Sequence And Series Question 49
Question: If $ a,\ b,\ c $ are in A.P., then $ \frac{{{(a-c)}^{2}}}{(b^{2}-ac)}= $
[Roorkee 1975]
Options:
A) 1
B) 2
C) 3
D) 4
Show Answer
Answer:
Correct Answer: D
Solution:
If $ a,\ b,\ c $ are in A.P.
$ \Rightarrow $ $ 2b=a+c $ So, $ \frac{{{(a-c)}^{2}}}{(b^{2}-ac)}=\frac{{{(a-c)}^{2}}}{{ {{( \frac{a+c}{2} )}^{2}}-ac }} $ $ =\frac{{{(a-c)}^{2}}4}{[a^{2}+c^{2}+2ac-4ac]}=\frac{4{{(a-c)}^{2}}}{{{(a-c)}^{2}}}=4 $ . Trick: Put $ a=1,\ b=2,\ c=3 $ , then the required value is $ \frac{4}{1}=4 $ .
BETA
