$$\begin{array}{|c|}\hline \sum _{i=1}^{n}1=n\\ \hline\end{array}$$
Another simple arithmetic sequence is when
${a}_{1}=1$ and
$d=1$ , which is the sequence of positive integers:
$$\begin{array}{ccc}\hfill {a}_{i}& =& {a}_{1}+d(i-1)\hfill \\ & =& 1+1\xb7(i-1)\hfill \\ & =& i\hfill \\ \hfill \left\{{a}_{i}\right\}& =& \{1;2;3;4;5;...\}\hfill \end{array}$$
If we wish to sum this sequence from
$i=1$ to any positive integer
$n$ , we would write
$$\sum _{i=1}^{n}i=1+2+3+...+n$$
This is an equation with a very important solution as it gives the answer to the sum of positive integers.
Interesting fact
Mathematician, Karl Friedrich Gauss, discovered this proof when he was only 8 years old. His teacher had decided to give his class a problem which would distract them for the entire day by asking them to add all the numbers from 1 to 100. Young Karl realised how to do this almost instantaneously and shocked the teacher with the correct answer, 5050.
We first write
${S}_{n}$ as a sum of terms in ascending order:
$${S}_{n}=1+2+...+(n-1)+n$$
We then write the same sum but with the terms in descending order:
$${S}_{n}=n+(n-1)+...+2+1$$
We then add corresponding pairs of terms from equations
[link] and
[link] , and we find that the sum for each pair is the same,
$(n+1)$ :
$$2{S}_{n}=(n+1)+(n+1)+...+(n+1)+(n+1)$$
We then have
$n$ -number of
$(n+1)$ -terms, and by simplifying we arrive at the final result:
$$\begin{array}{ccc}\hfill 2{S}_{n}& =& n\phantom{\rule{0.277778em}{0ex}}(n+1)\hfill \\ \hfill {S}_{n}& =& \frac{n}{2}(n+1)\hfill \end{array}$$
$${S}_{n}=\sum _{i=1}^{n}i=\frac{n}{2}(n+1)$$
Note that this is an example of a quadratic sequence.
If we wish to sum any arithmetic sequence, there is no need to work it out term-for-term. We will now determine the general formula to evaluate a finite arithmetic series. We start with the general formula for an arithmetic sequence and sum it from
$i=1$ to any positive integer
$n$ :
$$\begin{array}{ccc}\hfill \sum _{i=1}^{n}{a}_{i}& =& \sum _{i=1}^{n}[{a}_{1}+d(i-1)]\hfill \\ & =& \sum _{i=1}^{n}({a}_{1}+di-d)\hfill \\ & =& \sum _{i=1}^{n}[({a}_{1}-d)+di]\hfill \\ & =& \sum _{i=1}^{n}({a}_{1}-d)+\sum _{i=1}^{n}\left(di\right)\hfill \\ & =& \sum _{i=1}^{n}({a}_{1}-d)+d\sum _{i=1}^{n}i\hfill \\ & =& ({a}_{1}-d)n+\frac{dn}{2}(n+1)\hfill \\ & =& \frac{n}{2}(2{a}_{1}-2d+dn+d)\hfill \\ & =& \frac{n}{2}(2{a}_{1}+dn-d)\hfill \\ & =& \frac{n}{2}[\phantom{\rule{0.166667em}{0ex}}2{a}_{1}+d(n-1)]\hfill \end{array}$$
So, the general formula for determining an arithmetic series is given by
$${S}_{n}=\sum _{i=1}^{n}[{a}_{1}+d(i-1)]=\frac{n}{2}[2{a}_{1}+d(n-1)]$$
For example, if we wish to know the series
${S}_{20}$ for the arithmetic sequence
${a}_{i}=3+7(i-1)$ , we could either calculate each term individually and sum them:
$$\begin{array}{ccc}\hfill {S}_{20}& =& \sum _{i=1}^{20}[3+7(i-1)]\hfill \\ & =& 3+10+17+24+31+38+45+52+\hfill \\ & & 59+66+73+80+87+94+101+\hfill \\ & & 108+115+122+129+136\hfill \\ & =& 1390\hfill \end{array}$$
or, more sensibly, we could use equation
[link] noting that
${a}_{1}=3$ ,
$d=7$ and
$n=20$ so that
$$\begin{array}{ccc}\hfill {S}_{20}& =& \sum _{i=1}^{20}[3+7(i-1)]\hfill \\ & =& \frac{20}{2}[2\xb73+7(20-1)]\hfill \\ & =& 1390\hfill \end{array}$$
This example demonstrates how useful equation
[link] is.
Exercises
- The sum to
$n$ terms of an arithmetic series is
${S}_{n}=\frac{n}{2}(7n+15)$ .
- How many terms of the series must be added to give a sum of 425?
- Determine the 6
^{th} term of the series.
- The sum of an arithmetic series is 100 times its first term, while the last term is 9 times the first term. Calculate the number of terms in the series if the first term is not equal to zero.
- The common difference of an arithmetic series is 3. Calculate the values of
$n$ for which the
${n}^{th}$ term of the series is 93, and the sum of the first
$n$ terms is 975.
- The sum of
$n$ terms of an arithmetic series is
$5{n}^{2}-11n$ for all values of
$n$ . Determine the common difference.
- The sum of an arithmetic series is 100 times the value of its first term, while the last term is 9 times the first term. Calculate the number of terms in the series if the first term is not equal to zero.
- The third term of an arithmetic sequence is -7 and the 7
^{th} term is 9. Determine the sum of the first 51 terms of the sequence.
- Calculate the sum of the arithmetic series
$4+7+10+\cdots +901$ .
- The common difference of an arithmetic series is 3. Calculate the values of
$n$ for which the
${n}^{\mathrm{th}}$ term of the series is 93 and the sum of the first
$n$ terms is 975.