The sum of n terms of a series is `An^2+Bn`, then the `n^(th)` term is (A) A(2n-1)-B (B) A(1-2n)+B (C) A(1-2n)-B (D) A(2n-1)+B

Explanation

Here, we are given, `S_n = An^2+Bn`

`:. S_1 = A(1)^2+B(1) = A+B`

`S_2 = A(2)^2+B(2) = 4A+2B`

`S_3 = A(3)^2+B(3) = 9A+3B`

So, series will be,

`A+B, 3A+B,5A+B,...`

Here, first term, `a = A+B` and common difference `d = 2A`

So, `n^(th)` term will be, `T_n = a+(n-1)d = (A+B)+(n-1)2A`

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