Sequence And Series Question 220
Question: Given $ a^{x}=b^{y}=c^{z}=d^{u} $ and $ a,\ b,\ c,\ d $ are in G.P., then $ x,y,z,u $ are in
[ISM Dhanbad 1972; Roorkee 1984; RPET 2001]
Options:
A) A.P.
B) G.P.
C) H.P.
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ a^{x}=b^{y}=c^{z}=d^{u}=k $ (say) Then $ a={k^{1/x}},\ b={k^{1/y}},\ c={k^{1/z}},\ d={k^{1/u}} $ Since $ a,\ b,\ c $ are in G.P., therefore $ b^{2}=ac\Rightarrow {k^{2/y}}={k^{1/x}}.{k^{1/z}}={k^{1/x+1/z}} $
$ \Rightarrow \frac{2}{y}=\frac{1}{x}+\frac{1}{z}\Rightarrow \frac{1}{x},\ \frac{1}{y},\ \frac{1}{z} $ are in A.P.
$ \Rightarrow $ $ x,\ y,\ z $ are in H.P. Similarly it can be shown that $ y,\ z,\ u $ are also in H.P.
$ \therefore $ $ x,\ y,\ z $ and $ u $ are in H.P.
BETA
