Let be in an A.P. such that . If , then n is : [2025]

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Solution

Q.
Let $a_{1}, a_{2}, a_{3}, . . . . .$ be in an A.P. such that $\sum_{k = 1}^{12} a_{2 k - 1} = - \frac{72}{5} a_{1}, a_{1} \neq 0$ . If $\sum_{k = 1}^{n} a_{k} = 0$ , then n is : [2025]

1 17

2 18

3 10

4 11

Ans.
(4)

Let First term, $a_{1} = a$ , common difference = d

Given, $\sum_{k = 1}^{12} a_{2 k - 1} = \frac{- 72}{5} a, a_{1} \neq 0$

$\Rightarrow a_{1} + a_{3} + a_{5} + . . . . + a_{23} = - \frac{72}{5} a$

$\Rightarrow a + (a + 2 d) + (a + 4 d) + . . . . + (a + 22 d) = \frac{- 72}{5} a$

Now, $\sum_{k = 1}^{12} a_{2 k - 1}$ is an AP with common difference as 2d.

$∴ \frac{12}{2} [2 a + 11 \times 2 d] = - \frac{72}{5} a$

$\Rightarrow 12 a + 132 d = - \frac{72}{5} a$

$\Rightarrow 132 a + 132 \times 5 d = 0$

$\Rightarrow a = - 5 d$

Also, given $\sum_{k = 1}^{n} a_{k} = 0 \Rightarrow \frac{n}{2} [2 a + (n - 1) d] = 0$

$\Rightarrow n (- 10 d + n d - d) = 0 \Rightarrow n d (n - 11) = 0 \Rightarrow n = 11$ .

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