In an arithmetic sequence, the third term is 10 and the fifth term is 16. I'm just working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem.  

In an arithmetic sequence, the third term is $10$ and the fifth term is $16$.
  Find the common difference.
  Find the first term.
  Find the sum of the first $20$ terms in the sequence. 

So, the arithmetic formula is $a_n = a_1 + (n – 1)d$ right?
The common difference is the difference between the terms I think. So $16 - 10 = 6$, but there is a term between that so divided by $2$ it is $3$. 
How do I find the first term and the sum of the first 20 terms?
 A: The pedestrian approach is to use your formula $a_n=a_1+(n-1)d$ and plug in the two values you know
$$a_3=10=a_1+(3-1)d\\a_5=16=a_1+(5-1)d$$
This is two simultaneous equations that you can solve for $a_1,d$.  You have already found $d=3$ essentially by subtracting the two.  Now just plug that into one of them and evaluate $a_1$.  
Otherwise, since you know $d=3$ you can just count two steps down from $a_3=10$ to get $a_1$  
To get the sum of the first $20$ terms you need to know (and should have proved) that $$\sum_{i=1}^ni=\frac 12n(n+1)$$  If you write out the sum of the first $20$ terms you will need $d$ times this with $n=19$ plus to account for the $a$s
A: You have correctly found the common difference.
The $k$th term of an arithmetic sequence with initial term $a_1$ and common difference $d$ is 
$$a_k = a_1 + (k - 1)d$$
Since we are given that $a_3 = 10$ and $a_5 = 16$, we obtain
\begin{align*}
a_3 & = a_1 + (3 - 1)d = 10\\
a_5 & = a_1 + (5 - 1)d = 16
\end{align*}
which is a system of two linear equations in two variables.  Subtracting the first equation from the second yields $2d = 6 \implies d = 3$, as you found.  Substituting $d = 3$ in the equation for $a_3$ yields $a_1 = 4$, as you found in the comments.
The $n$th partial sum (the sum of the first $n$ terms) of an arithmetic series
\begin{align*}
S_n & = \sum_{k = 1}^{n} a_k\\
    & = \sum_{k = 1}^{n} [a_1 + (k - 1)d]\\
    & = \sum_{k = 1}^{n} a_1 + \sum_{k = 1}^{n} (k - 1)d\\
    & = a_1\sum_{k = 1}^{n} 1 + d\sum_{k = 1}^{n} (k - 1)\\
    & = na_1 + \frac{d(n - 1)n}{2}
\end{align*}
from which you can find $S_{20}$ by substituting $20$ for $n$, $3$ for $d$, and $4$ for $a_1$.
Alternatively, 
\begin{align*}
S_n & = \sum_{k = 1}^{n} a_k\\
    & = a_1 + a_2 + a_3 + \cdots + a_{n - 2} + a_{n - 1} + a_n\\
    & = a_1 + [a_1 + d] + [a_1 + 2d] + \cdots + [a_1 + (n - 3)d] + [a_1 + (n - 2)d] + [a_1 + (n - 1)d]
\end{align*}
Since we obtain the same sum if we write the terms in reverse order, we obtain
\begin{align*}
S_n & = a_n + a_{n - 1} + a_{n - 2} + \cdots + a_3 + a_2 + a_1\\
    & = [a_1 + (n - 1)d] + [a_1 + (n - 2)d] + [a_1 + (n - 3)d] + \cdots + [a_1 + 2d] + [a_1 + d] + a_1
\end{align*}
Adding the two expressions for $S_n$ yields
\begin{alignat*}{10}
S_n & = & a_1 & + & [a_1 + d] & + & \cdots & + & [a_1 + (n - 1)d] & + & [a_1 + (n - 1)d]\\
S_n & = & [a_1 + (n - 1)d] & + & [a_1 + (n - 2)d] & + & \cdots & + & [a_1 + d] & + & a_1\\ \hline
2S_n & = & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d] & + & \cdots & + & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d]
\end{alignat*} 
Since there are $n$ columns in that sum, we obtain
\begin{align*}
2S_n & = n[2a_1 + (n - 1)d]\\
2S_n & = n[a_1 + a_1 + (n - 1)d]\\
2S_n & = n(a_1 + a_n)\\
S_n & = \frac{n(a_1 + a_n)}{2}
\end{align*}
from which you can find $S_{20}$ by substituting $4$ for $a_1$ and using the formula $a_n = a_1 + (n - 1)d$ with $n = 20$ and $d = 3$ to find $a_{20}$. 
