Mathematics · Sequence and Series

JEE Main 2026 — 5 April, Morning Shift — Question 25

Let the sum of the first n terms of an A.P. be $3n² + 5n.$ Then the sum of squares of the first $10$ terms of the A.P. is:

Option A:
$10220$
Option B:
$12860$
Option C:
$15220$
Correct
Option D:
$19780$

Answer: C

Step-by-step solution

$S_{n}=3 n^{2}+5 n$ $\mathrm{T}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}$ $\left.\Rightarrow \mathrm{T}_{\mathrm{n}}=\left(3 \mathrm{n}^{2}+5 \mathrm{n}\right)-\left(3(\mathrm{n}-1)^{2}\right)+5(\mathrm{n}-1)\right)$ $\Rightarrow \mathrm{T}_{\mathrm{n}}=6 \mathrm{n}+2$ $\sum_{n=1}^{10} T_{n}^{2}=\sum_{n=1}^{10}(6 n+2)^{2}$ $=36 \sum_{\mathrm{n}=1}^{10} \mathrm{n}^{2}+24 \sum_{\mathrm{n}=1}^{10} \mathrm{n}+\sum_{\mathrm{n}=1}^{10} 4$ $=15220$

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Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Mathematics
Chapter: Sequence and Series
Topic: Arithmetic Progression

∑ n=1^10 528/n(n+1)(n+2) is equal to:

Answer: C

Step-by-step solution

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