Predicting What Comes Next: Exploring Sequences and ProgressionsClass 9 Mathematics NCERT Solutions

To Find: The $31^{\text{st}}$ term, $t_{31}$.

Formula: The $n^{\text{th}}$ term of an AP is $t_n = a + (n-1)d$.

Solution: From the given information, we form two linear equations:

Subtract Equation 1 from Equation 2: $(a + 15d) - (a + 10d) = 73 - 38$ $5d = 35$ $d = 7$

Substitute $d=7$ into Equation 1 to find $a$: $a + 10(7) = 38$ $a + 70 = 38$ $a = 38 - 70 = -32$

Now we have $a = -32$ and $d = 7$. We can find the $31^{\text{st}}$ term: $t_{31} = a + (31-1)d$ $t_{31} = -32 + (30)7$ $t_{31} = -32 + 210$ $t_{31} = 178$

Final Answer: The $31^{\text{st}}$ term is 178.

To Find: The AP.

Formula: $t_n = a + (n-1)d$.

Solution: First, use the second piece of information to find the common difference, $d$. $t_7 = t_5 + 12$ $(a + (7-1)d) = (a + (5-1)d) + 12$ $a + 6d = a + 4d + 12$ $6d = 4d + 12$ $2d = 12$ $d = 6$

Next, use the first piece of information ($t_3 = 16$) to find the first term, $a$. $t_3 = a + (3-1)d$ $16 = a + 2d$ Substitute $d=6$ into this equation: $16 = a + 2(6)$ $16 = a + 12$ $a = 16 - 12 = 4$

Now we have the first term $a=4$ and the common difference $d=6$. The AP is formed by starting with $a$ and repeatedly adding $d$. First term: 4 Second term: $4+6=10$ Third term: $10+6=16$ Fourth term: $16+6=22$ and so on.

Final Answer: The AP is 4, 10, 16, 22, ...

To Find: The count of three-digit numbers divisible by 7.

Method: The three-digit numbers divisible by 7 form an arithmetic progression (AP).

We now have an AP: 105, 112, ..., 994. We need to find the number of terms, $n$.

Formula: $l = a + (n-1)d$

Solution: $994 = 105 + (n-1)7$ $994 - 105 = 7(n-1)$ $889 = 7(n-1)$ $\frac{889}{7} = n-1$ $127 = n-1$ $n = 128$

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

To Find: The count of multiples of 4 that are strictly between 10 and 250.

Method: The multiples of 4 form an arithmetic progression (AP).

We now have an AP: 12, 16, ..., 248. We need to find the number of terms, $n$.

Formula: $l = a + (n-1)d$

Solution: $248 = 12 + (n-1)4$ $248 - 12 = 4(n-1)$ $236 = 4(n-1)$ $\frac{236}{4} = n-1$ $59 = n-1$ $n = 60$

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

To Find: The GP.

Formula: $t_n = ar^{n-1}$.

Solution: First, use the second condition to find the common ratio, $r$. $t_5 = 4t_3$ $ar^4 = 4(ar^2)$ Assuming $a \neq 0$ and $r \neq 0$, we can divide both sides by $ar^2$: $r^2 = 4$ $r = 2 \text{ or } r = -2$

We have two possible cases for $r$. We use the first condition ($a+ar=-4$) to find $a$ for each case.

Case 1: $r = 2$ $a + a(2) = -4$ $3a = -4$ $a = -\frac{4}{3}$ The GP is $a, ar, ar^2, ...$ $- \frac{4}{3}, -\frac{4}{3}(2), -\frac{4}{3}(2)^2, ...$ $- \frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, ...$

Case 2: $r = -2$ $a + a(-2) = -4$ $a - 2a = -4$ $-a = -4$ $a = 4$ The GP is $a, ar, ar^2, ...$ $4, 4(-2), 4(-2)^2, ...$ $4, -8, 16, ...$

Final Answer: There are two possible GPs that satisfy the conditions:

To Find: All sets of consecutive natural numbers that sum to 100.

Method: Let the sum be of $n$ consecutive natural numbers, starting with $k$. The sequence is $k, k+1, ..., k+n-1$. This is an arithmetic series with first term $a=k$, last term $l=k+n-1$, and $n$ terms.

Formula: Sum $S = \frac{n}{2}(a+l)$

Solution: $100 = \frac{n}{2}(k + (k+n-1))$ $200 = n(2k + n - 1)$ Here $n$ and $k$ must be natural numbers ($n, k \ge 1$). Since the sum is of consecutive numbers, $n$ must be at least 2. The term $(2k+n-1)$ must also be a natural number.

$n$ and $(2k+n-1)$ are factors of 200. Let's analyze the factors. One factor is $n$ and the other is $m = 2k+n-1$. Notice that their difference is $m-n = (2k+n-1) - n = 2k-1$, which is an odd number. This means one factor must be even and the other must be odd.

Let's find the prime factorization of 200: $200 = 20 \times 10 = (2^2 \times 5) \times (2 \times 5) = 2^3 \times 5^2 = 8 \times 25$. We need to find factor pairs $(n, m)$ of 200 where one is odd and one is even. The odd factor must be a factor of $5^2=25$. The possible odd factors are 1, 5, 25.

Case 1: The odd factor is 5. The even factor is $200/5 = 40$.

Case 2: The odd factor is 25. The even factor is $200/25 = 8$.

Case 3: The odd factor is 1. This would mean the other factor is 200. If $n=1$, it's not a sum of consecutive numbers. If $m=1$, then $1 = 2k+n-1$. Since $n \ge 2$, $2k+n-1 \ge 2(1)+2-1=3$, so $m$ cannot be 1.

We have found two possible ways.

Final Answer: There are two ways to express 100 as the sum of consecutive natural numbers:

Method: This is a geometric progression (GP).

To Find: The first three terms of the AP ($a, a+d, a+2d$).

Formula: $t_n = a + (n-1)d$.

Solution: Translate the given information into equations with $a$ and $d$.

First condition: $t_4 + t_8 = 24$ $(a + (4-1)d) + (a + (8-1)d) = 24$ $(a + 3d) + (a + 7d) = 24$ $2a + 10d = 24$ Dividing by 2, we get: $a + 5d = 12 \quad ...(\text{Eq } 1)$

Second condition: $t_6 + t_{10} = 44$ $(a + (6-1)d) + (a + (10-1)d) = 44$ $(a + 5d) + (a + 9d) = 44$ $2a + 14d = 44$ Dividing by 2, we get: $a + 7d = 22 \quad ...(\text{Eq } 2)$

Now we have a system of two linear equations. Subtract Equation 1 from Equation 2: $(a + 7d) - (a + 5d) = 22 - 12$ $2d = 10$ $d = 5$

Substitute $d=5$ into Equation 1 to find $a$: $a + 5(5) = 12$ $a + 25 = 12$ $a = 12 - 25 = -13$

So, the first term is $a = -13$ and the common difference is $d=5$.

The first three terms are:

Final Answer: The first three terms of the AP are -13, -8, -3.

To Find: The smallest natural number $n$ such that $1+2+3+...+n > 1000$.

Formula: The sum of the first $n$ natural numbers is $S_n = \frac{n(n+1)}{2}$.

Solution: We need to solve the inequality: $\frac{n(n+1)}{2} > 1000$ $n(n+1) > 2000$ $n^2 + n - 2000 > 0$

To find an approximate value for $n$, we can consider the equation $n^2 \approx 2000$. $n \approx \sqrt{2000} = \sqrt{100 \times 20} = 10\sqrt{20}$ Since $\sqrt{16}=4$ and $\sqrt{25}=5$, $\sqrt{20}$ is between 4 and 5, closer to 4.5. $4.4^2 = 19.36$, $4.5^2 = 20.25$. So $\sqrt{20} \approx 4.47$. $n \approx 10 \times 4.47 = 44.7$ Since $n$ must be a natural number, let's test values around 44 and 45.

Therefore, the smallest value of $n$ that satisfies the condition is 45.

Final Answer: The smallest value of $n$ is 45.

Given: The geometric progression (GP) 2, 8, 32, ...

Solution: First, identify the parameters of the GP.

Formula: $t_n = ar^{n-1}$

Calculation: Set $t_n = 131072$. $131072 = 2 \times 4^{n-1}$ $\frac{131072}{2} = 4^{n-1}$ $65536 = 4^{n-1}$ We need to express 65536 as a power of 4. We can use powers of 2: $4 = 2^2$. $65536 = 2^{16}$. So, the equation is $(2^2)^{n-1} = 2^{16}$, which is $2^{2(n-1)} = 2^{16}$. Equating exponents: $2(n-1) = 16 \implies n-1 = 8 \implies n=9$. Alternatively, using base 4: $4^1=4, 4^2=16, 4^3=64, 4^4=256, 4^5=1024, 4^6=4096, 4^7=16384, 4^8=65536$. So, $4^8 = 4^{n-1}$, which gives $n-1=8$, so $n=9$.

Final Answer (Part 1): 131072 is the $9^{\text{th}}$ term.

Formula: $t_n = ar^{n-1}$

Calculation: With $a=2$ and $r=4$, the explicit formula is $t_n = 2 \times 4^{n-1}$.

Final Answer (Part 2): The explicit formula is $t_n = 2 \times 4^{n-1}$.

Formula: $t_1 = a$, and $t_n = r \times t_{n-1}$ for $n \geq 2$.

Final Answer (Part 3): The recursive formula is $t_1 = 2, t_n = 4t_{n-1}$ for $n \geq 2$.

To Find: The common ratio ($r$) and the terms.

Method: It is convenient to represent the three terms as $\frac{a}{r}, a, ar$. This simplifies the product calculation.

Solution: Step 1: Use the product information. $(\frac{a}{r}) \times (a) \times (ar) = -1$ $a^3 = -1$ $a = -1$ So, the middle term is -1.

Step 2: Use the sum information. $\frac{a}{r} + a + ar = \frac{13}{12}$ Substitute $a=-1$: $\frac{-1}{r} - 1 - r = \frac{13}{12}$ Multiply the entire equation by $12r$ to clear fractions: $12r(\frac{-1}{r}) - 12r(1) - 12r(r) = 12r(\frac{13}{12})$ $-12 - 12r - 12r^2 = 13r$ Rearrange into a standard quadratic form ($Ax^2+Bx+C=0$): $12r^2 + 13r + 12r + 12 = 0$ $12r^2 + 25r + 12 = 0$

Step 3: Solve the quadratic equation for r. We can use the quadratic formula $r = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. Here, $a=12, b=25, c=12$. $r = \frac{-25 \pm \sqrt{25^2 - 4(12)(12)}}{2(12)}$ $r = \frac{-25 \pm \sqrt{625 - 576}}{24}$ $r = \frac{-25 \pm \sqrt{49}}{24}$ $r = \frac{-25 \pm 7}{24}$ This gives two possible values for $r$:

Step 4: Find the terms for each value of r. Remember the terms are $\frac{a}{r}, a, ar$ with $a=-1$.

Both cases give the same set of terms in a different order.

Final Answer: The common ratio can be either $- \frac{3}{4}$ or $- \frac{4}{3}$. The terms of the GP are $\frac{4}{3}, -1, \frac{3}{4}$.

To Prove: $x, y, z$ are in GP.

Method: To prove that three numbers $x, y, z$ are in GP, we need to show that the ratio of consecutive terms is constant, i.e., $\frac{y}{x} = \frac{z}{y}$, or equivalently, $y^2 = xz$.

Proof: First, express $x, y,$ and $z$ in terms of $a$ and $r$ using the formula $t_n = ar^{n-1}$.

Now, let's compute $y^2$ and $xz$ separately.

Calculate $y^2$: $y^2 = (ar^9)^2 = a^2(r^9)^2 = a^2r^{18}$

Calculate $xz$: $xz = (ar^3)(ar^{15})$ $xz = a \cdot a \cdot r^3 \cdot r^{15}$ $xz = a^2r^{3+15} = a^2r^{18}$

Since $y^2 = a^2r^{18}$ and $xz = a^2r^{18}$, we have shown that $y^2 = xz$.

Conclusion: Because the square of the middle term is equal to the product of the other two terms, $x, y, z$ are in a geometric progression.

To Find: The terms of the GP.

Method: Let the terms be $a, ar, ar^2$.

Solution: From the given information, we have two equations:

We can use the identity $1+r^2+r^4 = (1+r+r^2)(1-r+r^2)$. Substitute this into Eq 2: $a^2(1+r+r^2)(1-r+r^2) = 364 \quad ...(\text{Eq } 3)$

Now, let's square Equation 1: $(a(1+r+r^2))^2 = 26^2$ $a^2(1+r+r^2)^2 = 676 \quad ...(\text{Eq } 4)$

Divide Equation 4 by Equation 3: $\frac{a^2(1+r+r^2)^2}{a^2(1+r+r^2)(1-r+r^2)} = \frac{676}{364}$ Simplify the fraction on the right: $676/364 = (182 \times 3 + 130) / 182 = ...$ Let's simplify by dividing by common factors. Both are divisible by 4: $169/91$. Both are divisible by 13: $13/7$. So, $\frac{676}{364} = \frac{13}{7}$.

Simplifying the left side: $\frac{1+r+r^2}{1-r+r^2} = \frac{13}{7}$ Cross-multiply: $7(1+r+r^2) = 13(1-r+r^2)$ $7 + 7r + 7r^2 = 13 - 13r + 13r^2$ Rearrange into a quadratic equation: $(13-7)r^2 + (-13-7)r + (13-7) = 0$ $6r^2 - 20r + 6 = 0$ Divide by 2: $3r^2 - 10r + 3 = 0$ Factor the quadratic: $3r^2 - 9r - r + 3 = 0$ $3r(r-3) - 1(r-3) = 0$ $(3r-1)(r-3) = 0$ This gives two possible values for $r$: $r=3$ or $r=1/3$.

Case 1: $r=3$ Substitute into Eq 1: $a(1+3+3^2) = 26 \implies a(1+3+9) = 26 \implies 13a = 26 \implies a=2$. The terms are $a, ar, ar^2 \implies 2, 2(3), 2(3^2) \implies 2, 6, 18$. Check: Sum = $2+6+18=26$. Sum of squares = $4+36+324=364$. Correct.

Case 2: $r=1/3$ Substitute into Eq 1: $a(1+1/3+(1/3)^2) = 26 \implies a(1+1/3+1/9) = 26 \implies a(\frac{9+3+1}{9}) = 26 \implies a(\frac{13}{9}) = 26 \implies a = 26 \times \frac{9}{13} = 18$. The terms are $a, ar, ar^2 \implies 18, 18(1/3), 18(1/3)^2 \implies 18, 6, 2$. Check: Sum = $18+6+2=26$. Sum of squares = $324+36+4=364$. Correct.

Final Answer: The terms of the GP are 2, 6, 18 (or 18, 6, 2).

Final Answer (Part 1): The values are 1, 2, 4, 8, 16, 32, 64, 128.

Method: Consider the definition for $P_n$ and $P_{n-1}$ (for $n>3$): $P_n = P_1 + P_2 + ... + P_{n-2} + P_{n-1} + 1$ $P_{n-1} = P_1 + P_2 + ... + P_{n-2} + 1$ Notice that the expression for $P_{n-1}$ is part of the expression for $P_n$. We can write $P_n = (P_1 + P_2 + ... + P_{n-2} + 1) + P_{n-1}$. Substituting, we get: $P_n = P_{n-1} + P_{n-1} = 2P_{n-1}$. This holds for $n > 3$. Let's check for $n=3$: $P_3 = 2P_2 = 2(2) = 4$. This is correct. So the simpler recursion holds for $n > 2$.

Final Answer (Part 2): A simpler recursive formula is $P_1 = 1$, and $P_n = 2P_{n-1}$ for $n \geq 2$.

Method: Based on the values we calculated (1, 2, 4, 8, 16, ...), the sequence is a geometric progression starting from the second term.

Final Answer (Part 3): The explicit formula can be written in two parts: $P_1 = 1$ $P_n = 2^{n-1}$ for $n \geq 2$.

Final Answer (Part 1): The values are 1, 2, 3, 5, 8, 13, 21, 34.

Method: The sequence 1, 2, 3, 5, 8, 13, 21, 34, ... is the Virahānka-Fibonacci sequence. In this sequence, each term after the second is the sum of the two preceding ones. Let's verify if this holds for our sequence $W_n$.

For $n>3$, we have: $W_n = W_1 + ... + W_{n-3} + W_{n-2} + 2$ And for $n-1$, we have: $W_{n-1} = W_1 + ... + W_{n-3} + 2$

Subtracting the second equation from the first: $W_n - W_{n-1} = (W_1 + ... + W_{n-2} + 2) - (W_1 + ... + W_{n-3} + 2)$ $W_n - W_{n-1} = W_{n-2}$ $W_n = W_{n-1} + W_{n-2}$ This confirms that for $n>3$, the sequence follows the Fibonacci recurrence relation.

Final Answer (Part 2): Yes, this is the Virahānka-Fibonacci sequence.