JEE Main 2025 Apr 2 Shift 1, Mathematics Q8: Arithmetic, Geometric And Harmonic Progressions

JEE Main2025Apr 2, Shift 1Mathematics

Let $a_{1}, a_{2}, a_{3}, \dots$ 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:

A
11
B
10
C
18
D
17

Solution

Let the first term be $a$ and the common difference be $d$ . The odd-indexed terms $a_{1}, a_{3}, a_{5}, \dots, a_{23}$ themselves form an A.P. with first term $a$ and common difference $2 d$ , containing $12$ terms.

$k = 1 \sum 12 a_{2 k - 1} = \frac{12}{2} [2 a + 11 \cdot (2 d)] = 12 a + 132 d$

The given condition becomes:

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

Multiplying by $5$ :

$60 a + 660 d = - 72 a \Rightarrow 132 a + 660 d = 0 \Rightarrow a = - 5 d$

Find $n$ where the sum is zero:

$S_{n} = \frac{n}{2} [2 a + (n - 1) d] = 0 \Rightarrow 2 a + (n - 1) d = 0$

Substitute $a = - 5 d$ (with $d \neq = 0$ , otherwise the A.P. is constant and $a_{1} \neq = 0$ is violated):

$- 10 d + (n - 1) d = 0 \Rightarrow n - 1 = 10 \Rightarrow n = 11$

$∴$ Option (1) is correct.

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JEE Main 2025 Apr 2 Shift 1, Mathematics Q8: Arithmetic, Geometric And Harmonic Progressions

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