# The sum of the first and third of three consecutive odd integers is 131 less than three times the second integer. Find the three integers

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### 3 Answers

The sum of the first and third of three consecutive odd numbers is 131 less than 3 times the second. We have to find the three integers.

Let the integers be 2n + 1, 2n + 3 and 2n + 5

We have 2n + 1 + 2n + 5 = 3*(2n + 3) - 131

=> 4n + 6 = 6n + 9 - 131

=> 2n = 131 - 9 + 6

=> 2n = 128

2n + 1 = 128 + 1 = 129

2n + 3 = 128 + 3 = 131

2n + 5 = 128 + 5 = 133

**The required numbers are 129, 131 and 133.**

Let the first even integer be n.

Then the second even integer be n+2

The third even integer is x+ 4

Given that the sum of the first and the third is 131 less than 3 times the second.

==> n + (n+4) = 3(n+2) - 131

Now we will combine terms.

==> 2n + 4 = 3n + 6 -131

==> 4 - 6 + 131 = n

==> n= 129

==> n+2 = 131

==> n+4 = 133

**Then the integers are 129, 131, and 133**

Let three numbers be x, (x+2) and (x+4).

Now, x+x+4=3(x+2)-131

=2x+4=3x+6-131

125+4=3x-2x

129=x

The three numbers:

x=129; x+2=131; x+4=133

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