Sequence And Series Question 323
Question: The G.M. of the numbers $ 3,,3^{2},,3^{3},….,,3^{n} $ is
[DCE 2002]
Options:
A) $ {3^{\frac{2}{n}}} $
B) $ {3^{\frac{n+1}{2}}} $
C) $ {3^{\frac{n}{2}}} $
D) $ {3^{\frac{n-1}{2}}} $
Show Answer
Answer:
Correct Answer: B
Solution:
a = 3, r = 3 G.M. = $ {{({{3.3}^{2}}{{.3}^{3}}{{…..3}^{n}})}^{1/n}} $ $ ={{({3^{1+2+3……….+n}})}^{1/n}} $ $ ={{( {3^{\frac{n(n+1)}{2}}} )}^{1/n}}={3^{\frac{(n+1)}{2}}} $ .
BETA
