What are the 4 concecutive even integers such that the squar of the second term is equal to the product of the first and last term

Answer 1

The even integers are:

a1 = 2 n

a2 = 2 ( n + 1 )

a3 = 2 ( n + 2 )

a4 = 2 ( n + 3 )

a2 ^ 2 = a1 * a4

[ 2 ( n + 1 ) ] ^ 2 = 2 n * 2 ( n + 3 )

4 ( n + 1 ) ^ 2 = 4 n * ( n + 3 ) Divide both sides by 4

( n + 1 ) ^ 2 = n * ( n + 3 )

n ^ 2 + 2 * n * 1 + 1 ^ 2 = n * n + n * 3

n ^ 2 + 2 n + 1 = n ^ 2 + 3 n Subtract n ^ 2 to both sides

n ^ 2 + 2 n + 1 - n ^ 2 = n ^ 2 + 3 n - n ^ 2

2 n + 1 = 3 n Subtract 2 n to both sides

2 n + 1 - 2 n = 3 n - 2 n

1 = n

n = 1

Solution:

a1 = 2 n = 2 * 1 = 2

a2 = 2 ( n + 1 ) = 2 ( 1 + 1 ) = 2 * 2 = 4

a3 = 2 ( n + 2 ) = 2 * ( 1 + 2 ) = 2 * 3 = 6

a4 = 2 ( n + 3 ) = 2 * ( 1 + 3 ) = 2 * 4 = 8

Proof :

a2 ^ 2 = a1 * a4

4 ^ 2 = 2 * 8

16 = 16

Answer 2

Remark :

a1 = 2 n

a2 = 2 ( n + 1 )

a3 = 2 ( n + 2 )

a4 = 2 ( n + 3 )

n is an integer.

Answer 3

suppose we just define n to be an even number
(that way I don't have to worry about the 2n idea)

then the 4 consecutive even number are
n , n+2, n+4, and n+6

(n+2)^2 = n(n+6)
n^2 + 4n + 4 = n^2 + 6n
4 = 2n
n = 2

so the 4 numbers are
2 , 4, 6, and 8