Sequence And Series Question 383
Question: If $ \frac{1}{\sqrt{b}+\sqrt{c}},\frac{1}{\sqrt{c}+\sqrt{a}},\frac{1}{\sqrt{a}+\sqrt{b}} $ are in A.P. then $ {9^{ax+1}},{9^{bx+1}},{9^{cx+1}},x\ne 0 $ are in:
Options:
A) G.P
B) G.P. only if x<0
C) G.P. only if x>0
D) none of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \frac{2}{\sqrt{c}+\sqrt{a}}=\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{a}+\sqrt{b}} $ $ =\frac{2\sqrt{b}+\sqrt{a}+\sqrt{c}}{(\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b})} $
$ \Rightarrow ,2\sqrt{ab}+2b+2\sqrt{ac}+2\sqrt{bc} $ $ =2\sqrt{bc}+2\sqrt{ac}+c+2\sqrt{ab}+a\Rightarrow 2b=a+c $
$ \therefore $ a, b, c, are in A.P.
$ \Rightarrow ,ax,bx,cx, $ are in A.P.
$ \Rightarrow ,ax+1,,bx+1,cx+1, $ are in A.P.
$ \Rightarrow {9^{ax+1}},{9^{bx+1}},{9^{cx+1}} $ are in GP [See the properties of A.P & G.P.]
BETA
