If three numbers are in G.P., then the numbers obtained by adding the middle number to each of these numbers are in

Let the numbers be $a,ar$ and $a{r}^2\text{. }$ Then the 

number obtained by adding the middle number are

$a+ar,2ar,ar+a{r}^2$ 

Clearly, these numbers are neither in AP. nor in G.P. 

Now,

      $\frac{1}{2ar}-\mid \frac{1}{a+ar}=\frac{a-ar}{2ar(a+ar)}=\frac{1-r}{2ar(1+r)}$

and  $\frac{1}{ar+a{r}^2}-\frac{1}{2ar}=\frac{ar-a{r}^2}{2ar\left(ar+a{r}^2\right)}=\frac{1-r}{2ar(1+r)}$

We have,

      $\frac{1}{2ar}-\frac{1}{a+ar}=\frac{1}{ar+a{r}^2}-\frac{1}{2ar}$

So, $a+ar,2ar,ar+a{r}^2$ are in H.P.