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 $T_n = S_n - S_{n-1}$.For $n=1$,$T_1 = S_

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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 $T_n = S_n - S_{n-1}$.For $n=1$,$T_1 = S_1 = A(1) + (1)^2B = A + B$.For $n=2$,$S_2 = A(2) + (2)^2B = 2A + 4B$.Thus,$T_2 = S_2 - S_1 = (2A + 4B) - (A + B) = A + 3B$.The common difference $d$ is given by $d = T_2 - T_1$.$d = (A + 3B) - (A + B) = 2B$.Therefore,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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