Let the sum of the first $n$ terms of an A.P. be $3 n^2+5 n$. Then the sum of squares of the first $10$ terms of the A.P. is: [JEE Main 2026, 5 Apr (Shift 1)]

Question

YESPYQ · Accepted Answer

Correct answer: c) $15220$. $S_n=3 n^2+5 n$ $T_n=S_n-S_{n-1}$ $=3\left(n^2-(n-1)^2\right)+5(n-(n-1))$ $=3(2 n-1)+5$ $=6 n+2$ $\sum_{n=1}^{10}(6 n+2)^2=\sum 36 n^2+\sum 4+\sum 24 n$ $=36 \times \frac{10 \times 11 \times 21}{6}+4 \times 10+24 \frac{10 \times 11}{2}$ $=15220$

Let the sum of the first \(n\) terms of an A.P. be \(3 n^2+5 n\). Then the sum of squares of the first \(10\) terms of t…

Practice more JEE Maths PYQs