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Question

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


A

x + y - 1 / xy

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B

2xy / x + y - 1

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C

xy / x+y-1

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D

x+y-1 / 2xy

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Solution

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 = 11a -------------(1)

y = 11b --------------(2)

z = 11ab ----------(3)

From (1), 1 - a = 1x

a = 1 - 1x

= x1x

Similarly b = y1y

∴ z = 11(x1)x(y1)y

= xyxy(xyxy+1)

=xyx+y1


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