# If the sum of n terms of an A.P.

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, $a_{1}=S_{1}=P$

$a_{2}=S_{2}-S_{1}$

$=2 P+Q-P=P+Q$

$\therefore d=a_{2}-a_{1}=P+Q-P=Q$