Sequence And Series Question 620
Question: The sum of infinite terms of the geometric progression $ \frac{\sqrt{2}+1}{\sqrt{2}-1},\frac{1}{2-\sqrt{2}},\frac{1}{2}….. $ is
[Kerala (Engg.) 2002]
Options:
A) $ \sqrt{2}{{(\sqrt{2}+1)}^{2}} $
B) $ {{(\sqrt{2}+1)}^{2}} $
C) $ 5\sqrt{2} $
D) $ 3\sqrt{2}+\sqrt{5} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{\sqrt{2}+1}{\sqrt{2}-1},\frac{1}{\sqrt{2}(\sqrt{2}-1)},\frac{1}{2},…… $ Common ratio of the series $ =\frac{1}{\sqrt{2}(\sqrt{2}+1)} $
$ \therefore $ sum $ =\frac{a}{1-r}={( \frac{\sqrt{2}+1}{\sqrt{2}-1} )}/{( 1-\frac{1}{\sqrt{2}(\sqrt{2}+1)} )}; $ $ =\frac{(\sqrt{2}+1)}{(\sqrt{2}-1)}.,\frac{\sqrt{2},(\sqrt{2}+1)}{(1+\sqrt{2})} $ $ =\sqrt{2}{{(\sqrt{2}+1)}^{2}} $ .
BETA
