Suppose four distinct positive numbers $a_1,a_2,a_3,a_4$

are in G.P 

Let $b_1=a_1,b_2=b_1+a_2,b_3=b_2+a_3$and 

$b_4=b_3+{b_4}_{ }$

Statement-1: The numbers $b_1,b_2,b_3,b_4$

are neither in A.P. nor in G.P.

Statement-2: The numbers  $b_1,b_2,b_3,b_4$

are in H.P

Let $a_1=a,a_2=ar,a_3=a{r}^2$ and $a_4=a{r}^3$

As $a_1,a_2,a_3,a_4$ are distinct, $r\ne 1$

$\begin{array}{l}{b}_1=a,b_2=a+ar=a(1+r),\\ b_3=a\left(1+r+r^2\right)\end{array}$

and $b_4=a\left(1+r+r^2+r^3\right)$

As $b_2-b_1=ar\ne a{r}^2=b_3-b_2$

$\Rightarrow b_1,b_2,b_3,b_4$ are not in A.P.

$\frac{b_2}{b_1}=1+r\ne \frac{1+r+r^2}{1+r}=\frac{b_3}{b_2}$

$\Rightarrow b_1,b_2,b_3,b_4$ are not in G.P

Next, $\frac{1}{b_2}-\frac{1}{b_1}=\frac{-r}{a(1+r)}$

$\frac{1}{b_3}-\frac{1}{b_2}=\frac{-r^2}{a(1+r)\left(1+r+r^2\right)}$

Since $\frac{1}{b_2}-\frac{1}{b_1}\ne \frac{1}{b_3}-\frac{1}{b_2}$we get 

$b_1,b_2,b_3,b_4$ $b_1,b_2,b_3,b_4$

cannot be in H.P.