Question

Find four consecutive even integers such that seven times the first exceeds their sum by 18.

I can’t find the answer, I keep on getting 3. :(

Answer 1

10, 12, 14, 16

Explanation:

Given: Four consecutive even integers such that seven times the first exceeds their sum by 18.

Lets assume the first number be "x".

As it is even integers

∴ Second consecutive number will be
`(x+2)`

Third consecutive number will be
`(x+4)`

Fourth consecutive number will be
`(x+6)`

Now, as given seven times the first exceeds their sum by 18.

∴
`[x+(x+2)+(x+4)+(x+6)]+18 = 7x`

solving the equation to find the number.

⇒
`[x+(x+2)+(x+4)+(x+6)]+18 = 7x`

Opening parenthesis

`x+x+2+x+4+x+6+18 = 7x`

⇒
`4x+30= 7x`

subtracting both side by 4x

⇒
`30= 3x`

Dividing both side by 3

∴ x= 10.

Hence, subtituting the value x to find four consecutive even integers.

First number is 10

Second consecutive number will be
`(10+2)` = 12

Third consecutive number will be
`(10+4)`= 14

Fourth consecutive number will be
`(10+6)` = 16

∴ Four consecutive even integers are 10,12,14,16.