Sequence And Series Question 356
Question: The value of the infinite product $ {6^{\frac{1}{2}}}\times {6^{\frac{1}{2}}}\times {6^{\frac{3}{8}}}\times {6^{\frac{1}{4}}}…. $ is
Options:
A) 6
B) 36
C) 216
D) $ \infty $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ X={6^{( \frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{1}{4}+….. )}} $ $ ={6^{[ ( 1\times \frac{1}{2} )+( 2\times \frac{1}{4} )+( 3\times \frac{1}{8} )+( 4\times \frac{1}{16} )+…… ]}} $ $ \because $ It is arithmetic-geometric progression,
$ \therefore a=\frac{1}{2};,d=1 $ & $ r=\frac{1}{2} $
$ \Rightarrow X={6^{[ \frac{a}{1-r}+\frac{dr}{{{(1-r)}^{2}}} ]}}={6^{[ \frac{\frac{1}{2}}{1-\frac{1}{2}}+\frac{1\times \frac{1}{2}}{{{(1-\frac{1}{2})}^{2}}} ]}}=6^{3}=216 $
BETA
