If the sum of n terms of an A.P. is nA + n^2B | vedclass

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(D) Given that the sum of $n$ terms is $S_n = nA + n^2B$.To find the terms of the $A.P.$,we use the relation $T_n = S_n - S_{n-1}$.First,find $S_1$ an

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$2B$

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(D) Given that the sum of $n$ terms is $S_n = nA + n^2B$.To find the terms of the $A.P.$,we use the relation $T_n = S_n - S_{n-1}$.First,find $S_1$ and $S_2$:$S_1 = A(1) + B(1)^2 = A + B$$S_2 = A(2) + B(2)^2 = 2A + 4B$Now,calculate the first two terms:$T_1 = S_1 = A + B$$T_2 = S_2 - S_1 = (2A + 4B) - (A + B) = A + 3B$The common difference $d$ is given by $d = T_2 - T_1$:$d = (A + 3B) - (A + B) = 2B$Thus,the common difference is $2B$.

If the sum of $n$ terms of an $A.P.$ is $nA + n^2B$,where $A$ and $B$ are constants,then its common difference will be

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