Question

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

For $S_P$

$1$st term $\left(a\right)=P$

Common difference =$2P-1$

No of terms=$n$

$S_P=\frac{n}{2}[2_P+(n-1)(2_P-1)]=\frac{n}{2}[2_P+2_{nP}-2_P+1-n]=\frac{n}{2}[2_{nP}+1-n]S_1=\frac{n(n+1)}{2}{S}_2=\frac{n(3n+1)}{2}{S}_3=\frac{n(5n+1)}{2}{S}_2-S_1=\frac{3n^2+n}{2}-\frac{n^2+1}{2}=\frac{2n^2}{2}=n^2$

$S_1+S_2+S_3+{...+S}_P=\frac{p}{2}[n^2+n+(p-1\left)n^2\right]=\frac{p}{2}[n^2+n+n^{2}p-n^2]=\frac{np}{2}(np+1)$