If x-1-a- a 2 - a 3 - to - - a - -1 and y-1-b- b 2 - b 3 - to - - b - -1 then 1-ab- a 2 b 2 - a 3 b 3 - to - - xy x-y-1

The correct option is A True x = 1 1 a y = 1 1 b [summing infinite G.P's ]. ∴ a= x 1 x , b= y 1 y ∴ 1+ab+ a ^ 2 b ^ 2 +...∞ ...

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Byju's Answer

Standard VIII

Mathematics

Long Division

If x=1+a+ a...

Question

If $x = 1 + a + a^{2} + a^{3} + . . . .$ to $\infty (| a | < 1)$ and
$y = 1 + b + b^{2} + b^{3} + . . .$ to $\infty (| b | < 1)$ then
$1 + a b + a^{2} b^{2} + a^{3} b^{3} + . . .$ to $\infty = \frac{x y}{x + y - 1}$

True

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False

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Solution

The correct option is A True
$x = \frac{1}{1 - a}$ $y = \frac{1}{1 - b}$ [summing infinite $G . P^{'} s$ ].
$∴ a = \frac{x - 1}{x}$ , $b = \frac{y - 1}{y}$
$∴ 1 + a b + a^{2} b^{2} + . . . \infty$
$= \frac{1}{1 - a b} = \frac{1}{1 - \frac{(x - 1) (y - 1)}{x y}} = \frac{x y}{x + y - 1} .$

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Similar questions

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}

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 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

2 y - x = 8; 5 y - x = 14, y - 2 x = 1

intersect at

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(a_{1} + a_{2} + a_{3} + b_{1} + b_{2} + b_{3})

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+a3+.... to ∞(|a|<1) and y=1+b+b2+b3+... to ∞(|b|<1) then1+ab+a2b2+a3b3+... to ∞=xyx+y−1

The correct option is A Truex=11−a y=11−b [summing infinite G.P′s].∴a=x−1x, b=y−1y∴1+ab+a2b2+...∞=11−ab=11−(x−1)(y−1)xy=xyx+y−1.

If $x = 1 + a + a^{2} + a^{3} + . . . .$ to $\infty (| a | < 1)$ and
$y = 1 + b + b^{2} + b^{3} + . . .$ to $\infty (| b | < 1)$ then
$1 + a b + a^{2} b^{2} + a^{3} b^{3} + . . .$ to $\infty = \frac{x y}{x + y - 1}$

The correct option is A True
$x = \frac{1}{1 - a}$ $y = \frac{1}{1 - b}$ [summing infinite $G . P^{'} s$ ].
$∴ a = \frac{x - 1}{x}$ , $b = \frac{y - 1}{y}$
$∴ 1 + a b + a^{2} b^{2} + . . . \infty$
$= \frac{1}{1 - a b} = \frac{1}{1 - \frac{(x - 1) (y - 1)}{x y}} = \frac{x y}{x + y - 1} .$