JEE Main 2026 — Apr 5, Shift 1 — Mathematics solution

JEE Main2026Apr 5, Shift 1Mathematics

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 :

Solution

The general term is $T_{n} = S_{n} - S_{n - 1} = (3 n^{2} + 5 n) - (3 (n - 1)^{2} + 5 (n - 1)) = 6 n + 2$ .

Therefore $T_{n}^{2} = (6 n + 2)^{2} = 36 n^{2} + 24 n + 4$ , and

$n = 1 \sum 10 T_{n}^{2} = 36 n = 1 \sum 10 n^{2} + 24 n = 1 \sum 10 n + 40 = 36 (385) + 24 (55) + 40 = 13860 + 1320 + 40 = 15220$ .

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