3, 7, 11, 15, 19
is an arithmetic sequence with 4 being added to each term to get the next term.
Given any arithmetic sequence we can find an expression for the
\[n\]
th term. If \[d\]
is the number that is added each time (called the common difference) and \[a\]
is the first term, then the \[n\]
th term is \[a_n=a+(n-1)d\]
.For the sequence above,
\[a=3, \: d=4\]
.Hence
\[a_n=3+(n-1) \times 4=4n-1\]
We can also find a formula for the sum
\[S_n\]
of the first \[n\]
terms.\[S_n=a+(a+d)+...(a+(n-2)d)+ (a+(n-1)d)\]
Writing this sum backwards gives
\[S_n=(a+(n-1)d)+(a+(n-2)d)+...+ (a+d)+a\]
Now adding these two sums gives
\[\begin{equation} \begin{aligned} 2S_n &= \underbrace{(a+(n-1)d)+(a+(n-1)d)+...+ (a+(n-1)d)+(a+(n-1)d)}_{n \: terms} \\ &= n(2a+(n-1)d) \end{aligned} \end{equation}\]
Hence
\[S_n=\frac{n}{2}(2a+(n-1)d)\]
For the sequence above the sum of the first 20 terms is
\[S_{20}=\frac{20}{2} \times 3+(20-1) \times 4)=820\]