Question

$x=1+a+a^2+...\infty (a<1)$
$y=1+b+b^2+...\infty (b<1)$
Then the value of $1+ab+a^2{b}^2+.....\infty$ is
[MNR 1980; MP PET 1985]

• A. $\frac{xy}{x+y-1}$
• B. $\frac{xy}{x+y+1}$
• C. $\frac{xy}{x-y-1}$
• D. $\frac{xy}{x-y+1}$

Answer 1

The correct option is A $\frac{xy}{x+y-1}$

Since the series are G.P., therefore
$x=\frac{1}{1-a}\Rightarrow a=\frac{x-1}{x}andy=\frac{1}{1-b}\Rightarrow b=\frac{y-1}{y}$
$\therefore 1+ab+a^2{b}^2+......\infty =\frac{1}{1-ab}$
$=\frac{1}{1-\frac{x-1}{x}.\frac{y-1}{y}}=\frac{xy}{x+y-1}.$