Statistics And Probability Question 205
Question: The G.M. of the numbers $ 3,,3^{2},,3^{3},,……,,3^{n} $ is
[Pb. CET 1997]
Options:
A) $ {3^{2/n}} $
B) $ {3^{(n-1)/2}} $
C) $ {3^{n/2}} $
D) $ {3^{(n+1)/2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
G.M. $ ={{({{3.3}^{2}}{{.3}^{3}}{{…….3}^{n}})}^{1/n}} $ $ ={{({3^{1+2+…….n}})}^{1/n}} $ $ ={{( {3^{\frac{n,(n+1)}{2}}} )}^{1/n}} $ $ ={3^{\frac{n+1}{2}}} $ .
BETA
