JEE Main 2026 · 5 Apr · Shift 1 · Mathematics Q2 · Dronacharya Exams

My attemptsTake a paper

← JEE Main 2026 · 5 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 2 of 75

Let the sum of the first $n$ terms of an A.P. be $3n^2+5n$ . Then the sum of squares of the first 10 terms of the A.P. is:

A

10220

B

12860

C

15220

D

19780

Answer and worked solution

Official answer: C

Terms. $a_n=S_n-S_{n-1}=3(2n-1)+5=6n+2$ .

Sum of squares.
$\sum_{n=1}^{10}(6n+2)^2=36\sum n^2+24\sum n+4\cdot10=36(385)+24(55)+40=15220.$

(C)

Topics:Arithmetic progression Special series sums

← Previous question Next question →