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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calendarEducator since 2010
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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.
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calendarEducator since 2008
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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
You can check it if you want.
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