Sequence And Series Question 255
Question: If $ 2(y-a) $ is the H.M. between $ y-x $ and $ y-z $ , then $ x-a,\ y-a,\ z-a $ are in
Options:
A) A.P.
B) G.P.
C) H.P.
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ y-x,\ 2(y-a),\ (y-z) $ are in H.P.
$ \Rightarrow $ $ \frac{1}{y-x},\ \frac{1}{2(y-a)},\ \frac{1}{y-z} $ are in A.P.
$ \Rightarrow $ $ \frac{1}{2(y-a)}-\frac{1}{y-x}=\frac{1}{y-z}-\frac{1}{2(y-a)} $
$ \Rightarrow $ $ \frac{2a-y-x}{y-x}=\frac{y+z-2a}{y-z} $
$ \Rightarrow $
$ \therefore $
$ \Rightarrow $ $ \frac{x-a}{y-a}=\frac{y-a}{z-a} $ (Applying componendo and dividendo)
$ \Rightarrow $ $ x-a,\ y-a,\ z-a $ are in G.P.
BETA
