Sequence And Series Question 92
Question: If a, b, and c are in A.P., p, q, and r are in H.P., and ap, bq, and cr are in G.P., then $ \frac{p}{r}+\frac{r}{p} $ is equal to
Options:
A) $ \frac{a}{c}-\frac{c}{a} $
B) $ \frac{a}{c}+\frac{c}{a} $
C) $ \frac{b}{q}+\frac{q}{b} $
D) $ \frac{b}{q}-\frac{q}{b} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \frac{p}{r}+\frac{r}{p}=\frac{p^{2}+r^{2}}{pr}=\frac{{{(p+r)}^{2}}-2pr}{pr} $ $ =\frac{\frac{4p^{2}r^{2}}{q^{2}}-2pr}{pr} $ $ [ \begin{aligned} & \because p,q, r are,in,H.P. \\ & \therefore q=\frac{2pr}{p+r} \\ \end{aligned} ] $ $ =\frac{4pr}{q^{2}}-2=\frac{4b^{2}}{ac}-2 $ $ [\because ap,bq,cr,are,in,A.P\Rightarrow b^{2}q^{2}=acpr] $ $ =\frac{{{(a+c)}^{2}}}{ac}-2 $ $ [a,b,c,are,in,A.P\Rightarrow 2b=a+c] $ $ =\frac{a}{c}+\frac{c}{a} $
BETA
