SOLUTION: Jared wanted to find three consecutive even integers whose sum was 4 times the first of those integers. He let k represent the first integer, then wrote and solved this equation: k

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Jared wanted to find three consecutive even integers whose sum was 4 times the first of those integers. He let k represent the first integer, then wrote and solved this equation: k      Log On


   



Question 933763: Jared wanted to find three consecutive even integers whose sum was 4 times the first of those integers. He let k represent the first integer, then wrote and solved this equation: k+ (k+1) + (k+2) = 4k. Did he get the correct answer? Complete the explanation.
Jared did not get the correct answer. The solution to his equation is k = ____, giving ___,___, and ___ as the three integers. However 3 and 5 are not even integers. He should have used the equation k + (k + ___) + (k+___) = 4k, which gives k =___ and the correct answer ___,___,___.
Help would be much appreciated

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Instead of k, k+1, k+2, you should use k, k+2, k+4 with the constraint that k is even (i.e. if you return an odd integer for k, or a non-integer, then return no solution).

k + (k+2) + (k+4) = 4k
3k + 6 = 4k
k = 6

Indeed, 6+8+10 = 4*6.