Question

In a set of four numbers, the first three are in G.P and the last three are in A.P with a common difference of 6. If the first number is the same as the fourth, find the sum of four numbers.

Answer 1

Let the last three numbers in $A.P$ be $a,a+6,a+12$ ($\because 6$ is the common difference)
If first number is $b$

Hence the four numbers are
$b,a,a+6,a+12$
But given
$b=a+12\cdots \left(1\right)$

and first three numbers are in $G.P.$
Then, $a^2=b(a+6)$
$\Rightarrow a^2=(a+12)(a+6)$ from $\left(1\right)$
$\Rightarrow a^2=a^2+18a+72$
$\Rightarrow 18a+72=0$
$\Rightarrow a+4=0$
$\therefore a=-4$

from $\left(1\right),b=-4+12=8$

Hence the four numbers are $8,-4,2,8$

$\therefore$ Sum of 4 numbers is $\Rightarrow 8-4+2+8=14$