Question

In a set of four numbers the first three are in GP and the last three are in AP with common difference $6$. If the first number is the same as the fourth then third number of this set is

• A. $-4$
• B. $2$
• C. $6$
• D. $8$

Answer 1

Answer: (B)

$2$

Let the last three numbers in A.P with common difference $6$ are:
$q,q+6,q+12$
Hence, the four numbers are:- $p,q,q+6,q+12$
Given, first and fourth number are equal.
$\therefore p=q+12$
and first three numbers $p,q,q+6$ are in G.P.

$\therefore q^2=p(q+6)$
$\Rightarrow q^2=(q+12)(q+6)$ [$\because p=q+12$]
$\Rightarrow q^2=q^2+18q+72$
$\Rightarrow 18q=-72$
$\Rightarrow q=-4$
So, Third number $=q+6=-4+6=2$
Option B is correct.