Sequence And Series Question 583
Question: If $ \frac{1}{p+q},\ \frac{1}{r+p},\ \frac{1}{q+r} $ are in A.P., then
[RPET 1995]
Options:
A) $ p,\ ,q,\ r $ are in A.P.
B) $ p^{2},\ q^{2},\ r^{2} $ are in A.P.
C) $ \frac{1}{p},\ \frac{1}{q},\ \frac{1}{r} $ are in A.P.
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Since $ \frac{1}{p+q},\ \frac{1}{r+q} $ and $ \frac{1}{q+r} $ are in A.P.
$ \therefore $ $ \frac{1}{r+q}-\frac{1}{p+q}=\frac{1}{q+r}-\frac{1}{r+p} $
$ \Rightarrow $ $ \frac{p+q-r-p}{(r+p)(p+q)}=\frac{r+p-q-r}{(q+r)(r+p)} $
$ \Rightarrow $ $ \frac{q-r}{p+q}=\frac{p-q}{q+r} $ or $ q^{2}-r^{2}=p^{2}-q^{2} $
$ \therefore $ $ 2q^{2}=r^{2}+p^{2} $ Therefore $ p^{2},\ q^{2},\ r^{2} $ are in A.P.
BETA
