Question

In three consecutive even numbers, two times the first number is 2 more than the third number. Find the sum of these three numbers.
A) 24
B) 32
C) 18
D) 36

Answer 1

$\mathfrak{\huge{Answer:}}$

$\mathbb{WE'RE\:GIVEN\:THAT}$

The numbers are even and consecutive

Two times the first number is 2 more than the third one

$\mathbb{WE\:NEED\:TO\:FIND\:THAT}$

The sum of these numbers

$\mathbb{THE\:METHOD\:WILL\:BE}$

Let the numbers be = x , x + 2 and x + 4

According to the question, "Two times the first number is 2 more than the third one". This will algebraically mean :-

2x = x + 4 + 2

Solve this formed equation further

=》 2x = x + 6

Last one more step to go

=》 x = 6

The numbers will be = 6, 8, and 10

Sum of these numbers = 6 + 8 + 10 = 24

The answer will be $\sf{\huge{A) 24}}$

Answer 2

ANSWER:-------

Explanation:

{The smallest of the three consecutive integers:}

{Because these are consecutive even integers it means we can write the other two integers}

{ As:n+2n+4Now, we can write "three times the largest"}

{As:3(n+4)And if this is equal to "34 more than the sum of the two smaller integers" }

{ equation:3(n+4)=34+n+n+2We can now solve}

form:3(n+4)=34+1n+1n+23(n+4)

=1n+1n+2+343(n+4)=(1+1)n+363(n+4)

=2n+36(3⋅n)+(3⋅4)

=2n+363n+12

=2n+36−2n+3n+12−12

=−2n+2n+36−12(−2+3)n+0

=0+241n

=24n=24

{Therefore the three consecutive even integers}

hope it helps:------

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