Question

Given a sequence of $4$ members, first three of which are in G.P. and the last three are in A.P. with common difference six. If first and last terms of this sequence are equal, then the last term is:

• A. $8$
• B. $16$
• C. $2$
• D. $4$

Answer 1

The correct option is A $8$

Let for terms be $a,ar,a{r}^2,a$

$\because ar,a{r}^2$ and a are in A.P

$\therefore a{r}^2-ar=6$

$\Rightarrow ar(r-1)=6$

And $a-ar=2\times 6$

$\Rightarrow a(r-1)=-12$

$\Rightarrow \frac{ar(r-1)}{a(r-1)}=\frac{-6}{12}$

$\Rightarrow r=-\frac{1}{2}(\because r\ne 1)$

$\therefore a(1-r)=12$

$\Rightarrow a(1+\frac{1}{2})=12$

$\Rightarrow \frac{39}{2}=12$

$\Rightarrow a=8$

$\therefore$ Last term = $8$