The Sum of 5 Consecutive Integers is 505, what is the Third number in this Sequence? So I've never covered this and I'm a bit confused.
$$x + (x+1) + (x+2) + (x+3) + (x+4)$$
So $5x + 10 = 505$ which equals $x = 99$. So isn't the third integer $2$? Which would mean that it equals to $1 + 2 + (99) = 102$? Why is the correct answer $101$, only taking into account the third integer ignoring its sequence @_@?
 A: In overly pedantic detail, you might think of the problem this way:
We have five consecutive integers that add to 505.  Let the five numbers be $a$, $b$, $c$, $d$, and $e$.  Since they are consecutive integers, we may assume that they are ordered $a < b < c < d < e$, from which it follows that
\begin{align}
b &= a+1 \\
c &= b+1 = (a+1) + 1 = a+2 \\
d &= c+1 = (a+2) + 1 = a+3 \\
e &= d+1 = (a+3) + 1 = a+4.
\end{align}
We want to find the third integer in this sequence, which means that we are trying to find the value of $c$.  We already know that $c = a+2$, so we can, instead, try to find $a$, then add 2 to that in order to get $c$.  But the numbers add to 505, so we have
\begin{align}
505 &= a + b + c + d + e \\
&= a + (a+1) + (a+2) + (a+3) + (a+4) \\
&= 5a + 10.
\end{align}
This implies that
$$ 5a = 495 \implies a = 99. $$
But remember that we were trying to find $c$, not $a$, and that $c=a+2$.  Therefore the third integer in the sequence is
$$ c = a+2 = 99+2 = 101.$$

It might be simpler to work this way:  $c$ is the third integer in the sequence.  Since we want to find $c$, we should write the other four integers in terms of $c$.  As above
\begin{align}
d &= c+1 \\
e &= d+1 = (c+1) + 1 = c+2,
\end{align}
since $d$ and $e$ are the next two consecutive integers.  To get $a$ and $b$, we have to count down, instead of up, which gives us
\begin{align}
b &= c-1 \\
a &= b-1 = (c-1)-1 = c-2.
\end{align}
Then we have
\begin{align}
505 &= a + b + c + d + e \\
&= (c-2) + (c-1) + c + (c+1) + (c+2) \\
&= 5c,
\end{align}
which implies that $c = 101$.
A: A very simple variant:
Let $a_1$ the first integer, $a_5$ the fifth. The sum $S$ of these integers is
$$S=5\cdot\frac{a_1+a_5}2=505, \quad\text{whence }\quad a_3=\frac{a_1+a_5}2=101.$$
A: Based on how you've setup the problem, we know that the first integer is $x$, the second is $x+1$, the third is $x+2$, and so forth. Since you found $x$ to be $99$, then the third integer is
$$x+2=99+2=101.$$
