JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

Let

x=1+a+a^{2}+\cdots \infty(a < 1)

y=1+b+b^{2} \cdots \infty(b < 1)

Then the value of

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

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because, given the infinite GPs for x and y, the infinite GP

1 + ab + a^2b^2 + ...

simplifies to this expression.

More Questions from SEQUENCE OF SERIES

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A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}

{A} \cup({B} \cap {C})

A-(B \cap C)

(B \cup C)-A

a,b,c

a,x,b

b,y,c

{x^2},{b^2},{y^2}

P(x)=1-x+x^2-x^3+\ldots+x^{2012}

y

Q(y)=a_0+a_1 y+\ldots+a_{2012} y^{2012}

y=x-2

\sum\limits_{{i=0}}^{2012} a_i

\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)

\log \left(\frac{a}{3 b}\right)

A P

a, b

G P

a, b, c

\frac{3}{4}