Question

3 consecutive numbers of a series are in A.P with d=2, smallest number?

Answer 1

the smallest number is 1

Answer 2

3 consecutive numbers: n, n+1, n+2

Their sum is 3n+3.

Their product is n(n+1)(n+2) = n^3 + 3n^2 + 2n

sum = product, therefore 3n+3 = n^3 + 3n^2 + 2n

Now rearrange that equation to n^3 + 3n^2 - n - 3 = 0

Factorise the cubic equation in order to solve it: (n + 3) (n + 1) (n - 1) = 0

Therefore, the three solutions are n= -3; n=-1; n=1