Let the sum of the first n terms of an A.P. be . Then the sum of squares of the first 10 terms of the A.P. is: [2026]

Question

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: [2026]

Smart Achievers · Accepted Answer

(3)

We have, $S_{n} = 3 n^{2} + 5 n$

$S_{n - 1} = 3 {(n - 1)}^{2} + 5 (n - 1)$

Also, $T_{n} = S_{n} - S_{n - 1}$

$= 3 n^{2} + 5 n - (3 (n^{2} - 2 n + 1) + 5 n - 5)$

$= 3 n^{2} + 5 n - (3 n^{2} - 6 n + 3 + 5 n - 5)$

$\Rightarrow T_{n} = 6 n + 2$

Now, $\sum_{n = 1}^{10} {(6 n + 2)}^{2} = 36 \sum_{n = 1}^{10} n^{2} + 24 \sum_{n = 1}^{10} n + \sum_{n = 1}^{10} 1$

$= 36 \frac{n (n + 1) (2 n + 1)}{6} + 24 \frac{n (n + 1)}{2} + 4 \times 10$

$= 6 (10) (11) (21) + 12 (10 \times 11) + 40$

=15220

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