1. Arithmetic Sequences & Series

Video coming soon3 worked examples

Theory

An arithmetic sequence increases by a constant common difference $d$ each step. With first term $a$ , the $n$ th term is:

u_n = a + (n-1)d

The sum of the first $n$ terms is:

S_n = \frac{n}{2}\bigl[2a + (n-1)d\bigr] = \frac{n}{2}(a + l)

where $l$ is the last term. Use the second form when you already know the last term.

The sum formula comes from a neat trick worth knowing, since you may be asked to derive it. Write the series out, then write it again in reverse underneath:

\begin{aligned} S_n &= a + (a+d) + \cdots + \bigl(a+(n-1)d\bigr) \\ S_n &= \bigl(a+(n-1)d\bigr) + \cdots + (a+d) + a \end{aligned}

Adding the two lines pairs the first term of one with the last of the other. Every pair sums to the same total, $2a + (n-1)d$ , and there are $n$ pairs:

2S_n = n\bigl[2a + (n-1)d\bigr] \quad\implies\quad S_n = \frac{n}{2}\bigl[2a + (n-1)d\bigr]

The Golden Rule: almost every arithmetic question reduces to finding $a$ and $d$ — pin those down first, then substitute.

⚠️ Common Examiner Traps

The $(n-1)$ slip: the $n$ th term uses $(n-1)d$ , not $nd$ — the first term already counts as one.
Wrong sum formula: only use $\frac{n}{2}(a+l)$ when the last term $l$ is actually known.
Counting terms: the number of terms from the $p$ th to the $q$ th is $q - p + 1$ .

Worked examples

Example 1

An arithmetic sequence has first term $5$ and common difference $3$ . Find the 20th term and the sum of the first 20 terms.

Step 1: Apply the $n$ th term formula with $a=5$ , $d=3$ , $n=20$ :

u_{20} = 5 + (20-1)(3) = 5 + 57 = 62

Step 2: Apply the sum formula:

S_{20} = \frac{20}{2}\bigl[2(5) + (19)(3)\bigr] = 10\,[10 + 57] = 670

Example 2

The 4th term of an arithmetic sequence is $11$ and the 9th term is $26$ . Find the first term and common difference, then the sum of the first 12 terms.

Step 1: Write each term with the formula and subtract to eliminate $a$ :

a + 3d = 11 \qquad a + 8d = 26 \implies 5d = 15 \implies d = 3

Step 2: Substitute back to find $a$ :

a + 3(3) = 11 \implies a = 2

Step 3: Find the sum of the first 12 terms:

S_{12} = \frac{12}{2}\bigl[2(2) + (11)(3)\bigr] = 6\,[4 + 33] = 222

Example 3

How many terms are there in the arithmetic sequence $7,\ 11,\ 15,\ \ldots,\ 111$ ? Hence find the sum of the sequence.

Step 1: Identify $a$ and $d$ , then write a formula for the $n$ th term:

a = 7, \quad d = 11 - 7 = 4 \quad\implies\quad u_n = 7 + (n-1)(4) = 4n + 3

Step 2: The last term is $111$ , so set $u_n = 111$ and solve for $n$ :

4n + 3 = 111 \implies 4n = 108 \implies n = 27

There are 27 terms.

Step 3: Since the last term is known, the quicker sum formula applies:

S_{27} = \frac{27}{2}(a + l) = \frac{27}{2}(7 + 111) = \frac{27}{2}(118) = 27 \times 59 = 1593