Question

If x = 1 + a + $a^2$ .......... to $\infty$ (|a|<1), y = 1 + b + $b^2$ ......... to $\infty$(|b| < 1),

then Z = 1 + ab + $a^2$ $b^2$ + $a^3$ $b^3$..... to $\infty$ is

• A. x + y - 1 / xy
• B. 2xy / x + y - 1
• C. xy / x+y-1
• D. x+y-1 / 2xy

Answer 1

The correct option is C

xy / x+y-1

We want to find Z in this problem. It is going to be in terms of a and b.

But the options are in terms of x and y.

We can find a and b in terms of x and y from the first two infinite series.

x = $\frac{1}{1-a}$ -------------(1)

y = $\frac{1}{1-b}$ --------------(2)

z = $\frac{1}{1-ab}$ ----------(3)

From (1), 1 - a = $\frac{1}{x}$

a = 1 - $\frac{1}{x}$

= $\frac{x-1}{x}$

Similarly b = $\frac{y-1}{y}$

∴ z = $\frac{1}{1-\frac{(x-1)}{x}\frac{(y-1)}{y}}$

= $\frac{xy}{xy-(xy-x-y+1)}$

=$\frac{xy}{x+y-1}$