Question

If the sum of three numbers in A.P. is 24 and their product is 440, find the numbers.

Answer 1

Let 3 numbers in A.P. be
a - d, a and a + d
$\Rightarrow (a-d)+\left(a\right)+(a+d)=24\Rightarrow 3a=24\Rightarrow a=8$
and product =
$\Rightarrow (a-d)\left(a\right)(a+d)=440(a^2-d^2)a=440(8^2-d^2)\times 8=4408^2-d^2=55\Rightarrow d^2=9d=\pm 3$
With a = 8, d = 3, we have 5, 8, 11
With a = 8, d = -3, we have 11, 8, 5