Question

If $p,q,r$ are in A.P. and $p^2,q^2,r^2$ are in G.P, where $p<q<r$ and $p+q+r=\frac{3}{2},$ then the value of $p$ is

• A. $\frac{1}{2\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $\frac{1}{2}-\frac{1}{\sqrt{3}}$
• D. $\frac{1}{2}-\frac{1}{\sqrt{2}}$

Answer 1

The correct option is D $\frac{1}{2}-\frac{1}{\sqrt{2}}$

Given : $p,q,r$ are in A.P.
$\Rightarrow p+r=2q\cdots \left(1\right)$
$p^2,q^2,r^2$ are in G.P
$\Rightarrow q^4=p^2{r}^2\Rightarrow q^2=\pm pr\cdots \left(2\right)$

Now,
$p+q+r=\frac{3}{2}$
Using equation $\left(1\right)$, we get
$\Rightarrow q=\frac{1}{2}$

So,
$p+r=1,pr=\pm \frac{1}{4}$

When $p+r=1,pr=\frac{1}{4}$
$\Rightarrow 4p^2-4p+1=0\Rightarrow p=\frac{1}{2}=r$
This is not possible as $p<q<r$

When $p+r=1,pr=-\frac{1}{4}$
$\Rightarrow 4p^2-4p-1=0\Rightarrow p=\frac{4\pm \sqrt{32}}{8}$
As $p<q<r$, so
$p=\frac{1}{2}-\frac{1}{\sqrt{2}},r=\frac{1}{2}+\frac{1}{\sqrt{2}}$