If S_n = nP + frac{1}{2}n(n - 1)Q,where S_n d | vedclass

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(D) Given the sum of the first $n$ terms of an $A.P.$ is $S_n = nP + \frac{1}{2}n(n - 1)Q$.Comparing this with the standard formula for the sum of an

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(D) Given the sum of the first $n$ terms of an $A.P.$ is $S_n = nP + \frac{1}{2}n(n - 1)Q$.Comparing this with the standard formula for the sum of an $A.P.$,$S_n = \frac{n}{2}\{2a + (n - 1)d\}$,we have:$S_n = n \cdot a + \frac{n(n - 1)}{2} \cdot d$.By comparing the coefficients of the terms,we get $a = P$ and $d = Q$.Alternatively,the common difference $d$ can be calculated as $d = S_2 - 2S_1$.$S_1 = P(1) + \frac{1}{2}(1)(0)Q = P$.$S_2 = P(2) + \frac{1}{2}(2)(1)Q = 2P + Q$.$d = (2P + Q) - 2(P) = Q$.

If $S_n = nP + \frac{1}{2}n(n - 1)Q$,where $S_n$ denotes the sum of the first $n$ terms of an $A.P.$,then the common difference is

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