An arithmetic sequence goes up or down by the same amount every step. This note covers the formula for the \(n\)th term, the formula for the sum of the first \(n\) terms, and sigma notation. These ideas model steady growth, such as regular savings, seating plans and repeated training increases.
Key ideas
A sequence is an ordered list of numbers called terms. We write \(u_1\) for the first term, \(u_2\) for the second, and \(u_n\) for the \(n\)th term. A sequence is arithmetic if the difference between consecutive terms is constant. This constant is the common difference \(d\):
\[d = u_{n+1} - u_n\]
If \(d \gt 0\) the terms increase. If \(d \lt 0\) they decrease. For example, \(5, 9, 13, 17, \dots\) has \(u_1 = 5\) and \(d = 4\). The sequence \(20, 17, 14, \dots\) has \(d = -3\).
Each term is found by starting at \(u_1\) and adding \(d\) a total of \(n-1\) times. That gives the formula for the general term. Notice that you add \(d\) one fewer time than the position number, because the first term needs no addition.
An arithmetic series is the sum of the terms of an arithmetic sequence. The sum of the first \(n\) terms is written \(S_n\). A clean way to see the formula is to write the sum forwards and backwards. Each pair of terms (first and last, second and second-last, and so on) adds to the same total \(u_1 + u_n\). There are \(n\) terms, which make \(\tfrac{n}{2}\) such pairs, so \(S_n = \tfrac{n}{2}(u_1 + u_n)\).
Use the version with \(u_n\) when you know the last term. Use the version with \(d\) when you do not.
Sigma notation. The symbol \(\sum\) means "add up". The expression \(\sum_{k=1}^{20} (3k+2)\) means: put \(k = 1, 2, 3, \dots, 20\) into \(3k+2\) and add the results. The numbers below and above the sigma are the first and last values of \(k\). To use the formulae, find \(u_1\) (put in the first value of \(k\)), find \(d\) (the coefficient of \(k\) when the expression is linear in \(k\)), and count the terms. The number of terms from \(k=a\) to \(k=b\) is \(b-a+1\).
To find how many terms a list has, solve \(u_n = \) last term for \(n\). The result must be a positive whole number. To find the first \(n\) for which a sum passes a target, use the table or graph feature on your GDC, or solve the resulting quadratic inequality, and always give a whole number of terms.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(d = u_{n+1} - u_n\) | Common difference. Constant for an arithmetic sequence. |
| \(u_n = u_1 + (n-1)d\) | The \(n\)th term, given the first term and the common difference. |
| \(S_n = \dfrac{n}{2}\left(2u_1 + (n-1)d\right)\) | Sum of the first \(n\) terms when \(u_1\) and \(d\) are known. |
| \(S_n = \dfrac{n}{2}(u_1 + u_n)\) | Sum of the first \(n\) terms when the first and last terms are known. |
| \(\sum_{k=1}^{n} u_k\) | Sigma notation for the sum of the first \(n\) terms. |
| Number of terms | Solve \(u_1 + (n-1)d = \) last term for \(n\). |
Worked examples
Example 1 A stadium has 30 rows of seats. The first row has 24 seats and each row has 3 more seats than the one in front of it. Find (a) the number of seats in the last row and (b) the total number of seats.
This is arithmetic with \(u_1 = 24\), \(d = 3\), \(n = 30\).
(a) \(u_{30} = 24 + (30-1) \times 3 = 24 + 87 = 111\).
(b) \(S_{30} = \dfrac{30}{2}(24 + 111) = 15 \times 135 = 2025\).
Answer: 111 seats in the last row and 2025 seats in total.
Example 2 An arithmetic sequence has \(u_5 = 17\) and \(u_{12} = 45\). Find \(u_1\), \(d\) and \(u_{30}\).
Write each fact using \(u_n = u_1 + (n-1)d\).
\(u_5\): \(u_1 + 4d = 17\). \(u_{12}\): \(u_1 + 11d = 45\).
Subtract the first equation from the second: \(7d = 28\), so \(d = 4\).
Substitute: \(u_1 + 16 = 17\), so \(u_1 = 1\).
\(u_{30} = 1 + 29 \times 4 = 117\).
Answer: \(u_1 = 1\), \(d = 4\), \(u_{30} = 117\).
Example 3 (a) Evaluate \(\sum_{k=1}^{20}(3k+2)\). (b) A runner trains for a marathon. In week 1 she runs 5 km and she increases her weekly distance by 4 km each week. Find the smallest number of weeks after which her total distance first exceeds 1000 km.
(a) The terms are \(3k+2\). First term: \(k=1\) gives 5. Last term: \(k=20\) gives 62. There are 20 terms.
\(S_{20} = \dfrac{20}{2}(5 + 62) = 10 \times 67 = 670\).
(b) For the runner, \(u_1 = 5\) and \(d = 4\).
\(S_n = \dfrac{n}{2}(2 \times 5 + (n-1) \times 4) = \dfrac{n}{2}(4n + 6) = n(2n+3) = 2n^2 + 3n\).
We need \(2n^2 + 3n \gt 1000\). Use the table feature on your GDC to compare values: \(S_{21} = 2(441) + 63 = 945\), which is too small. \(S_{22} = 2(484) + 66 = 1034\), which is greater than 1000.
Answer: (a) 670. (b) 22 weeks.
Common errors
- Common error: using \(u_n = u_1 + nd\). The correct factor is \((n-1)\), because the first term has no added difference.
- Common error: checking only the first two terms. A sequence is arithmetic only if every consecutive difference is the same.
- Common error: miscounting terms in a sum. From \(k=3\) to \(k=10\) there are \(10-3+1 = 8\) terms, not 7.
- Common error: giving a non-integer value of \(n\). The number of terms must be a whole number, so when a total must first pass a target, give the smallest whole number that works.
- Common error: losing the sign of a negative common difference when substituting into a formula.
Exam tip: When two terms are given, write one equation for each and subtract to eliminate \(u_1\). Check your answer by substituting back into both original equations.
Check yourself
- An arithmetic sequence has \(u_1 = 7\) and \(d = -2\). Find \(u_{10}\) and \(S_{10}\).
Answer
\(u_{10} = 7 + 9(-2) = -11\). \(S_{10} = \dfrac{10}{2}(7 + (-11)) = 5 \times (-4) = -20\). - How many terms are in the sequence \(3, 8, 13, \dots, 198\)?
Answer
\(d = 5\). Solve \(3 + 5(n-1) = 198\): \(5(n-1) = 195\), \(n - 1 = 39\), so \(n = 40\). - Evaluate \(\sum_{k=1}^{15}(4k-1)\).
Answer
First term \(= 3\), last term \(= 4(15)-1 = 59\), 15 terms. \(S_{15} = \dfrac{15}{2}(3 + 59) = \dfrac{15}{2} \times 62 = 465\). - In an arithmetic sequence \(u_3 = 20\) and \(u_7 = 8\). Find \(u_1\) and \(d\).
Answer
\(u_1 + 2d = 20\) and \(u_1 + 6d = 8\). Subtracting: \(4d = -12\), so \(d = -3\). Then \(u_1 = 20 + 6 = 26\).