VCE Specialist Mathematics (Year 11) · Arithmetic and Number · 18m read

Arithmetic and Geometric Sequences

Ever wondered how your loan repayment shrinks each month, or how bacteria multiply in a lab? Those patterns follow sequences — lists of numbers that change in predictable ways. You'll learn to spot arithmetic sequences (where numbers jump by the same amount each time) and geometric sequences (where each number multiplies by the same factor), then use formulas to find any term or add them all up. Mastering sequences unlocks real applications in finance, biology, and beyond — plus it's essential for Specialist Maths.

Learning objective: Identify, describe and use arithmetic and geometric sequences and their general terms and partial sums.

By the end of this lesson you can:

I state the common difference or common ratio of a sequence.
I find the $n$-th term of arithmetic ($t_n = a + (n-1)d$) and geometric ($t_n = ar^{n-1}$) sequences.
I calculate partial sums $S_n$ for both types using the correct formula.
I determine whether an infinite geometric series converges and find its sum.

Sequences are staircases. Arithmetic staircases rise by the same step each time; geometric ones multiply — gentle at first, then towering.

Arithmetic sequences — constant difference

An arithmetic sequence (AP) is one where each term differs from the previous by a fixed amount called the common difference $d$:

$$t_1, \ t_1 + d, \ t_1 + 2d, \ t_1 + 3d, \ldots$$

Using first term $a = t_1$, the general (n-th) term is:

$$t_n = a + (n-1)d$$

This is a linear function of $n$ — plotting $t_n$ vs $n$ gives a straight line with gradient $d$.

### Partial sum of an AP

The sum of the first $n$ terms, $S_n$, can be found two ways:

Formula 1 (when $a$ and $d$ are known):
$$S_n = \frac{n}{2}[2a + (n-1)d]$$

Formula 2 (when first and last terms are known — Gauss's method):
$$S_n = \frac{n}{2}(a + \ell) \qquad \text{where } \ell = t_n \text{ is the last term}$$

> Gauss's insight: Pair the first and last term, second and second-last, etc. Each pair sums to $a + \ell$. There are $n/2$ such pairs. This gives $S_n = \frac{n}{2}(a+\ell)$.

### Identifying an AP

Sequence is AP $\iff$ consecutive differences are constant:
$$t_2 - t_1 = t_3 - t_2 = \cdots = d \quad \text{(constant)}$$

Geometric sequences — constant ratio

A geometric sequence (GP) is one where each term is obtained by multiplying the previous term by a fixed common ratio $r$:

$$a, \ ar, \ ar^2, \ ar^3, \ldots$$

The general term is:

$$t_n = a r^{n-1}$$

This is an exponential function of $n$ when $r > 0$.

### Identifying a GP

Sequence is GP $\iff$ ratios of consecutive terms are constant:
$$\frac{t_2}{t_1} = \frac{t_3}{t_2} = \cdots = r \quad \text{(constant)}$$

Note: $r$ can be negative (alternating signs) or fractional (decay).

### Partial sum of a GP

$$S_n = \frac{a(1 - r^n)}{1 - r} \qquad (r \neq 1)$$

Equivalently (useful when $r > 1$):

$$S_n = \frac{a(r^n - 1)}{r - 1}$$

### Infinite geometric series

If $|r| < 1$, the terms approach 0 and the series converges:

$$S_\infty = \frac{a}{1 - r} \qquad (|r| < 1)$$

If $|r| \geq 1$, the series diverges (sum is infinite or oscillates).

> Why $|r| < 1$? As $n \to \infty$, $r^n \to 0$ only when $|r| < 1$. Substituting into $S_n$ gives $S_\infty = \frac{a}{1-r}$.

Sequences and Series Formulas

AP: $n$-th term: t_n = a + (n-1)d ($a$ = first term, $d$ = common difference)
AP: partial sum: S_n = \dfrac{n}{2}[2a + (n-1)d] (Or $S_n = \frac{n}{2}(a + \ell)$)
GP: $n$-th term: t_n = ar^{n-1} ($a$ = first term, $r$ = common ratio)
GP: partial sum: S_n = \dfrac{a(1-r^n)}{1-r} ($r \neq 1$)
GP: infinite sum: S_\infty = \dfrac{a}{1-r} ($|r| < 1$ required)
Common difference: d = t_2 - t_1 (Must be constant throughout)
Common ratio: r = \dfrac{t_2}{t_1} (Must be constant throughout)

Sequence problem strategy

Identify AP or GP: Check: are differences constant (AP) or ratios constant (GP)?
Extract $a$ and $d$ or $r$: State the first term $a$ and the common difference/ratio explicitly.
Use correct formula: AP: $t_n = a + (n-1)d$. GP: $t_n = ar^{n-1}$.
Check convergence: For infinite GP: write $|r| < 1$ before applying $S_\infty$.
Reverse engineering: Given $t_n$ or $S_n$, set up an equation in $n$ and solve algebraically.

Quick reference

| Type | nth term | Sum |
|---|---|---|
| Arithmetic | a + (n−1)d | n/2 [2a + (n−1)d] |
| Geometric | arⁿ⁻¹ | a(rⁿ − 1)/(r − 1) |
| Infinite geometric | — | a/(1 − r), −1 < r < 1 |

Worked example (Easy)

An AP has first term $a = 3$ and common difference $d = 5$. Find: (a) the 20th term, and (b) the sum of the first 20 terms.

(a) 20th term. Use $t_n = a + (n-1)d$ with $n = 20$: t_{20} = 3 + (20-1)(5) = 3 + 95 = 98
(b) Sum of 20 terms. Use $S_n = \frac{n}{2}(a + \ell)$ where $\ell = t_{20} = 98$: S_{20} = \frac{20}{2}(3 + 98) = 10 \times 101 = 1010

Answer: $t_{20} = 98$; $S_{20} = 1010$.

Worked example (Medium)

A GP has first term $a = 6$ and common ratio $r = \frac{1}{2}$. Find (a) the sum to infinity, and (b) the smallest $n$ such that $t_n < 0.1$.

(a) Check convergence. $|r| = \frac{1}{2} < 1$ ✓ — converges. S_\infty = \frac{a}{1-r} = \frac{6}{1 - \frac{1}{2}} = \frac{6}{\frac{1}{2}} = 12
(b) Find smallest $n$ with $t_n < 0.1$. Use $t_n = ar^{n-1} = 6 \cdot (\frac{1}{2})^{n-1} < 0.1$: \left(\frac{1}{2}\right)^{n-1} < \frac{0.1}{6} = \frac{1}{60}
Take log (both sides). Since $r = \frac{1}{2} < 1$, taking $\log$ reverses the inequality: (n-1)\log\left(\frac{1}{2}\right) < \log\left(\frac{1}{60}\right)
Divide by $\log(\frac{1}{2})$ — negative, so flip inequality. n - 1 > \frac{\log(1/60)}{\log(1/2)} = \frac{-\log 60}{-\log 2} = \frac{\log 60}{\log 2} \approx \frac{1.778}{0.301} \approx 5.91
Conclude. $n - 1 > 5.91 \Rightarrow n > 6.91 \Rightarrow n \geq 7$ (since $n$ is an integer).

Answer: (a) $S_\infty = 12$. (b) Smallest $n = 7$.

Worked example (Hard)

The 5th term of a GP is 48 and the 8th term is 6. Find the first term and common ratio. Hence find $S_{10}$.

Write two equations. $t_5 = ar^4 = 48$ and $t_8 = ar^7 = 6$. ar^4 = 48 \quad \cdots (1), \qquad ar^7 = 6 \quad \cdots (2)
Divide (2) by (1). This eliminates $a$: \frac{ar^7}{ar^4} = \frac{6}{48} \implies r^3 = \frac{1}{8} \implies r = \frac{1}{2}
Find $a$ from (1). Substitute $r = \frac{1}{2}$: a \left(\frac{1}{2}\right)^4 = 48 \implies \frac{a}{16} = 48 \implies a = 768
Find $S_{10}$. S_{10} = \frac{768(1 - (1/2)^{10})}{1 - 1/2} = \frac{768(1 - 1/1024)}{1/2} = 1536 \times \frac{1023}{1024} = \frac{1536 \times 1023}{1024}
Simplify. $\frac{1536}{1024} = \frac{3}{2}$, so: S_{10} = \frac{3}{2} \times 1023 = \frac{3069}{2} = 1534.5

Answer: $a = 768$, $r = \frac{1}{2}$, $S_{10} = 1534.5$.

Common mistakes

Applying $S_\infty = \frac{a}{1-r}$ when $|r| \geq 1$ The formula $S_\infty = \frac{a}{1-r}$ is only valid when $|r| < 1$. If $r = 1$, all terms are equal and the sum grows without bound. If $r = 2$, terms double each time — obviously diverging. Fix: Always write and check "$|r| < 1$" before applying the infinite sum formula. If $|r| \geq 1$, state the series diverges.
Using $t_n = ar^n$ instead of $t_n = ar^{n-1}$ The first term is $t_1 = ar^0 = a$ (since $r^0 = 1$). If you use $ar^n$, then $t_1 = ar^1 = ar \neq a$ — every answer is off by one factor of $r$. Fix: Verify: set $n = 1$. You should get $t_1 = a$. With $ar^{n-1}$: $a \cdot r^0 = a$ ✓. With $ar^n$: $a \cdot r^1 = ar$ ✗.
Confusing $S_n$ and $t_n$ $t_n$ is the $n$-th term (one value). $S_n$ is the sum of the first $n$ terms. They are related by $t_n = S_n - S_{n-1}$ for $n \geq 2$, but they are not the same. Fix: If the question asks "find the 10th term", use $t_{10}$. If it asks "find the sum of the first 10 terms", use $S_{10}$. Read carefully.

Proof of the GP partial sum formula

Let $S_n = a + ar + ar^2 + \cdots + ar^{n-1}$.

Multiply both sides by $r$:

$$rS_n = ar + ar^2 + ar^3 + \cdots + ar^n$$

Subtract the second equation from the first:

$$S_n - rS_n = a - ar^n$$
$$S_n(1-r) = a(1-r^n)$$
$$S_n = \frac{a(1-r^n)}{1-r} \quad (r \neq 1)$$

This "multiply and subtract" trick is a classic algebraic proof technique. For the infinite sum, let $n \to \infty$: when $|r| < 1$, $r^n \to 0$, leaving $S_\infty = \frac{a}{1-r}$.

Practice questions

The common ratio of the GP $4, -2, 1, -\frac{1}{2}, \ldots$ is:
1. $-\frac{1}{2}$
2. $\frac{1}{2}$
3. $-2$
4. $2$
(1 marks)
Find the sum of the first 15 terms of the AP: $7, 12, 17, 22, \ldots$
1. $697$
2. $690$
3. $712$
4. $600$
(2 marks)
A GP has $a = 3$ and $r = \frac{2}{3}$. What is its sum to infinity?
1. $9$
2. $6$
3. $3$
4. It diverges
(2 marks)
The 4th term of an AP is 19 and the 10th term is 43. Find the first term and common difference.
(3 marks)
Express $0.\overline{27} = 0.272727\ldots$ as a fraction using the infinite GP formula.
(4 marks)

Summary

AP: $t_n = a + (n-1)d$; $S_n = \frac{n}{2}[2a+(n-1)d]$.
GP: $t_n = ar^{n-1}$; $S_n = \frac{a(1-r^n)}{1-r}$; $S_\infty = \frac{a}{1-r}$ when $|r|<1$.
Find $a$ and $d$ or $r$: two equations from two given terms, then solve simultaneously.
Infinite sum exists only if $|r| < 1$ — always state this condition.
Recurring decimals = infinite GPs: write the decimal expansion as $a + ar + ar^2 + \cdots$.