Question:
If the sum of $n$ terms of an A.P. is $n P+\frac{1}{2} n(n-1) Q$, where $P$ and $Q$ are constants, find the common difference.
Solution:
We have:
$S_{n}=n P+\frac{1}{2} n(n-1) Q$
For $n=1, S_{1}=P+0=P$
For $n=2, S_{2}=2 P+Q$
Also, $a_{1}=S_{1}=P$
$a_{2}=S_{2}-S_{1}$
$=2 P+Q-P=P+Q$
$\therefore d=a_{2}-a_{1}=P+Q-P=Q$
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