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

The first three of four given numbers are in G.P. and their last three are in A.P. with common difference 6. If first and fourth numbers are equal, then the first number is

(a) 2

(b) 4

(c) 6

(d) 8

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 $\mathrm{p}=\mathrm{q}+12$ :

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