Question

If $a,b,c$ are in AP, $p,q,r$ are in HP, and $ap,bq,cr$ are in GP, then $\frac{p}{q}+\frac{r}{p}$ is equal to

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

Answer 1

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

Given,

a,b,c are in A.P

p,q,r are in H.P

ap,bq, cr are in G.P

To find the value of $\frac{p}{q}+\frac{r}{p}$

We know that if a,b,c are in A.P then

$2b=a+c$ (1)

We know that p,q,r are in HP then

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

We know that ap,bq,cr are in G.p then

$(bq{)}^2=(ap\left)\right(cr)$

$b^2{q}^2=apcr$ (3)

Put the value of q from (2) to (3)

$\frac{4b^2{p}^2{r}^2}{{(p+r)}^2}=apcr$ (4)

Put value of b from (1) to (4)

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

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

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

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

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