Question

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

Answer 1

Answer:

10, 12, 14, 16

Step-by-step 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.