Answer :
[tex]n+(n+2)+(n+4)=153[/tex] -- combine like terms
[tex]3n+6=153[/tex] -- subtract 6 from both sides
[tex]3n=147[/tex] -- divide by 3
[tex]n=49[/tex]
So the numbers are 49, (49+2) and (49+4)
I.e. 49, 51, and 53
But the question only wants the smallest (i.e. the first of the consecutive odd numbers).
The smallest of the 3 consecutive odd numbers that sum to 153 is 49
[tex]3n+6=153[/tex] -- subtract 6 from both sides
[tex]3n=147[/tex] -- divide by 3
[tex]n=49[/tex]
So the numbers are 49, (49+2) and (49+4)
I.e. 49, 51, and 53
But the question only wants the smallest (i.e. the first of the consecutive odd numbers).
The smallest of the 3 consecutive odd numbers that sum to 153 is 49
[tex]a+a+2+a+4=153 \\ \\ 3a+6=153 \\ \\ 3a=153-6 \\ \\ 3a=147 \\ \\ \boxed{a=\frac{147}{3}=49- \ the \ smallest} \\ \\ a+2=49+2=51 \\ a+4=49+4=53[/tex]
