How do you find three consecutive odd integers such that twice the sum of the first and third integers is 21 more than the second integer?

1 Answer
Alan P.
Oct 20, 2015

The three consecutive odd numbers are 5, 7, 9

Explanation:

Let
color(white)("XXX")first odd number = color(red)(2n-1)
color(white)("XXX")second odd number =color(blue)(2n+1)
color(white)("XXX")third odd number =color(brown)(2n+3)

We are told
color(white)("XXX")2( color(red)((2n-1)) + color(brown)((2n+3)) ) = color(blue)((2n+1)) + 21

Simplifying
color(white)("XXX")8n+4 = 2n+22

color(white)("XXX")6n = 18

color(white)("XXX")n=3

So the three odd numbers are
color(white)("XXX")first = 2n-1 = 2(3)-1 = 5
color(white)("XXX")second = 2(3)+1 = 7
color(white)("XXX")third = 2(3)+3 = 9

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