Question

If $1,2,3,.....$ are first terms, $1,3,5,......$ are common differences and $S_1,S_2,S_3,...$ are sums of $n$ terms of given AP'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 $\frac{np(np+1)}{2}$

Let $a_1=1,a_2=2,a_3=3,.....,a_p=p$ be the first terms of APs

and $d_1=1,d_2=3,d_3=5,......,d_p=2p-1;$ be the common differences of the respective AP

$S_1+S_2+S_3+....S_p=\frac{n}{2}(2a_1+(n-1\left)d_1\right)+\frac{n}{2}(2a_2+(n-1\left)d_2\right)+....+\frac{n}{2}(2a_p+(n-1\left)d_p\right)=n(a_1+a_2+a_3+....+a_p)+\frac{n(n-1)}{2}(d_1+d_2+d_3+....+d_p)=n\frac{p(p+1)}{2}+\frac{n(n-1)}{2}{p}^2=\frac{np}{2}(p+1-pn-p)=\frac{np}{2}(pn+1)$