Question

The first three of four given numbers are in G.P. and last three are in A.P. whose common difference is $6$. If the first and last numbers are same, then first will be?

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is D $8$

Last $3$ of the $4$ numbers are in AP.

Let they are, $a-d,a,a+d$. Also the first number is same as $4th,a+d$.

Therefore the $4$ numbers are $a+d,a-d,a,a+d$

The first $3$ of these are in G.P.

$\therefore (a-d{)}^2=a(a+d)$ But $d=6$

$\therefore (a-6{)}^2=a(a+6)$

$\therefore a^2-2a+36=0$

Solving the above quadratic equation, we get,

$a=2$

Therefore the series is:

$8,-4,2,8$