Find 3 consecutive integers such that the sum of the three is equal to 2 times the sum of the first two integers. ?

Let (n - 1), n and (n + 1) be the three consecutive integers.

(n - 1) + n + (n + 1) = 2 × [(n - 1) + n]
3n = 4n - 2
n = 2

The 3 consecutive integers are 1, 2 and 3.

Find 3 consecutive integers such that the sum of the three is equal to 
2 times the sum of the first two integers.3x = 2(2x - 1)
x = 2
The 3 consecutive integers are 1, 2 and 3.

You must first set up your formula. Let x= lowest number 

X + (x+1) + (x+2)= 2(x + [x+1])
3x + 3= 2(2x + 1)
3x+3=4x+2
1=x

Call the numbers n - 1, n, and n + 1

Then the problem says n-1 + n + n+1 = 3n = 2(n-1 + n)

3n = 4n - 2

n = 2 - the middle number, so the

numbers are 1, 2, and 3

n + (n + 1) + ( n+2) = 2(n + (n + 1))
Hence 
3n + 3 = 4n + 2 
4n - 3n = 3 - 2 
n = 1 
n + 1 = 2 
n + 2 = 3 

Letting the integers be n, n + 1 and n + 2 we have,

sum of three => n + n + 1 + n + 2 = 3n + 3

sum of first two => n + n + 1 = 2n + 1

Hence, 2(2n + 1) = 3n + 3

i.e. 4n + 2 = 3n + 3

so, n = 1

Therefore, the numbers are 1, 2 and 3

checking gives, 2(1 + 2) = (1 + 2 + 3) = 6

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