Question

Question 21 (iii)
Find the sum:
$\frac{a-b}{a+b}+\frac{3a-2b}{a+b}+\frac{5a-3b}{a+b}+\cdots to11terms$

Answer 1

Here, first term, $a=\frac{a-b}{a+b}$
Common difference, $d=\frac{3a-2b}{a+b}-\frac{a-b}{a+b}=\frac{2a-b}{a+b}$
Sum of n terms of an AP, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$\Rightarrow S_n=\frac{n}{2}\{2\frac{(a-b)}{(a+b)}+(n-1\left)\frac{(2a-b)}{(a+b)}\right\}$
$=\frac{n}{2}\left\{\frac{2a-2b+2an-2a-bn+b}{a+b}\right\}$
$=\frac{n}{2}\left(\frac{2an-bn-b}{a+b}\right)$
$\therefore S_{11}=\frac{11}{2}\left\{\frac{2a\left(11\right)-b\left(11\right)-b}{a+b}\right\}$
$=\frac{11}{2}\left(\frac{22a-12b}{a+b}\right)=\frac{11(11a-6b)}{a+b}$