Question

find the three numbers in arithmetic progression whose sum and product are 6 and 6 respectively​

Answer 1

let three numbers be

a-b, a , a+b

sum is 6

so 3a = 6

a= 2.

now product is 6

a(a²-b²) = 6

a²-b²= 3

4-b²= 3

b= +/- 1

so numbers are 1,2,3.