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? Explain

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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.

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erato erato · Answer 2 · 2019-09-10T06:31:51+02:00

Answer with Step-by-step explanation:

Let k be the first even integer

then, next even integer will be k+2

and the next consecutive even integer will be k+4

Sum of three consecutive even integers was 4 times the first of those integers.

i.e. k+(k+2)+(k+4)=4k

3k+6=4k

6=4k-3k

i.e. k=6

Jared wrote and solved the equation:

k+ (k+1) + (k+2) = 4k

i.e. 3k+3=4k

i.e. 4k-3k=3

i.e. k=3

No, he didn't get the correct answer

(since, he wrote 3 consecutive integers as k,k+1 and k+2 instead of k,k+2 and k+4)