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JEE Main 202524 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Sequences and Series

Let S_n=(1)/(2)+(1)/(6)+(1)/(12)+(1)/(20)+ upto n terms. If the sum...

Let

S_n=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\ldots

upto

n

terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is

\sqrt{2026 \mathrm{ S}_{2025}}

, then the absolute difference betwen

20^{\text{th }}

and

15^{\text{th }}

terms of the A.P. is

Options

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

Correct answer

D. $25$

Step-by-step solution

\begin{aligned} & \mathrm{Sn}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20} \ldots . \mathrm{N} \text{ terms } \\ & \mathrm{S}_{2025}=\sum_{\mathrm{n}=1}^{2025} \frac{1}{\mathrm{n}(\mathrm{n}+1)}=\sum_{\mathrm{n}=1}^{2025}\left(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1}\right) \\ & =\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right) \quad \ldots \cdots\left(\frac{1}{2025}-\frac{1}{2026}\right)\end{aligned}

\begin{aligned} & =\frac{2025}{2026} \\ & \sqrt{2026 . \mathrm{S}_{2025}}=\sqrt{2025}=45 \end{aligned}

Given :

\frac{6}{2}[-2 p+(6-1) p]=45

\begin{aligned} & 9 \mathrm{p}=45 \\ & \mathrm{p}=5 \\ & \left|\mathrm{ A}_{20}-\mathrm{A}_{15}\right|=|-5+19 \times 5|-[-5+14 \times 5] \\ & =|90-65| \\ & =25 \end{aligned}

$

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