Question

Sum of 3 numbers is 24 and their product is 440 in an A.P. Find the numbers.

Answer 1

Let the numbers be

$a-d,a,a+d$

Sum = $(a-d)+a+(a+d)=24$

$3a=24\Rightarrow a=8$

product = $(a-d)a(a+d)=440$

$a(a^2-d^2)=440$

$8(64-d^2)=440$

$64-d^2=55$

$d^2=9$ Number

$d=3$ $5,8,11$