Question

If $x=1+a+a^2+....................$to infinity and $y=1+b+b^2+...................$to infinity, where a, b are proper fractions, then $1+ab+a^2{b}^2+.....$ to infinity is equal :

• 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

Answer: (A)

If $a,ar,a{r}^2,a{r}^3.........$ are in G.P., then
sum of infinite $GP=a+ar+.....+\infty =\frac{a}{1-r}$
where 'a' is the first term and 'r' is the common ratio of G.P.
Given $x=1+a+a^2+.....\infty$
This is a GP, with common ratio 'a'.
$\Rightarrow x=\frac{1}{1-a}\Rightarrow x-ax=1\Rightarrow a=\frac{x-1}{x}$
Again, $y=1+b+b^2+......\infty$ This is also a G.P., with common ratio 'b'.
$\Rightarrow y=\frac{1}{1-b}\Rightarrow b=\frac{y-1}{y}$
Now, consider $1+ab+a^2{b}^2+.....\infty$
which is again a GP with common ratio 'ab'.
$\therefore$ Sum $-\frac{1}{1-ab}=\frac{1}{1-\frac{x-1}{x}.\frac{y-1}{y}}$
$=\frac{xy}{xy-xy+x+y-1}=\frac{xy}{x+y-1}$