If the sum of n terms of an AP is

Let the first term be a and common difference be d

To Find: d

Given: Sum of $n$ terms of $A P=n P+\frac{n}{2}(n-1) Q$

$\Rightarrow \frac{n}{2}[2 a+(n-1) d]=n P+\frac{n}{2}(n-1) Q$

$\Rightarrow 2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}=2 \mathrm{P}+(\mathrm{n}-1) \mathrm{Q}$

$\Rightarrow 2(\mathrm{a}-\mathrm{P})=(\mathrm{n}-1)(\mathrm{Q}-\mathrm{d})$

Put n = 1 to get the first term as sum of 1 term of an AP is the term itself.

$\Rightarrow P=a$

$\Rightarrow(n-1)(Q-d)=0$

For $n$ not equal to $1 Q=d$

Common difference is $Q$.