Sequence-And-Series Question 657
Question: Let a, b, c, be in A.P. with a common difference d. Then $ {e^{1/c}},{e^{b/ac}},{e^{1/a}} $ are in:
Options:
A) G.P. with common ratio $ e^{d} $
B) G.P. with common ratio $ {e^{1/d}} $
C) G.P. with common ratio $ {e^{d/(b^{2}-d^{2})}} $
D) A.P.
Show Answer
Answer:
Correct Answer: C
Solution:
[c] a, b, c are in A.P.
$ \Rightarrow 2b=a+c $ Now, $ {e^{1/c}}\times {e^{1/a}}={e^{(a+c)/ac}}={e^{2b/ac}}={{({e^{b/ac}})}^{2}} $
$ \therefore ,{e^{1/c}},{e^{b/ac}},{e^{1/a}} $ in GP. with common ratio $ =\frac{{e^{b/ac}}}{{e^{1/c}}}={e^{(b-a)/ac}}={e^{d/(b-d)(b+d)}} $ $ ={e^{d/(b^{2}-d^{2})}} $ [ $ \because $ a, b, c are in A.P. with common difference d
$ \therefore b-a=c-b=d $ ]
BETA
