Find five consecutive even integers if the sum of the first and fifth is 2 less than 3 times the fourth.

2 a , 2 a + 2 , 2 a + 4 , 2 a + 6 , 2 a + 8 − f i v e co n sec u t i v e e v e n in t e g ers 2 a + 2 a + 8 + 2 = 3 ( 2 a + 6 ) 2 a + 2 a + 10 = 6 a + 18 4 a + 10 = 6 a + 18 ∣ S u b t r a c t 4 a 10 = 2 a + 18 ∣ S u b t r a c t 18 10 − 18 = 2 a − 8 = 2 a ∣ D i v i d e b y 2 a = − 4 N u mb ers : − 4 , − 2 , 0 , 2 , 4 ​

Answered by luana | 2024-06-10

The five consecutive even integers are -8, -6, -4, -2, and 0. The integers were found by setting up an equation based on the conditions provided. We defined the first integer and expressed the others in terms of a variable, then solved the resulting equation.
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Answered by luana | 2024-10-09