Can someone please help me solve this?

3 2 and 1.... 3+2+1= 6 and 3 x 2= 6 lol i think i did it wrong

Ahhh editt its 3, 4 and 5....

Hello Andy, let the three consecutive integers be k-1, k and k+1

Their sum is 3k

Product of the first two = k^2 - k

Given k^2 - k = 3k

==> k^2 -4k = 0

===> k(k-4) = 0

Hence k=0 or k=4

So the three integers could be -1, 0, +1 OR 3,4,5

You can verify yourself.

n,(n+1),(n+2)

n+(n+1)+(n+2) = n*(n+1)

3n+3 = n^2+n

n^2-2n-3 = 0

(n-3)(n+1) = 0

n = 3 or -1

n must be positive then n = 3

sequence is 3,4,5

let the nos are n-1 , n , n+1 the from question 3n= n*(n-1) or 3*n=n^2-n or n^2-4*n=0 or n=0 or n=4so the answer is -1,0,1 or 3,4,5

hope this helps