Sequence And Series Question 501
Question: $ x=1+a+a^{2}+….\infty ,(a<1) $ $ y=1+b+b^{2}…….\infty ,(b<1) $ Then the value of $ 1+ab+a^{2}b^{2}+……….\infty $ is
[MNR 1980; MP PET 1985]
Options:
A) $ \frac{xy}{x+y-1} $
B) $ \frac{xy}{x+y+1} $
C) $ \frac{xy}{x-y-1} $
D) $ \frac{xy}{x-y+1} $
Show Answer
Answer:
Correct Answer: A
Solution:
Since the series are G.P., therefore $ x=\frac{1}{1-a}\Rightarrow a=\frac{x-1}{x} $ and $ y=\frac{1}{1-b}\Rightarrow b=\frac{y-1}{y} $
$ \therefore $ $ 1+ab+a^{2}b^{2}+……….\infty =\frac{1}{1-ab} $ $ =\frac{1}{1-\frac{x-1}{x}.\frac{y-1}{y}}=\frac{xy}{x+y-1} $ .
BETA
