Question

If $S_1,S_2,S_3,$________$S_m$ are the sums of n terms of m A.P.'s whose first terms are $1,2,3$,___m and common differences are $1,3,5$_______, $2m-1$ respectively. Show that $S_1+S_2+S_3+$________$+S_m=\frac{1}{2}mn(mn+1)$.

Answer 1

Here $a=1,2,3$, m respectively
and $d=1,3,5$, . $2m-1$, and $n=n$
To find $S_1+S_2+S_3+...+S_m$
$S_1=\frac{n}{2}[2.1+(n-1\left).1\right]$
$S_2=\left(n/2\right)[2.2+(n-1\left).3\right]$
$S_m=\left(n/2\right)[2.m+(n-1\left)\right(2m-1\left)\right]$
$\therefore S_1+S_2+S_3+...+S_n$ $=n(1+2+3+...m)+\frac{n(n-1)}{2}[1+3+5+...+(2m-1\left)\right]$
$=n\cdot \frac{m(m+1)}{2}+\frac{n(n-1)}{2}\cdot \frac{m}{2}[1+2m-1]$, using $S=\frac{n}{2}[a+l]$
$=\frac{1}{2}mn[m+1+m(n-1\left)\right]=\frac{1}{2}mn(mn+1)$.