## Arithmetic Sequences

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

\[S_{20}=\frac{20}{2} \times 3+(20-1) \times 4)=820\]