Question

If $\frac{1}{(p+q)},\frac{1}{(r+p)},\frac{1}{(q+r)}$ are in $A.P$, then

• A. $p,q,r$ are in $AP$
• B. $p^2,q^2,r^2$ are in $AP$
• C. $\frac{1}{p},\frac{1}{q},\frac{1}{r}$ are in $AP$
• D. None of these

Answer 1

The correct option is B

$p^2,q^2,r^2$ are in $AP$

Explanation for the correct option:

Given that $\frac{1}{(p+q)},\frac{1}{(r+p)},\frac{1}{(q+r)}$ are in $A.P$

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

$$\begin{gathered} \frac{2}{(r+p)}=\frac{(q+r+p+q)}{(p+q)(q+r)} \\ \frac{2}{(r+p)}=\frac{(p+2q+r)}{(p+q)(q+r)} \end{gathered}$$

Cross multiply

$\Rightarrow$ $2(p+q)(q+r)=(p+2q+r)(r+p)$

$\Rightarrow$ $2(pq+q^2+pr+qr)=pr+2qr+r^2+p^2+2pq+pr$

$\Rightarrow$ $2q^2=r^2+p^2$

So ${\mathit{p}}^{\mathbf{2}}\mathbf{,}{\mathit{q}}^{\mathbf{2}}\mathbf{,}{\mathit{r}}^{\mathbf{2}}$ are in $\mathit{A}\mathbf{.}\mathit{P}$

Hence, Option ‘B’ is Correct.