Can someone please help me solve this?
Find three consecutive integers whose sum is equal to the product of the first two.
Find three consecutive integers whose sum is equal to the product of the first two.
Anonymous
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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....
trueprober
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.
Anonymous
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