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

• 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 $\frac{a}{c}+\frac{c}{a}$

Given,
$a,b,c$ are in A.P., $p,q,r$ are in H.P and $ap,bq,cr$ are in G.P
$\Rightarrow 2b=a+c,$ $q=\frac{2pr}{p+r},$ ${\left(bq\right)}^2=acpr$
Writing the second equation as $\frac{q}{r}=\frac{2x}{x+1}$ where $x$ is $\frac{p}{r}$
Now substituting $\frac{q}{r}$ in third equation and also using the first equation, we end up at a quadratic equation in $x$
$x^2-(\frac{a}{c}+\frac{c}{a})x+1=0$
So our required problem is to find $x+\frac{1}{x}$
i.e nothing but $\frac{x^2+1}{x}=\frac{a}{c}+\frac{c}{a}$