Find the sum:
$ \frac{a-b}{a+b}+\frac{3 a-2 b}{a+b}+\frac{5 a-3 b}{a+b}+\ldots $ to 11 terms.

Given:

\( \frac{a-b}{a+b}+\frac{3 a-2 b}{a+b}+\frac{5 a-3 b}{a+b}+\ldots \) 

To do:

We have to find the sum of the given sequence to 11 terms.

Solution:

In the given sequence,

First term $a_1=\frac{a-b}{a+b}$

Common difference $d=\frac{3 a-2 b}{a+b}-\frac{a-b}{a+b}$

$=\frac{2 a-b}{a+b}$

Sum of $n$ terms of an AP $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{n} =\frac{n}{2}[2 \frac{(a-b)}{(a+b)}+(n-1) \frac{(2 a-b)}{(a+b)}]$

$=\frac{n}{2}[\frac{2 a-2 b+2 a n-2 a-b n+b}{a+b}]$

$=\frac{n}{2}(\frac{2 a n-b n-b}{a+b})$

$S_{11}=\frac{11}{2}[\frac{2 a(11)-b(11)-b}{a+b}]$

$=\frac{11}{2}(\frac{22 a-12 b}{a+b})$

$=\frac{11(11 a-6 b)}{a+b}$

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Updated on: 10-Oct-2022

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