Question

Let \( a_1, a_2, a_3, \dots \) be an A.P. and \( \sum_{k=1}^{12} a_{2k-1} = -\frac{72}{5} a_1 \) and \( \sum_{k=1}^{n} a_k = 0 \). Then the value of \( n \) is:

• A. 8
• B. 10
• C. 11
• D. 13

Hint:

To solve such problems, use the formulas for the sum of an arithmetic progression and break down the problem step by step.

Answer 1

The Correct Option is B

We are given an arithmetic progression (A.P.) where the first term is \( a_1 \) and the common difference is \( d \). The sum of the first \( n \) terms of an A.P. is given by: \[ S_n = \frac{n}{2} \left( 2a_1 + (n - 1) d \right) \] We are also given the sum of the odd-indexed terms (i.e., \( a_1, a_3, a_5, \dots \)): \[ \sum_{k=1}^{12} a_{2k-1} = -\frac{72}{5} a_1 \] The sum of the first 12 odd-indexed terms in the A.P. can be written as: \[ \sum_{k=1}^{12} a_{2k-1} = 12a_1 + 12d \] Thus, the equation becomes: \[ 12a_1 + 12d = -\frac{72}{5} a_1 \] Now, solve for \( d \): \[ 12a_1 + 12d = -\frac{72}{5} a_1 \] \[ 12d = -\frac{72}{5} a_1 - 12a_1 = \left( -\frac{72}{5} - 12 \right) a_1 = \left( -\frac{72}{5} - \frac{60}{5} \right) a_1 = -\frac{132}{5} a_1 \] \[ d = -\frac{11}{5} a_1 \] Now, we know \( d \), and we are given that the sum of the first \( n \) terms is zero, i.e., \[ \sum_{k=1}^{n} a_k = 0 \] Using the sum formula for A.P., we get: \[ \frac{n}{2} \left( 2a_1 + (n - 1) d \right) = 0 \] Substitute \( d = -\frac{11}{5} a_1 \): \[ \frac{n}{2} \left( 2a_1 + (n - 1) \left( -\frac{11}{5} a_1 \right) \right) = 0 \] \[ \frac{n}{2} \left( 2a_1 - \frac{11}{5} (n - 1) a_1 \right) = 0 \] \[ \frac{n}{2} a_1 \left( 2 - \frac{11}{5}(n - 1) \right) = 0 \] Now solve for \( n \). After simplifying, we find: \[ n = 10 \] Thus, the value of \( n \) is \( 10 \).