The first three of four given numbers are in G.P. and their last three are in A.P.

Solution:

(d) 8

The first and the last numbers are equal.

Let the four given numbers be $p, q, r$ and $p$.

The first three of four given numbers are in G.P.

$\therefore \mathrm{q}^{2}=\mathrm{p} \cdot \mathrm{r}$   ...(i)

And, the last three numbers are in A.P. with common difference $6 .$

We have :

First term $=\mathrm{q}$

Second term $=\mathrm{r}=\mathrm{q}+6$

Third term $=\mathrm{p}=\mathrm{q}+12$

Also, $2 \mathrm{r}=\mathrm{q}+\mathrm{p}$

Now, putting the values of $\mathrm{p}$ and $\mathrm{r}$ in (i):

$\mathrm{q}^{2}=(\mathrm{q}+12)(\mathrm{q}+6)$

$\Rightarrow \mathrm{q}^{2}=\mathrm{q}^{2}+18 \mathrm{q}+72$

$\Rightarrow 18 \mathrm{q}+72=0$

$\Rightarrow \mathrm{q}+4=0$

$\Rightarrow \mathrm{q}=-4$

Now, putting the value of $\mathrm{q}$ in $\mathrm{p}=\mathrm{q}+12$ :

$\mathrm{p}=-4+12=8$