Mathematics · Sequence and Series

JEE Main 2025 — 24 January, Morning Shift — Question 4

Let $\mathrm{S}_{\mathrm{n}}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\ldots$ 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 between $20^{\text {th }}$ and $15^{\text {th }}$

terms of the A.P. is

Option A:
25
Correct
Option B:
90
Option C:
20
Option D:
45

Answer: A

Step-by-step solution

$\mathrm{Sn}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20} \quad \ldots . \mathrm{n}$ terms

\begin{aligned} & \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) \\& \quad=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right) \quad \cdots \cdot\left(\frac{1}{2025}-\frac{1}{2026}\right) \end{aligned}

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

$\sqrt{2026 \cdot \mathrm{~S}_{2025}}=\sqrt{2025}=45$

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

$9 p=45$

$\mathrm{p}=5$

\begin{aligned} & \left|A_{20}-A_{15}\right|=|-5+19 \times 5|-[-5+14 \times 5] \\& =|90-65| \\& =25 \end{aligned}

Answer key and solution verified before publishing.

Practise Sequence and Series

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 24 January, Morning Shift
Subject: Mathematics
Chapter: Sequence and Series
Topic: Arithmetic Progression