Question

If $\frac{1}{q+r}$, $\frac{1}{r+p}$ and $\frac{1}{p+q}$ are in AP, then $p^2$, $q^2$ and $r^2$ are in

• A. AP
• B. GP
• C. HP
• D. AGP

Answer 1

The correct option is A AP

Since the given terms are in AP, the common difference will be same.
$=>SecondTerm-FirstTerm=ThirdTerm-SecondTerm$
$=>\frac{1}{r+p}-\frac{1}{q+r}=\frac{1}{p+q}-\frac{1}{r+p}$
$=>\frac{q+r-r-p}{(r+p)(q+r)}=\frac{r+p-p-q}{(p+q)(r+p)}$
$=>\frac{q-p}{(q+r)}=\frac{r-q}{(p+q)}$
$=>(q-p)(p+q)=(q+r)(r-q)$
$=>q^2-p^2=r^2-q^2$
Hence, we can see that $p^2,q^2,r^2$ are in AP.