Question

If $a,b,c$ are in $A.P.,p,q,r$ are in H.P. and $ap,bq,cr$ and in G.P., then $\frac{p}{r}+\frac{r}{p}$ is equal to

• A. $\frac{a}{c}+\frac{c}{a}$
• B. $\frac{a}{c}-\frac{c}{a}$
• C. $\frac{b}{q}+\frac{q}{b}$
• D. $\frac{b}{q}-\frac{a}{p}$

Answer 1

The correct option is A $\frac{a}{c}+\frac{c}{a}$

$\frac{p}{r}+\frac{r}{p}=\frac{p^2+q^2}{pr}$

$p,q,r$ are in H.P $\Rightarrow \frac{2}{q}=\frac{p+r}{pr}\Rightarrow \frac{4}{q^2}=\frac{p^2+r^2+2pr}{p^2{r}^2}$

$\frac{4pr}{q^2}-2=\frac{p^2+r^2}{p^2{r}^2}\dots \dots \left(1\right)$

$ap,bq,cr$ are in G.P.$\Rightarrow \frac{b^2}{ac}=\frac{pr}{q^2}\dots \dots \left(2\right)$

From (1) and (2)

$\frac{p}{r}+\frac{r}{p}=\frac{4b^2-2ac}{ac}=\frac{a}{c}+\frac{c}{a}(\because 2b=a+c)$