Question

If $S_n=n^{2}p+\frac{n(n-1)}{4}q$ be the sum to ${}^{\prime }{n}^{\prime }$ terms of an A.P., then the common difference of the A.P. is

• A. $p+2q$
• B. $2p+\frac{q}{2}$
• C. $2p+q$
• D. $\frac{p+q}{2}$

Answer 1

The correct option is B $2p+\frac{q}{2}$

Given $S_n=n^{2}p+\frac{n(n-1)}{4}q$ $...\left(1\right)$
Also, we know that $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ $...\left(2\right)$
Comparing the equations $\left(1\right)$ and $\left(2\right)$, we get
$n^{2}p+\frac{n(n-1)}{4}q=\frac{n}{2}[2a+(n-1\left)d\right]$
$\Rightarrow na+\frac{n^{2}d}{2}-\frac{nd}{2}=n^2(p-\frac{q}{4})-n\frac{q}{4}$
By comparing coefficients of $n^2$ and $n$, we get
$\frac{d}{2}=p-\frac{q}{4}$ and $a-\frac{d}{2}=-\frac{q}{4}$
$d=2p-\frac{q}{2}$ and $d=2p+\frac{q}{2}$