Question
If $x=a+\frac{a}{r}+\frac{a}{r^2}+\ldots+\infty, y=b-\frac{b}{r}+\frac{b}{r^2}-\ldots+\infty$ and $z=c+\frac{c}{r^2}+\frac{c}{r^4}+\ldots+\infty$ Prove that $\frac{x . y}{z}=\frac{a b}{c}$

Answer

Given series $x=a+\frac{a}{r}+\frac{a}{r^2}+\ldots+\infty$ is in G.P. with common ratio $\frac{1}{r}$.
and series $y=b-\frac{b}{r}+\frac{b}{r^2}-\ldots+\infty$ is in G.P. with common ratio $-\frac{1}{r}$.
and series $z=c+\frac{c}{r^2}+\frac{c}{r^4}+\ldots+\infty$ is in G.P. with common ratio $\frac{1}{r^2}$
Sum of infinite terms of series$
x=a+\frac{a}{r}+\frac{a}{r^2}+\ldots+\infty \text { is } x=\frac{a}{1-\frac{1}{r}}
$
Sum of infinite term of series $y=b-\frac{b}{r}+\frac{b}{r^2}-\ldots+\infty$ is
$
y=\frac{b}{1-\left(-\frac{1}{r}\right)}=\frac{b}{1+\frac{1}{r}}
$
Sum of infinite terms of series $z=c+\frac{c}{r^2}+\frac{c}{r^4}+\ldots+\infty$ is
$
z=\frac{c}{1-\frac{1}{r^2}}
$
Now, LHS $=\frac{x y}{z}$
$=\frac{\left\{\frac{a}{\left(1-\frac{1}{r}\right)}\right\}\left\{\frac{b}{\left(\left(1+\frac{1}{r}\right)\right.}\right\}}{\frac{c}{1-\frac{1}{r^2}}}$
$\begin{array}{l}=\frac{\frac{a b}{\left(1-\frac{1}{r^2}\right)}}{\frac{c}{\left(1-\frac{1}{r^2}\right)}} \\ =\frac{a b}{c}=\text { R.H.S }\end{array}$
Hence Proved

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