If x - 1 - a - a2 - to - -a-1- y - 1 - b - b2 - to - -b- - 1- then Z - 1 - ab - a2 b2 - a3 b3- to - is

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Sum of Infinite Terms of a GP

If x = 1 + a ...

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 ∞ is

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

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$\frac{2 x y}{x + y - 1}$

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

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$\frac{x + y - 1}{2 x y}$

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Solution

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

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

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

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

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Similar questions

Q. If

x = 1 + a + a^{2} + a^{3} + . . . .

\infty (| a | < 1)

and

y = 1 + b + b^{2} + b^{3} + . . .

\infty (| b | < 1)

then

1 + a b + a^{2} b^{2} + a^{3} b^{3} + . . .

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

Q. If

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

and

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

; prove that

1 + a b + a^{2} b^{2} + . . . . \infty = \frac{x y}{x + y - 1}

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

Q. If

x = 1 + a + a^{2} + a^{3} + . . .

and

y = 1 + b + b^{2} + b^{3} + . . . .

, then show that

1 + a b + a^{2} b^{2} + a^{3} b^{3} + . . . = \frac{x y}{x + y - 1}

If a, b, c are in G.P., prove that:

(i) $a (b^{2} + c^{2}) = c (a^{2} + b^{2})$

(ii) $a^{2} b^{2} c^{2} (\frac{1}{a^{3}} + \frac{1}{b^{3}} + \frac{1}{c^{3}}) = a^{3} + b^{3} + c^{3}$

(iii) $\frac{(a + b + c)^{2}}{a^{2} + b^{2} + c^{2}} = \frac{a + b + c}{a - b + c}$

(iv) $\frac{1}{a^{2} - b^{2}} + \frac{1}{b^{2}} = \frac{1}{b^{2} - c^{2}}$

(v) $(a + 2 b + 2 c) (a - 2 b + 2 c) = a^{2} + 4 c^{2}$ .

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

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