7 consecutive integers are written the sum of the smallest 3 is 33 what is the sum of the largest three
Answers
First off, 33/3 =11, or: 11+11+11=33
since we were asked for consecutive integers all we have to do is substract 1 from one of the 11s and add 1 to another:
(11–1)+11+(11+1)=33
10+11+12=33
Answer:
Sum of the largest three is 45.
Step-by-step explanation:
Let the 7 consecutive integers are x , x + 1 , x + 2 , x + 3 , x + 4 , x + 5 , x + 6
from this x , x + 1 & x + 2 are the smallest 3 integers
And x + 4 , x + 5 , x + 6 are largest three integers
According to the question,
x + ( x + 1 ) + ( x + 2 ) = 33
x + x + 1 + x + 2 = 33
3x + 3 = 33
3x = 33 - 3
3x = 30
x = 10
Sum of Largest 3 = ( x + 4 ) + ( x + 5 ) + ( x + 6 )
= ( 10 + 4 ) + ( 10 + 5 ) + ( 10 + 6 )
= 14 + 15 + 16
= 45
Therefore, Sum of the largest three is 45.