Let the sum of the first n terms of an A.P.

Question

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

Accepted Answer

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

$$\sum_{n=1}^{10}a_n^2 = \sum_{n=1}^{10}(6n+2)^2 = 36\sum n^2+24\sum n+4(10) = 36(385)+24(55)+40 = 13860+1320+40 = 15220.$$

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