Question

Find four numbers in A.P. such that the sum of the first and last number is $8$ and the product of second and third number is $12$.

Answer 1

Let the numbers be $a-2d,a-d,a+d$ and $a+2d$

According to the question it is give that

$a-2d+a+2d=8$

$2a=8$

$a=4$

it is also given that,

$(a-d)(a+d)=12$

$a^2+ad-ad-d^2=12$

$a^2-d^2=12$

$a^2-12=d^2$

$4^2-12=d^2$

$16-12=d^2$

$d=\sqrt{4}$

$d=2$

$a-2d=4-4=0$

$a-d=4-2=2$

$a+d=4+2=6$

$a+2d=4+4=8$

Therefore the four numbers in A.P are $0,2,6,8$