Sequence And Series Question 457
Question: If $ 3+3\alpha +3{{\alpha }^{2}}+………\infty =\frac{45}{8} $ , then the value of $ \alpha $ will be
[Pb. CET 1989]
Options:
A) 15/23
B) 7/15
C) 7/8
D) 15/7
Show Answer
Answer:
Correct Answer: B
Solution:
$ 3+3\alpha +3{{\alpha }^{2}}+3{{\alpha }^{3}}+…………..\infty =\frac{45}{8} $
$ \Rightarrow $ $ 3[ \frac{1}{1-\alpha } ]=\frac{45}{8} $
$ \Rightarrow $ $ 8=15(1-\alpha ) $
$ \Rightarrow $ $ \alpha =\frac{7}{15} $ .
BETA
