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
Oct 20, 2015

Answer:

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#