Question

If $a,b,c$ are in $A.P$. $,p,q,r$ are in $H.P$. and $ap,bq,cr$ are 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. $\frac{b}{q}-\frac{q}{b}$

Answer 1

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

a,b,c are in A.P

p,q,r are in H.P

ap, bq, cr are in G.P

then $\frac{p}{r}+\frac{r}{p}$

$\Rightarrow \frac{p^2+r^2}{pr}$

$=\frac{(p^2+r^2)}{\left(pqr\right)}$

$[\frac{2}{q}=\frac{1}{p}+\frac{1}{r}]$ ___ (1)

$[\frac{bq}{ap}=\frac{cr}{bq}]$ ___ (2)

$\Rightarrow [p^2{q}^2=acpr]$

$\to$ from (1) $\frac{4}{q^2}=\frac{(p+r{)}^2}{(pr{)}^2}$ ___ (3)

Multiply $\left(3\right)\times \left(4\right)\Rightarrow$

$4b^2=\frac{ac(p+r{)}^2}{\left(pr\right)}$

$\frac{4b^2}{ac}=\frac{r}{p}+\frac{p}{r}+2$

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

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

$\Rightarrow \frac{a^2+c^2+2ac-2}{ac}=\frac{r}{p}+\frac{p}{r}$

$\frac{a}{c}+\frac{c}{a}+2-2=\frac{r}{p}+\frac{p}{r}$

$\therefore \frac{p}{r}+\frac{r}{p}\Rightarrow \frac{a}{c}+\frac{c}{a}$

Option (B) is correct