JEE Main 2025 Jan 24 Shift 1, Mathematics Q4: Summation of Series

JEE Main2025Jan 24, Shift 1Mathematics

Let $S_{n} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \dots$ upto $n$ terms. If the sum of the first six terms of an A.P. with first term $- p$ and common difference $p$ is $2026 S_{2025}$ , then the absolute difference between $2 0^{t h}$ and $1 5^{t h}$ terms of the A.P. is

A
$25$
B
$90$
C
$20$
D
$45$

Solution

The $n$ -th term of $S_{n}$ is $\frac{1}{n ( n + 1 )}$ , which telescopes:

$S_{n} = k = 1 \sum n (\frac{1}{k} - \frac{1}{k + 1}) = 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$

So $S_{2025} = \frac{2025}{2026}$ and $2026 S_{2025} = 2025 = 45$ .

Sum of first six A.P. terms with $a = - p$ , $d = p$ :

$S_{6} = \frac{6}{2} [2 (- p) + 5 p] = 3 (3 p) = 9 p$

Setting $9 p = 45$ gives $p = 5$ .

Difference between the $2 0^{t h}$ and $1 5^{t h}$ terms is $5 d = 5 p = 25$ .

$∴$ Option (1).

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JEE Main 2025 Jan 24 Shift 1, Mathematics Q4: Summation of Series

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