84.1k views
4 votes
Find four consecutive integers such that the sum of the first three is 44 more than the fourth.

1 Answer

4 votes

We will have the following:

We will have to sum 3 consecutive numbers, so we will name those numbers:

n

n+1

n+2

and the fourth number n+3

We can see that they are consecutive since the sepatation is of 1 unit each, we will also have that they must follow:


n+(n+1)+(n+2)=(n+3)+44

So, we solve for n:


\Rightarrow3n+3=n+47\Rightarrow2n=44
\Rightarrow n=22

So, the three consecutive numbers are: 22, 23, & 24:

And we prove this by adding them and then subctacting 44 nd we should get 25:


22+23+24=69-44=25

User David Sickmiller
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories