Sequence And Series Question 314
Question: If $ (y-x),2(y-a) $ and $ (y-z) $ are in H.P., then $ x-a, $ $ y-a, $ $ z-a $ are in
[RPET 2001]
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.
Þ $ \frac{1}{y-x},\frac{1}{2(y-a)},\frac{1}{y-z} $ are in A.P.
Þ $ \frac{1}{2(y-a)}-\frac{1}{(y-x)}=\frac{1}{y-z}-\frac{1}{2(y-a)} $
Þ $ \frac{y-x-2y+2a}{(y-x)}=\frac{2y-2a-y+z}{(y-a)-(z-a)} $
$ \Rightarrow \frac{-x-y+2a}{(y-x)}=\frac{y+z-2a}{(y-z)} $
$ \Rightarrow \frac{(x-a)+(y-a)}{(x-a)-(y-a)}=\frac{(y-a)+(z-a)}{(y-a)-(z-a)} $
Þ $ \frac{(x-a)}{(y-a)}=\frac{(y-a)}{(z-a)} $ $ (x-a),(y-a),(z-a) $ are in G.P.
BETA
