Sequence And Series Question 105
Question: Let $ S=\frac{4}{19}+\frac{44}{19^{2}}+\frac{444}{19^{3}}+… $ up to $ \infty $ . Then S is equal to
Options:
A) 40/9
B) 38/81
C) 36/171
D) none of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ S=\frac{4}{19}+\frac{44}{19^{2}}+\frac{444}{19^{3}}+… $ …(1)
$ \Rightarrow \frac{1}{19}S=\frac{4}{19^{2}}+\frac{44}{19^{3}}+… $ …(2) Subtracting (2) from (1), we get $ \frac{18}{19}S=\frac{4}{19}+\frac{40}{19^{2}}+\frac{400}{19^{3}}+… $ $ =\frac{\frac{4}{19}}{1-\frac{10}{19}}=\frac{4}{9} $
$ \Rightarrow S=\frac{38}{81} $
BETA
