If the sum of the first n terms of a sequence | vedclass

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(A) We are given $S_n = An^2 + Bn$.$S_{n - 1} = A(n - 1)^2 + B(n - 1)$$= A(n^2 - 2n + 1) + B(n - 1)$$= An^2 - 2An + A + Bn - B$The $n^{th}$ term $a_n

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Arithmetic Progression

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(A) We are given $S_n = An^2 + Bn$.$S_{n - 1} = A(n - 1)^2 + B(n - 1)$$= A(n^2 - 2n + 1) + B(n - 1)$$= An^2 - 2An + A + Bn - B$The $n^{th}$ term $a_n = S_n - S_{n - 1} = (An^2 + Bn) - (An^2 - 2An + A + Bn - B)$$= An^2 + Bn - An^2 + 2An - A - Bn + B$$= 2An + B - A$Now,$a_{n - 1} = 2A(n - 1) + B - A = 2An - 2A + B - A = 2An + B - 3A$.The common difference $d = a_n - a_{n - 1} = (2An + B - A) - (2An + B - 3A) = 2A$.Since the difference $d = 2A$ is a constant independent of $n$,the sequence is an Arithmetic Progression.

If the sum of the first $n$ terms of a sequence is of the form $An^2 + Bn$,where $A$ and $B$ are constants independent of $n$,then the sequence is a ........

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