Ifx -1- a - a 2- - - to - a -1- y -1- b - b 2- to - b -1- then Z -1- ab - a 2 b 2- a 3 b 3- to - to isA- x-y-1-x yB- 2 x y-x-y1-1C- x y-x-y-1D- x-y-1-2 x y

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Sum of N Terms of an AP

Ifx =1+ a + a...

Question

If $x = 1 + a + a^{2} . . . . . . . . . . t o \infty$ $(| a | < 1), y = 1 + b + b^{2}$ ......... to $\infty$ (|b| < 1), then $Z = 1 + a b + a^{2}$ $b^{2}$ + $a^{3}$ $b^{3} . . . . . t o \infty$ to is

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C

We want to find Z in this problem. It is going to be in terms of 1 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)

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

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

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

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

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

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

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

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

Suggest Corrections

Similar questions

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

Q. If

x = 1 + a + a^{2} + . . . . \infty

and

y = 1 + b + b^{2} + . . . . \infty,

then the value of

1 + a b + + a^{2} b^{2} + . . . . \infty

(0 < a < 1, 0 < b < 1)

Q. If

x = 1 + a + a^{2} + . . . \infty, (| a | < 1), y = 1 + b + b^{2} + . . . \infty, (| b | < 1),

then

z = 1 + a b + a^{2} b^{2} + a^{3} b^{3} + . . . \infty

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 ∞ is

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

Join BYJU'S Learning Program

Ifx = 1+a+a2..........to ∞(|a|<1),y=1+b+b2 ......... to ∞ (|b| < 1), then Z = 1+ab+a2 b2 + a3 b3.....to ∞ to is

If $x = 1 + a + a^{2} . . . . . . . . . . t o \infty$ $(| a | < 1), y = 1 + b + b^{2}$ ......... to $\infty$ (|b| < 1), then $Z = 1 + a b + a^{2}$ $b^{2}$ + $a^{3}$ $b^{3} . . . . . t o \infty$ to is