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If the sum of $n$ terms of an A.P. is $\left(p n+q n^2\right)$, where $p$ and $q$ are constants, find the common difference.
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Verified Answer
Let $S_n$ be the sum of $n$ term $S_n=p n+q n^2$ Putting $n=1,2$
$S_1=T_1=p .1+q \cdot 1^2=p+q \text {, }$
$S_2=T_1+T_2=p .2+q \cdot 2^2=2 p+4 q$
$T_2=\left(T_1+\bar{T}_2\right)-T_1=S_2-S_1$
$=[(2 p+4 q)-(p+q)]=p+3 q$
$d=T_2-T_1=(p+3 q)-(p+q)=2 q$
$\therefore$ Common difference of the series is $2 q$.
$S_1=T_1=p .1+q \cdot 1^2=p+q \text {, }$
$S_2=T_1+T_2=p .2+q \cdot 2^2=2 p+4 q$
$T_2=\left(T_1+\bar{T}_2\right)-T_1=S_2-S_1$
$=[(2 p+4 q)-(p+q)]=p+3 q$
$d=T_2-T_1=(p+3 q)-(p+q)=2 q$
$\therefore$ Common difference of the series is $2 q$.
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