Find five consecutive odd integers such that their sum is $55$. So my professor asked us to do an Olympiad exercise which says that the sum of five consecutive odd integers is $55$, find those integers. But I've never seen such an exercise so it is quite new and if I learn how then I will be able to solve similar exercises.
I tried to think of it in terms of geometry but that didn't help me.
 A: Avoiding explicit algebra computation, their average would be $11$ and they would have to surround that average symmetrically.
A: We have
$$\sum_{k=p}^{p+4} (2k+1)=55\iff2\times\frac{5(p+4+p)}{2}+5=55\iff p=3$$
so the first odd number is $7$.
A: As consecutive  odd numbers differ by $2$ 
and we have odd number of integer terms (namely $5$), its beneficial take the numbers to be $n,n\pm2,n\pm4$
Had the number of odd integers been even, we could take $n\pm1,n\pm3,\cdots$
A: I did it this way:
We need 5 consecutive odd numbers who add up to 55, so their mid/centre or the 3rd (arithmetic mean) number must be 55/5 which is 11(as the sequence is symmetric about this number). The previous two and the next two odd numbers complete the needed sequence.
A: let k be the smallest number. Then you have $k+(k+2)+(k+4)+(k+6)+(k+8)=55.$ So you get $5k+20=55\rightarrow 5k=35\rightarrow k=7$
A: Let's take the numbers as $x-4,x-2,x,x+2,x+4$. Now, our equality takes the form
$$(x-4) + (x-2) + x + (x+2) + (x+4) = 55,$$
so $5x = 55$, i.e., $x = 11$.
We conclude that the numbers are $7,9,11,13,15$.
A: For odd numbers the answer is :$(x-4)+ (x-2)+x+(x+2)+(x+4)=55$ i.e. $5x= 55$ therefore $x=11$, and the series would be: $7+9+11+13+15=55$.
A: The sum of five consecutive odd integers is $55$.
Let $n$ = any integer.
Then $2n$ = an even integer 
and  $2n + 1$ = an odd integer.
Thus $\color{red}{\underbrace{2n +1}},\underbrace{(2n+1)+\color{blue}{2}},\underbrace{ (2n+1)+\color{blue}{4}},\underbrace{(2n+1)+\color{blue}{6}},\underbrace{(2n+1)+\color{blue}8} $
are five consecutive odd integers. The sum is $$10n + 25 =55$$ $$10n=30$$$$n=3$$ This gives $\color{red}{\overbrace{2(3)+1}}=7$ as the $\color{red}{first}$ of the five consecutive odd integers whose sum is $55$.
A: Hint: Choose one of those equations and solve it:
$$\begin{align}
& \text{The first integer:} & \qquad x+\color{blue}{x+2+x+4+x+6+x+8}=55 \\\,\\
& \text{The second integer:} & \qquad \color{red}{x-2}+x+\color{blue}{x+2+x+4+x+6}=55 \\\,\\
& \text{The third integer:} & \qquad \color{red}{x-4+x-2}+x+\color{blue}{x+2+x+4}=55 \\\,\\
& \text{The fourth integer:} & \qquad \color{red}{x-6+x-4+x-2}+x+\color{blue}{x+2}=55 \\\,\\
& \text{The fifth integer:} & \qquad \color{red}{x-8+x-6+x-4+x-2}+x=55
\end{align}$$
Obviously the simplest choice is to solve for the third integer, since the $2$s and $4$s cancel out, and you're left with the simple equation, $5x=55$.
A: Hint $\, $ The sum of an odd number $k$ of terms in arithmetic progression is $k$ times the middle term, since the differences $\,\color{#c00}{\pm\Delta}\,$ from the middle term $\,n\,$ cancel out on addition, e.g. for $\,k=7$
$$\begin{eqnarray} n &\,+\,& n\!\color{#c00}{+\!a} &\,+\,& n\!\color{#c00}{+\!2a} &\,+\,& n\!\color{#c00}{+\!3a} \\
&\,+\,& n\!\color{#c00}{-\!a} &\,+\,& n\!\color{#c00}{-\!2a} &\,+\,& n\!\color{#c00}{-\!3a} \\
\hline
=\ 7n \end{eqnarray}$$
Thus, in your case we have $\ 5n = 55\ \Rightarrow\ n =\,\ldots$
