Question

The sum of the four numbers which are in arithmetic progression is 20 and product is 625. Find the numbers?

Answer 1

let us take the the four values as a-2d,a-d,a+d,a+2d
Sum of these numbers is
a-2d+a-d+a+d+a+2d=20
we can find a values by cancelling d's so a=5
now product of these values is 
(a-2d)*(a-d)*(a+d)*(a+2d)=625
taking a common we get
a[(1-2d)*(1-d)*(1+d)*(1+2d)]=625
since a=5;
after computing we get
1-5d^2+4d^4=125
4d^4-5d^2=124
we can find the d value and on substituting that we can get the four values due to time i cant complete the problem but this is the concept.