Have to find 3 consecutive odd integers

Question

find 3 consecutive odd integers such that four times the middle integer is twice the sum of the other two integers

Answer 1

x, x+2, x+4 are the integers

4(x+2)=2[(x)+(x+4)]

4(x+2)=2(2x+4)

4x+8=4x+8

any three consecutive odd integers will work

examples:

 

5,7,9

4*7=2*(9+5)

28=2*(14)

28=28

why ???

 

take any 3 odd integers, say, 3,5,7

the mean is the middle number; 3+5+7=15 and 15/3=5

having 3,5,7 is the same as having 5,5,5

4*5 is the same as 2*(5+5)

4*5 is the same as 2*(2*5)

4*5 is the same as 2*2*5

4*5 is the same as 4*5

look at it this way...

x-2,x, and x+2 are the integers

4*x is the same as 2*(x-2+x+2)

4*x is the same as 2*(2x)

4x is the same as 4x

 

example:

21, 23, 25

4*23=2*(21+25)

4*23=2*(46)

4*23=2*(2*23)

4*23=(2*2)*23

4*23=4*23