Sequence And Series Question 413
Question: lf $ x=1+a+a^{2}+……………….. $ to infinity and $ y=1+b+b^{2}+……………….. $ to infinity, where a, b are proper fractions, then $ 1+ab+a^{2}b^{2}+….. $ to infinity is equal:
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:
[a] If a, $ ar,,ar^{2},ar^{3}………. $ are in GP, then sum of infinite GP. $ =a+ar+…..+\infty =\frac{a}{1-r} $ where ‘a’ is the first term and V is the common ratio of GP. Given $ x=1+a+a^{2}+….\infty $ This is a GP, with common ratio ‘a’.
$ \Rightarrow x=\frac{1}{1-a}\Rightarrow x-ax=1\Rightarrow a=\frac{x-1}{x} $ Again, $ y=1+b+b^{2}+….\infty $ This is also a GP, with common ratio ‘b?.
$ \Rightarrow ,y=\frac{1}{1-b}\Rightarrow b=\frac{y-1}{y} $ Now, consider $ 1+ab+a^{2}b^{2}+….\infty $ which is again a GP with common ratio ‘ab?.
$ \therefore $ Sum $ =\frac{1}{1-ab}=\frac{1}{1-\frac{x-1}{x}.\frac{y-1}{y}} $ $ =\frac{xy}{xy-xy+x+y-1}=\frac{xy}{x+y-1} $
BETA
