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 @_@?
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$\begingroup$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 $c = 101$.
$\endgroup$ 1 $\begingroup$A very simple variant:
Let $a_1$ the first integer, $a_5$ the fifth. The sum $S$ of these integers is $$S=5\cdot\frac{a_1+a_5}2=505, \quad\text{whence }\quad a_3=\frac{a_1+a_5}2=101.$$
$\endgroup$ $\begingroup$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 $$x+2=99+2=101.$$
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