Use the formulae for the nth term and the sum of the first n terms of an arithmetic progression, and solve problems involving arithmetic sequences and series

What this dot point is asking

SEAB wants you to use the standard formulae for an arithmetic progression (AP) - the $n$th term and the sum of the first $n$ terms - to find unknowns from given conditions and to solve worded problems. This is the foundation for sigma notation and series work.

The answer

Definition and the nth term

An arithmetic progression has a constant common difference $d$ between consecutive terms. With first term $a$:

The terms form a straight-line pattern: plotting $u_n$ against $n$ gives points on a line of gradient $d$.

The sum of the first n terms

The sum $S_n = u_1 + u_2 + \cdots + u_n$ has two equivalent formulae:

where $l = u_n$ is the last term. The second form is handy when you know the first and last terms.

Finding a and d from conditions

Most AP problems give you two pieces of information (two terms, or a term and a sum). Write each as an equation in $a$ and $d$ using the formulae, then solve the simultaneous equations.

Recovering terms from the sum

If you are given $S_n$ as a formula in $n$, the $n$th term is $u_n = S_n - S_{n-1}$. For an AP this expression is always linear in $n$, and the coefficient of $n$ is the common difference.

Summing a slice of an AP

To add the terms from the $p$th to the $q$th of an arithmetic progression, the cleanest method is to subtract two partial sums: $\sum_{p}^{q} = S_q - S_{p-1}$. For the AP $3, 7, 11, \ldots$, the sum of the $5$th to the $10$th terms is $S_{10} - S_4$. An alternative is to treat the slice as a new AP whose first term is $u_p$, last term is $u_q$, and number of terms is $q - p + 1$, then apply $\tfrac{n}{2}(a + l)$. Both routes work, but writing the slice as a difference of partial sums is less prone to the off-by-one error in counting the terms, which is the usual pitfall here.

Recognising an AP hidden in a word problem

Many H2 problems describe an AP without naming it, so the first skill is spotting the constant common difference. Any situation where a quantity increases or decreases by the same fixed amount each step, equal monthly repayments, seats increasing by a fixed number per row, a salary rising by a set raise each year, is arithmetic. Once you identify $a$ (the starting value) and $d$ (the fixed change), the whole problem reduces to substituting into the two AP formulae. Translating the words into $a$ and $d$ before reaching for a formula is what turns a wordy question into a routine calculation.

Examples in context

Example 1. A savings plan. Saving $50 in month one and$10 more each month gives an AP with $a = 50$, $d = 10$. After $12$ months the total saved is $S_{12} = \dfrac{12}{2}(2 \times 50 + 11 \times 10) = 6 \times 210 = 1260$ dollars.

Example 2. Stacked logs. Logs stacked with $20$ on the bottom row and one fewer each row up form an AP. The number in a full stack down to a single top log is $S = \dfrac{20}{2}(20 + 1) = 210$ logs, using the first-plus-last form.

Try this

Q1. Find the $12$th term of the AP $7, 10, 13, \ldots$ [2 marks]

• Cue. $a = 7$, $d = 3$, $u_{12} = 7 + 11 \times 3 = 40$.

Q2. The sum of the first $n$ terms of an AP is $S_n = 3n^2 - n$. Find the first term and the common difference. [3 marks]

• Cue. $u_1 = S_1 = 2$; $u_n = S_n - S_{n-1} = 6n - 4$, so $d = 6$.

Q3. An AP has $a = 100$ and $d = -4$. Find how many terms are positive. [3 marks]

• Cue. $u_n = 100 - 4(n-1) > 0$ gives $n < 26$, so $25$ terms are positive (the $26$th is zero).