a + (n-1)d

. The sum of the first

n

terms of an AP is given by

S_n = \frac{n}{2}[2a + (n-1)d]

.

Solution: According to the problem: The third term is

a_3 = a + (3-1)d = a + 2d

. The seventh term is

a_7 = a + (7-1)d = a + 6d

.

From the first condition, the sum of the third and seventh terms is 6:

a_3 + a_7 = 6

(a + 2d) + (a + 6d) = 6

2a + 8d = 6

Dividing by 2, we get:

a + 4d = 3 \quad ...(1)

From the second condition, the product of the third and seventh terms is 8:

a_3 \times a_7 = 8

(a + 2d)(a + 6d) = 8 \quad ...(2)

From equation (1), we can express

a

in terms of

d

:

a = 3 - 4d

Substitute this expression for

a

into equation (2):

((3 - 4d) + 2d)((3 - 4d) + 6d) = 8

(3 - 2d)(3 + 2d) = 8

Using the identity

(x-y)(x+y) = x^2 - y^2

:

3^2 - (2d)^2 = 8

9 - 4d^2 = 8

4d^2 = 9 - 8

4d^2 = 1

d^2 = \frac{1}{4}

d = \pm \sqrt{\frac{1}{4}}

d = \pm \frac{1}{2}

We have two possible cases for the common difference

d

.

Case 1: When $d = \frac{1}{2}$ Substitute

d = \frac{1}{2}

into equation (1):

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

So, for this AP,

a = 1

and

d = \frac{1}{2}

. The sum of the first 16 terms is

S_{16}

:

S_{16} = \frac{16}{2}[2a + (16-1)d]

S_{16} = 8[2(1) + 15(\frac{1}{2})]

S_{16} = 8[2 + \frac{15}{2}]

S_{16} = 8[\frac{4+15}{2}]

S_{16} = 8[\frac{19}{2}] = 4 \times 19 = 76

Case 2: When $d = -\frac{1}{2}$ Substitute

d = -\frac{1}{2}

into equation (1):

a = 3 - 4(-\frac{1}{2}) = 3 + 2 = 5

So, for this AP,

a = 5

and

d = -\frac{1}{2}

. The sum of the first 16 terms is

S_{16}

:

S_{16} = \frac{16}{2}[2a + (16-1)d]

S_{16} = 8[2(5) + 15(-\frac{1}{2})]

S_{16} = 8[10 - \frac{15}{2}]

S_{16} = 8[\frac{20-15}{2}]

S_{16} = 8[\frac{5}{2}] = 4 \times 5 = 20

Final Answer: The sum of the first sixteen terms of the AP can be either 76 or 20.

Q3EXERCISE 5.4 (Optional)

A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are $2 \frac{1}{2} \mathrm{~m}$ apart, what is the length of the wood required for the rungs? [Hint : Number of rungs $=\frac{250}{25}+1$ ] \begin{figure} \includegraphics[width=\textwidth]{https://cdn.mathpix.com/cropped/ef221bdc-fc6c-45ad-9119-3f04c686f63f-23.jpg?height=678&width=415&top_left_y=397&top_left_x=964} \captionsetup{labelformat=empty} \caption{Fig. 5.7} \end{figure}

Solution

Given: Distance between consecutive rungs = 25 cm. Length of the bottom rung = 45 cm. Length of the top rung = 25 cm. Total distance between the top and bottom rungs =

2 \frac{1}{2} \text{ m} = 2.5 \text{ m} = 250 \text{ cm}

.

To Find: The total length of the wood required for the rungs.

Solution: First, we need to find the total number of rungs. The rungs are placed 25 cm apart. Number of gaps between the rungs =

\frac{\text{Total distance}}{\text{Distance between rungs}} = \frac{250}{25} = 10

. The total number of rungs is one more than the number of gaps. Number of rungs,

n = 10 + 1 = 11

.

The lengths of the rungs decrease uniformly, so they form an Arithmetic Progression (AP). Let the AP be the sequence of the lengths of the rungs from bottom to top. First term (length of the bottom rung),

a = 45

cm. Last term (length of the top rung),

l = a_{11} = 25

cm. Number of terms (rungs),

n = 11

.

We need to find the total length of wood required, which is the sum of the lengths of all 11 rungs. This is the sum of the AP,

S_{11}

.

Formula: The sum of an AP when the first and last terms are known is

S_n = \frac{n}{2}(a+l)

.

Calculation: Substituting the values

n=11

,

a=45

, and

l=25

into the formula:

S_{11} = \frac{11}{2}(45 + 25)

S_{11} = \frac{11}{2}(70)

S_{11} = 11 \times 35

S_{11} = 385

So, the total length of wood required for the 11 rungs is 385 cm.

Final Answer: The length of the wood required for the rungs is 385 cm.

Q4EXERCISE 5.4 (Optional)

The houses of a row are numbered consecutively from 1 to 49 . Show that there is a value of $x$ such that the sum of the numbers of the houses preceding the house numbered $x$ is equal to the sum of the numbers of the houses following it. Find this value of $x$ . [Hint : $\mathrm{S}_{x-1}=\mathrm{S}_{49}-\mathrm{S}_{x}$ ]

Solution

Given: Houses in a row are numbered consecutively from 1 to 49. There is a house numbered

x

such that the sum of the numbers of the houses preceding it is equal to the sum of the numbers of the houses following it.

To Find: The value of

x

.

Solution: The numbers on the houses form an Arithmetic Progression (AP): 1, 2, 3, ..., 49. For this AP: First term,

a = 1

. Common difference,

d = 1

. Total number of houses,

n = 49

.

The houses preceding the house numbered

x

are numbered from 1 to

x-1

. The sum of these numbers is

S_{x-1}

. The houses following the house numbered

x

are numbered from

x+1

to 49. The sum of these numbers can be found by taking the sum of all house numbers from 1 to 49 (

S_{49}

) and subtracting the sum of house numbers from 1 to

x

(

S_x

).

According to the problem statement:

S_{x-1} = S_{49} - S_x

Formula: The sum of the first

n

terms of an AP is

S_n = \frac{n}{2}[2a + (n-1)d]

. For an AP of natural numbers starting from 1,

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

.

Using this simplified formula: Sum of numbers up to

x-1

:

S_{x-1} = \frac{(x-1)((x-1)+1)}{2} = \frac{(x-1)x}{2}

. Sum of numbers up to 49:

S_{49} = \frac{49(49+1)}{2} = \frac{49 \times 50}{2} = 49 \times 25 = 1225

. Sum of numbers up to

x

:

S_x = \frac{x(x+1)}{2}

.

Now, substitute these expressions into the condition

S_{x-1} = S_{49} - S_x

:

\frac{x(x-1)}{2} = 1225 - \frac{x(x+1)}{2}

To solve for

x

, bring all terms involving

x

to one side:

\frac{x(x-1)}{2} + \frac{x(x+1)}{2} = 1225

\frac{x(x-1) + x(x+1)}{2} = 1225

\frac{x^2 - x + x^2 + x}{2} = 1225

\frac{2x^2}{2} = 1225

x^2 = 1225

x = \sqrt{1225}

To find the square root of 1225: Since the number ends in 25, its square root must end in 5. We can test values like 35.

35^2 = (30+5)^2 = 30^2 + 2(30)(5) + 5^2 = 900 + 300 + 25 = 1225

. So,

x = 35

.

Since

x

represents a house number, it must be a positive integer. The value

x=35

is a positive integer and lies between 1 and 49. Thus, a value for

x

exists and is 35.

Final Answer: The value of

x

is 35.

Q5EXERCISE 5.4 (Optional)

A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of $\frac{1}{4}$ m and a tread of $\frac{1}{2}$ m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint : Volume of concrete required to build the first step = $\frac{1}{4} \times \frac{1}{2} \times 50 \mathrm{~m}^{3}$ ]

Solution

Given: A small terrace has 15 steps. Length of each step = 50 m. Rise of each step =

\frac{1}{4}

m. Tread of each step =

\frac{1}{2}

m.

To Find: The total volume of concrete required to build the terrace.

Solution: Let's calculate the volume of concrete required for each step. Each step can be considered a cuboid.

The volume of a cuboid is given by the formula: Volume = length

\times

breadth

\times

height.

For the first (bottom) step: Length = 50 m Breadth (tread) =

\frac{1}{2}

m Height (rise) =

\frac{1}{4}

m Volume of the first step,

V_1 = 50 \times \frac{1}{2} \times \frac{1}{4} = \frac{50}{8} = \frac{25}{4} \text{ m}^3

.

For the second step: The second step is placed on top of the first. The total height from the ground to the top of the second step is

2 \times \frac{1}{4}

m. The breadth (tread) is

\frac{1}{2}

m and the length is 50 m. So the volume of concrete for the second step is the volume of a cuboid with height

2 \times \frac{1}{4}

m, but we only need the volume of the concrete used for that step itself. Let's think about the volume of each block of concrete.

The first step is a block of concrete with dimensions

50 \times \frac{1}{2} \times \frac{1}{4}

. The second step is another block of concrete with dimensions

50 \times \frac{1}{2} \times \frac{1}{4}

which is placed on the first. However, the problem structure implies a cumulative volume. The volume of the concrete structure for the first step is

V_1

. The volume of the structure for the second step is the volume of a cuboid of height

2 \times \frac{1}{4}

m, breadth

1 \times \frac{1}{2}

m and length 50 m. This is incorrect.

Let's follow the hint and the structure. The total volume is the sum of the volumes of 15 individual cuboids, where the height of each cuboid represents the total height from the ground. Volume of concrete for the 1st step: Height is

1 \times \frac{1}{4}

m.

V_1 = (\frac{1}{4} \times \frac{1}{2} \times 50) \text{ m}^3

.

Volume of concrete for the 2nd step: This step has a total height of

2 \times \frac{1}{4}

m from the ground.

V_2 = (\frac{2}{4} \times \frac{1}{2} \times 50) \text{ m}^3

.

Volume of concrete for the 3rd step: Height is

3 \times \frac{1}{4}

m.

V_3 = (\frac{3}{4} \times \frac{1}{2} \times 50) \text{ m}^3

.

This interpretation is also incorrect. Let's re-read.

Arithmetic ProgressionsClass 10 Mathematics NCERT Solutions

In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km .

The amount of air present in a cylinder when a vacuum pump removes 14\frac{1}{4}41​ of the air remaining in the cylinder at a time.

The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.

The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8%8\%8% per annum.

Solution

Write first four terms of the AP, when the first term aaa and the common difference ddd are given as follows:

a=10,d=10a=10, \quad d=10a=10,d=10

a=−2,d=0a=-2, \quad d=0a=−2,d=0

a=4,d=−3a=4, \quad d=-3a=4,d=−3

a=−1,d=12a=-1, d=\frac{1}{2}a=−1,d=21​

a=−1.25,d=−0.25a=-1.25, d=-0.25a=−1.25,d=−0.25

Solution

For the following APs, write the first term and the common difference:

3,1,−1,−3,…3,1,-1,-3, \ldots3,1,−1,−3,…

−5,−1,3,7,…-5,-1,3,7, \ldots−5,−1,3,7,…

13,53,93,133,…\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots31​,35​,39​,313​,…

0.6,1.7,2.8,3.9,…0.6,1.7,2.8,3.9, \ldots0.6,1.7,2.8,3.9,…

Solution

Which of the following are APs ? If they form an AP, find the common difference ddd and write three more terms.

2,4,8,16,…2,4,8,16, \ldots2,4,8,16,…

2,52,3,72,…2, \frac{5}{2}, 3, \frac{7}{2}, \ldots2,25​,3,27​,…

−1.2,−3.2,−5.2,−7.2,…-1.2,-3.2,-5.2,-7.2, \ldots−1.2,−3.2,−5.2,−7.2,…

−10,−6,−2,2,…-10,-6,-2,2, \ldots−10,−6,−2,2,…

3,3+2,3+22,3+32,⋯3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \cdots3,3+2​,3+22​,3+32​,⋯

0.2,0.22,0.222,0.2222,…0.2,0.22,0.222,0.2222, \ldots0.2,0.22,0.222,0.2222,…

0,−4,−8,−12,…0,-4,-8,-12, \ldots0,−4,−8,−12,…

−12,−12,−12,−12,…-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots−21​,−21​,−21​,−21​,…

1,3,9,27,…1,3,9,27, \ldots1,3,9,27,…

Solution

Solution

Choose the correct choice in the following and justify :

30 th term of the AP: 10,7,4,…10,7,4, \ldots10,7,4,…, is

11th term of the AP: −3,−12,2,…-3,-\frac{1}{2}, 2, \ldots−3,−21​,2,…, is

Solution

In the following APs, find the missing terms in the boxes :

2,□,262, \square, 262,□,26

□,13,□,3\square, 13, \square, 3□,13,□,3

5,□,□,9125, \square, \square, 9 \frac{1}{2}5,□,□,921​

−4,□,□,□,□,6-4, \square, \square, \square, \square, 6−4,□,□,□,□,6

□,38,□,□,□,−22\square, 38, \square, \square, \square,-22□,38,□,□,□,−22

Solution

Which term of the AP: 3,8,13,18,…3,8,13,18, \ldots3,8,13,18,…, is 78 ?

Solution

Find the number of terms in each of the following APs :

7,13,19,…,2057,13,19, \ldots, 2057,13,19,…,205

18,1512,13,…,−4718,15 \frac{1}{2}, 13, \ldots,-4718,1521​,13,…,−47

Solution

Check whether -150 is a term of the AP : 11,8,5,2…11,8,5,2 \ldots11,8,5,2…

Solution

Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Solution

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106 . Find the 29th term.

Solution

If the 3 rd and the 9 th terms of an AP are 4 and -8 respectively, which term of this AP is zero?

Solution

The 17 th term of an AP exceeds its 10 th term by 7 . Find the common difference.

Solution

Which term of the AP: 3,15,27,39,…3,15,27,39, \ldots3,15,27,39,… will be 132 more than its 54 th term?

Solution

Two APs have the same common difference. The difference between their 100 th terms is 100, what is the difference between their 1000th terms?

Solution

How many three-digit numbers are divisible by 7 ?

Solution

How many multiples of 4 lie between 10 and 250 ?

Solution

For what value of nnn, are the nnnth terms of two APs: 63,65,67,…63,65,67, \ldots63,65,67,… and 3,10,17,…3,10,17, \ldots3,10,17,… equal?

Solution

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12 .

Solution

Find the 20 th term from the last term of the AP : 3,8,13,…,2533,8,13, \ldots, 2533,8,13,…,253.

Solution

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution

Subba Rao started work in 1995 at an annual salary of ₹5000₹ 5000₹5000 and received an increment of ₹200₹ 200₹200 each year. In which year did his income reach ₹7000₹ 7000₹7000?

Solution

Ramkali saved ₹5₹ 5₹5 in the first week of a year and then increased her weekly savings by ₹1.75₹ 1.75₹1.75. If in the nnnth week, her weekly savings become ₹20.75₹ 20.75₹20.75 , find nnn.

Solution

Find the sum of the following APs:

The amount of air present in a cylinder when a vacuum pump removes $\frac{1}{4}$ of the air remaining in the cylinder at a time.

The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at $8\%$ per annum.

Write first four terms of the AP, when the first term $a$ and the common difference $d$ are given as follows:

$a=10, \quad d=10$

$a=-2, \quad d=0$

$a=4, \quad d=-3$

$a=-1, d=\frac{1}{2}$

$a=-1.25, d=-0.25$

$3,1,-1,-3, \ldots$

$-5,-1,3,7, \ldots$

$\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots$

$0.6,1.7,2.8,3.9, \ldots$

Which of the following are APs ? If they form an AP, find the common difference $d$ and write three more terms.

$2,4,8,16, \ldots$

$2, \frac{5}{2}, 3, \frac{7}{2}, \ldots$

$-1.2,-3.2,-5.2,-7.2, \ldots$

$-10,-6,-2,2, \ldots$

$3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \cdots$

$0.2,0.22,0.222,0.2222, \ldots$

$0,-4,-8,-12, \ldots$

$-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots$

$1,3,9,27, \ldots$

30 th term of the AP: $10,7,4, \ldots$ , is

11th term of the AP: $-3,-\frac{1}{2}, 2, \ldots$ , is

$2, \square, 26$

$\square, 13, \square, 3$

$5, \square, \square, 9 \frac{1}{2}$

$-4, \square, \square, \square, \square, 6$

$\square, 38, \square, \square, \square,-22$

Which term of the AP: $3,8,13,18, \ldots$ , is 78 ?

$7,13,19, \ldots, 205$

$18,15 \frac{1}{2}, 13, \ldots,-47$

Check whether -150 is a term of the AP : $11,8,5,2 \ldots$

Which term of the AP: $3,15,27,39, \ldots$ will be 132 more than its 54 th term?

For what value of $n$ , are the $n$ th terms of two APs: $63,65,67, \ldots$ and $3,10,17, \ldots$ equal?

Find the 20 th term from the last term of the AP : $3,8,13, \ldots, 253$ .

Subba Rao started work in 1995 at an annual salary of $₹ 5000$ and received an increment of $₹ 200$ each year. In which year did his income reach $₹ 7000$ ?

Ramkali saved $₹ 5$ in the first week of a year and then increased her weekly savings by $₹ 1.75$ . If in the $n$ th week, her weekly savings become $₹ 20.75$ , find $n$ .

$2,7,12, \ldots$ , to 10 terms.

$-37,-33,-29, \ldots$ , to 12 terms.

$0.6,1.7,2.8, \ldots$ , to 100 terms.

$\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots$ , to 11 terms.

$7+10 \frac{1}{2}+14+\ldots+84$

$34+32+30+\ldots+10$

$-5+(-8)+(-11)+\ldots+(-230)$

In an $A P$ :

given $a=5, d=3, a_{n}=50$ , find $n$ and $\mathrm{S}_{n}$ .

given $a=7, a_{13}=35$ , find $d$ and $\mathrm{S}_{13}$ .

given $a_{12}=37, d=3$ , find $a$ and $\mathrm{S}_{12}$ .

given $a_{3}=15, \mathrm{~S}_{10}=125$ , find $d$ and $a_{10}$ .

given $d=5, S_{9}=75$ , find $a$ and $a_{9}$ .

given $a=2, d=8, \mathrm{~S}_{n}=90$ , find $n$ and $a_{n}$ .

given $a=8, a_{n}=62, \mathrm{~S}_{n}=210$ , find $n$ and $d$ .

given $a_{n}=4, d=2, \mathrm{~S}_{n}=-14$ , find $n$ and $a$ .

given $a=3, n=8, \mathrm{~S}=192$ , find $d$ .

given $l=28, \mathrm{~S}=144$ , and there are total 9 terms. Find $a$ .

How many terms of the AP : $9,17,25, \ldots$ must be taken to give a sum of 636 ?

Find the sum of first 22 terms of an AP in which $d=7$ and 22nd term is 149 .

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289 , find the sum of first $n$ terms.

Show that $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ form an AP where $a_{n}$ is defined as below:

$a_{n}=3+4 n$

$a_{n}=9-5 n$ Also find the sum of the first 15 terms in each case.

If the sum of the first $n$ terms of an AP is $4 n-n^{2}$ , what is the first term (that is $\mathrm{S}_{1}$ )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the $n$ th terms.

Which term of the AP : $121, 117, 113, \ldots$ , is its first negative term? [Hint : Find $n$ for $a_{n}<0$ ]