Find five consecutive odd integers such that their sum is $55$.
Avoiding explicit algebra computation, their average would be $11$ and they would have to surround that average symmetrically.
let k be the smallest number. Then you have $k+(k+2)+(k+4)+(k+6)+(k+8)=55.$ So you get $5k+20=55\rightarrow 5k=35\rightarrow k=7$