Arithmetic Series (Summing an AP)
Arithmetic series add the terms of an arithmetic sequence, with sum \(S_n=\dfrac{n}{2}\big(2a+(n-1)d\big)=\dfrac{n}{2}(a+l)\).
Part of the NSW Year 12 Mathematics Advanced course, in the Sequences and series focus area of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.
Theory
An arithmetic series is the sum of the terms of an arithmetic sequence. This Year 12 Mathematics Advanced topic (MAV-12-03) uses \(S_n=\tfrac{n}{2}(a+\ell)\) and \(S_n=\tfrac{n}{2}[2a+(n-1)d]\) to find sums.
The sum of the first \(n\) terms is \(S_n\). There are two equivalent formulas: use \(S_n=\dfrac{n}{2}(a+\ell)\) when you know the last term \(\ell\), and \(S_n=\dfrac{n}{2}[\,2a+(n-1)d\,]\) when you know the common difference \(d\).
\(S_n\) is a quadratic function of \(n\). If given the last term, first find \(n\) with \(a_n=a+(n-1)d\).
Method
- Identify \(a\), and either the last term \(\ell\) or the difference \(d\).
- Find \(n\) first if you are given the last term, using \(a_n=a+(n-1)d\).
- Substitute into the matching sum formula.
| \(S_{20}\) | \(=\) | \(\dfrac{20}{2}[2(3)+19(4)]\) |
| \(=\) | \(10(82)=820\) |
| \(n\) | \(=\) | \(16\) |
| \(S_{16}\) | \(=\) | \(\dfrac{16}{2}(5+50)=440\) |
| \(S_{100}\) | \(=\) | \(\dfrac{100}{2}(1+100)\) |
| \(=\) | \(50\times101=5050\) |
| \(S_{10}\) | \(=\) | \(\dfrac{10}{2}[4+27]=155\) |
| \(n\) | \(=\) | \(10\) |
Common pitfalls
Frequently asked questions
What is the formula for the sum of an arithmetic series?
Use S n equals n over 2 times (a plus the last term), or S n equals n over 2 times (2a plus (n minus 1) d), where a is the first term and d the common difference.
Which sum formula should I use?
Use the first form, with a plus the last term, when you know the last term. Use the second form, with 2a plus (n minus 1) d, when you know the common difference.
How do you find the number of terms in an arithmetic series?
Use the nth term formula a n equals a plus (n minus 1) d, set it equal to the last term, and solve for n.
What is the sum of the first 100 whole numbers?
It is 100 over 2 times (1 plus 100), which equals 50 times 101, which is 5050.