Question

If $1,2,\mathrm{3....}$ are first terms; $1,3,5....$ are common differences and $S_1,S_2,S_3....$ are sums of $n$ terms of given $pA{P}^{\prime }s$; then $S_1+S_2+S_3+.....+S_p$ is equal to

• A. $\frac{np(np+1)}{2}$
• B. $\frac{n(np+1)}{2}$
• C. $\frac{np(p+1)}{2}$
• D. $\frac{np(np-1)}{2}$

Answer 1

The correct option is A $\frac{np(np+1)}{2}$

We know,

$\sum a=1+2+3+...+p=\frac{p(p+1)}{2}$ ......( 1 )

$\sum d=1+3+5+...+(2p-1)=\frac{p}{2}.[2.1+(p-1\left)2\right]=p^2$ ......( 2 )

$S_p=\frac{n}{2}.[2a_p+(n-1\left)d_p\right]$ ......( 3 )

From eq $\left(1\right),\left(2\right),\left(3\right):$

$\therefore S_1+S_2+...+S_p=\frac{n}{2}.[2\sum a+(n-1)\sum d]$

$=\frac{n}{2}.[2\frac{p(p+1)}{2}+(n-1).p^2]$

$=\frac{n}{2}.[p^2+p+n{p}^2-p^2]$

$=\frac{np(np+1)}{2}$