Find the sum of the following arithmetic progressions:$ \frac{x-y}{x+y}, \frac{3 x-2 y}{x+y}, \frac{5 x-3 y}{x+y}, \ldots $ to $ n $ terms

Given:

Given A.P. is \( \frac{x-y}{x+y}, \frac{3 x-2 y}{x+y}, \frac{5 x-3 y}{x+y}, \ldots \)

To do:

We have to find the sum of the given A.P. to $n$ terms.

Solution:

Here,

$a=\frac{x-y}{x+y}, d=\frac{3 x-2 y}{x+y}-\frac{x-y}{x+y}=\frac{3 x-2 y-x+y}{x+y}$ 

\( \Rightarrow d=\frac{2 x-y}{x+y} \)

\( \therefore \mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d] \)

\( =\frac{n}{2}\left[2 \times \frac{x-y}{x+y}+(n-1)\left(\frac{2 x-y}{x+y}\right)\right] \)

\( =\frac{n}{2}\left[\frac{2(x-y)}{x+y}+\frac{(n-1)(2 x-y)}{x+y}\right] \)

\( =\frac{n}{2(x+y)}[2(x-y)+(n-1)(2 x-y)] \)

\( =\frac{n}{2(x+y)}[2 x-2 y+2 n x-n y-2 x+y] \)

\( =\frac{n}{2(x+y)}[2 n x-n y-y] \)

\( =\frac{n}{2(x+y)}[n(2 x-y)-y] \)

The sum of the given A.P. to $n$ terms is $\frac{n}{2(x+y)}[n(2x-y)-y]$.