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Question

six consecutive integers are given. the sum of the first three is 27. what is the sum of the last three?

sorry, ik people on here dont like to do homework for other people, but i really dont understand this and my mom hasnt done these in a looong time! sorry peoplee=[

is it 36? because 8,9,and 10 equal 27.. so 11,12,and13 would equal 36..?

Answer 1

Integers are numbers like 1, 6, 7, 42, etc.

Consecutive integers means the sequence doesn't skip over any integers. 5, 6, 7, 8, 9 would be a sequence of consecutive integers.

So, if you take the list of integers in consecutive order:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20

Which three consecutive (adjacent) integers add up to 27?
8 + 9 + 10 = 27

If 8, 9, 10 are the first three numbers in the sequence of six integers, the last three must be 11, 12, 13.

The sum of these is
11 + 12 + 13 = 36

Answer 2

integer 1 = x
integer 2 = x+1
integer 3 = x+2
integer 4 = x+3
integer 5 = x+4
integer 6 = x+5

x+(x+1)+(x+2)=27
3x+3=27
x+1=9
x=8


Since x equals 8, you can fine integer 4,5, and 6 by adding
3,4, and 5 to 8. You get 11,12, and 13.

Therefore the sum of the last three integers is 36.

Answer 3

x = first integer
x + 1 = second consecutive integer
x + 2 = third consecutive integer

sum means addiction

so (x) + (x + 1) + (x + 2) = 27
3x + 3 = 27
3x = 24
x = 8

so the six integers are:
8, 9, 10, 11, 12, and 13

the sum of the last 3 is:
(11 + 12 + 13) = 36 <== answer

hope it helps
Rec

Answer 4

n +n+1 + n +2 = 27
3 n + 3 = 27
3n = 24
n = 8 ,
the # s are
8 , 9 , 10 , 11 , 12 , 13

Answer 5

two ways.
1. if x-1,x,x+1 add up to 27, the next three numbers x+2,x+3,x+4 are each 3 more than the previous ones, so they will add up to 36.

2. let the numbers be x-1,x,x+1
x-1 +x + x+1 = 27
3x = 27
x = 9 (check 8+9+10 = 27)
so the next three are 11, 12, 13
11+12+13 = 36

btw, we don't mind helping with homework. what we don't like is people putting all their homework on the site.

Answer 6

You are correct!
My approach is:
Let n = the middle number of 3 consecutive numbers. Then
(n - 1) + n + (n + 1) = 3n = 27
n = 9
so the numbers are 8, 9, 10
The next 3 consecutive numbers are
11, 12, 13
11 + 12 + 13 = 3*12 = 36

Answer 7

Answer is 36.

x+(x+1)+(x+2) = 27
=> 3x =24 => x=8

so (x+3)+(x+4)+(x+5) = 36

Answer 8

You could either use six consecutive numbers to see the difference between the sum of the first set of three and the second set of three (and add that amount to 27), or you could make it into an algebra problem.

Answer 9

n1+n2+n3+n4+n5+n6

n1+n2+n3=27
Since the numbers are consecutive, you can say that the second number is the first number plus one.
n1+1=n2
n1+2=n3
(n1)+(n1+1)+(n1+2)=27
3n1+3=27
3n1=24
n1=8
This means that the first integer is eight, the second is nine, et cetera. The fourth, fifth, and sixth digits are respectively eleven, twelve, and thirteen, so the sum of the last three is 36.

Answer 10

start with x being the lowest number. x+(x+1)+(X+2)=27.. x = 6

so your 6 numbers are 6,7,8,9,10,11

the sum would be 20