Sequences and Series

Q: When is an infinite geometric series convergent?

An infinite geometric series converges if and only if the absolute value of the common ratio is less than 1: . If , the series diverges and has no finite sum.

Q: What is the difference between a sequence and a series?

A sequence is an ordered list of individual terms, while a series is the sum of the terms of a sequence.

IB Mathematics: Analysis and Approaches HL· 1.2 Sequences and series· 15 min read

1. Arithmetic Sequences and Series★★☆☆☆⏱ 5 min

📘 Definition

Arithmetic Progression (AP)

First term $u_{1}$ , common difference $d$

An arithmetic progression is a sequence where each term after the first is obtained by adding a constant common difference to the previous term.

Example:

$2, 5, 8, 11, ...$ has $u_{1} = 2, d = 3$

The nth term of an AP follows a linear formula derived from extending the sequence pattern:

u_n = u_1 + (n-1)d

The sum of the first $n$ terms of an AP is derived by pairing first and last terms to get constant sums, giving two equivalent formulas:

S_n = \frac{n}{2} \left( 2u_1 + (n-1)d \right) = \frac{n}{2} (u_1 + u_n)

📐 Worked Example

The 3rd term of an AP is 10, and the 7th term is 34. Find the first term and the sum of the first 10 terms.

1
Write the nth term formula for the two given terms:
2
$u_3 = u_1 + 2d = 10 \\ u_7 = u_1 + 6d = 34$
3
Subtract the first equation from the second to eliminate $u_{1}$ :
4
$(u_1 + 6d) - (u_1 + 2d) = 34 - 10 \\ 4d = 24 \\ d = 6$
5
Substitute $d = 6$ back to find $u_{1}$ :
6
$u_1 + 2(6) = 10 \\ u_1 = -2$
7
Calculate the sum of the first 10 terms:
8
$S_{10} = \frac{10}{2} \left( 2(-2) + 9(6) \right) = 5 (-4 + 54) = 250$

Exam tip:

Always confirm whether the question asks for the nth term or the sum of the first n terms — mixing these up is an extremely common avoidable error.

2. Geometric Sequences and Series★★★☆☆⏱ 5 min

📘 Definition

Geometric Progression (GP)

First term $u_{1}$ , common ratio $r$

A geometric progression is a sequence where each term after the first is obtained by multiplying the previous term by a constant common ratio.

Example:

$3, 6, 12, 24, ...$ has $u_{1} = 3, r = 2$

The nth term of a GP follows an exponential formula, from repeated multiplication by the common ratio:

u_n = u_1 r^{n-1}

The sum of the first $n$ terms of a GP uses the standard finite geometric series formula, which works for all $r \neq = 1$ . When $r = 1$ , all terms are equal to $u_{1}$ , so $S_{n} = n u_{1}$ .

S_n = u_1 \frac{1 - r^n}{1 - r} = u_1 \frac{r^n - 1}{r - 1}, \quad r \neq 1

📐 Worked Example

The 2nd term of a GP is 6, and the 5th term is -162. Find the common ratio and the sum of the first 4 terms.

1
Write the nth term for each given term:
2
$u_2 = u_1 r = 6 \\ u_5 = u_1 r^4 = -162$
3
Divide the second equation by the first to eliminate $u_{1}$ :
4
$\frac{u_1 r^4}{u_1 r} = \frac{-162}{6} \\ r^3 = -27 \\ r = -3$
5
Find $u_{1}$ by substituting back:
6
$u_1 (-3) = 6 \\ u_1 = -2$
7
Calculate the sum of the first 4 terms:
8
$S_4 = (-2) \frac{1 - (-3)^4}{1 - (-3)} = (-2) \frac{1 - 81}{4} = 40$

3. Infinite Geometric Series and Convergence★★★☆☆⏱ 5 min

For an infinite geometric series, the sum of all terms only approaches a finite value if the common ratio satisfies a specific condition, called convergence. If the series does not converge, it diverges and has no finite sum.

📘 Definition

Convergent Infinite Geometric Series

An infinite geometric series converges to a finite sum if and only if $∣ r ∣ < 1$ , where $r$ is the common ratio. The sum to infinity is given by the formula below.

Example:

$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...$ converges to 2

S_\infty = \frac{u_1}{1 - r}, \quad |r| < 1

📐 Worked Example

A convergent infinite geometric series has a first term of 12 and a sum to infinity of 18. Find the common ratio and the 4th term of the series.

1
Use the sum to infinity formula to solve for $r$ :
2
$S_\infty = \frac{u_1}{1 - r} \\ 18 = \frac{12}{1 - r}$
3
Rearrange to find $r$ :
4
$1 - r = \frac{12}{18} = \frac{2}{3} \\ r = 1 - \frac{2}{3} = \frac{1}{3}$
5
Confirm the convergence condition is satisfied: $∣ r ∣ = \frac{1}{3} < 1$ , which is valid.
6
Calculate the 4th term:
7
$u_4 = u_1 r^3 = 12 \left( \frac{1}{3} \right)^3 = 12 \times \frac{1}{27} = \frac{4}{9}$

4. Financial Applications: Compound Interest and Depreciation★★★☆☆⏱ 6 min

✓ Calculator OK

Compound interest and depreciation are direct applications of geometric sequences: each period the amount is multiplied by a fixed factor. This material is on AA SL 1.4, and as HL is a superset of SL you are expected to master it. Because interest can be added more than once a year, the formula includes the compounding frequency $k$ .

📘 Definition

Compound Interest

FV = PV\left(1 + \frac{r}{100k}\right)^{kn}

$P V$ is the present value (initial amount), $F V$ is the future value, $r$ is the nominal annual interest rate as a percentage, $n$ is the number of years, and $k$ is the number of compounding periods per year.

Example:

Yearly $k = 1$ , half-yearly $k = 2$ , quarterly $k = 4$ , monthly $k = 12$ .

The value forms a geometric sequence with common ratio $(1 + \frac{r}{100 k})$ over each compounding period. For a fixed nominal rate, increasing the frequency $k$ increases the future value slightly, because interest starts earning interest sooner. On a GDC you can also use the built-in finance solver (TVM), but you must know the formula.

📐 Worked Example

$5000 is invested at a nominal annual rate of 4%. Find the value after 3 years if interest is compounded (i) yearly, (ii) quarterly, (iii) monthly. Give answers to 2 decimal places.

1
Yearly ( $k = 1$ ): apply the formula with $r = 4$ , $n = 3$ :
2
$FV = 5000\left(1 + \frac{4}{100}\right)^{3} = 5000(1.04)^3 = 5624.32$
3
Quarterly ( $k = 4$ ): the period rate is $\frac{4}{400} = 0.01$ over $4 \times 3 = 12$ periods:
4
$FV = 5000\left(1 + \frac{4}{400}\right)^{12} = 5000(1.01)^{12} = 5634.13$
5
Monthly ( $k = 12$ ): the period rate is $\frac{4}{1200}$ over $12 \times 3 = 36$ periods:
6
$FV = 5000\left(1 + \frac{4}{1200}\right)^{36} = 5636.36$
7
As expected, more frequent compounding gives a slightly larger future value.

📘 Definition

Annual Depreciation

FV = PV\left(1 - \frac{r}{100}\right)^{n}

When an asset loses a fixed percentage $r$ of its value each year, its value forms a geometric sequence with common ratio $(1 - \frac{r}{100})$ . Here $n$ is the number of years.

Example:

A car worth $18000 d e p r ec ia t in g a t 15$ 18000(0.85)^n $a f t er$ n$ years.

Money also loses purchasing power over time through inflation. To compare amounts fairly, the real value of a future amount is found by discounting for inflation. If the inflation rate is $i %$ per year, an amount $F V$ received in $n$ years has real value $\frac{F V}{( 1 + \frac{i}{100} ) ^{n}}$ in today's money.

📐 Worked Example

A savings account grows $8000 t o$ 9724.05 over 5 years. During the same period inflation averages 3% per year. Find the real value of the final amount in today's money, to the nearest dollar.

1
Discount the final amount by the inflation factor over 5 years:
2
$\text{Real value} = \frac{9724.05}{\left(1 + \frac{3}{100}\right)^{5}} = \frac{9724.05}{(1.03)^5}$
3
Evaluate the denominator and divide:
4
$= \frac{9724.05}{1.159274} \approx 8388$
5
So although the balance grew to $9724.05, in t o d a y^{'} s m o n ey i t i s w or t hab o u t$ 8388, still a real-terms gain on the original $8000.

Exam tip:

Read the compounding frequency carefully: 'quarterly' means $k = 4$ and the exponent is $k n$ , not $n$ . A common slip is to use the annual rate over quarterly periods.

5. Common Pitfalls

Wrong move:

Using $n$ instead of $n - 1$ in the nth term formula for AP/GP

Why:

The first term corresponds to $n = 1$ , so the exponent/multiplier must be $1 - 1 = 0$ to get $u_{1}$ as expected. Using $n$ shifts all terms by one position.

Correct move:

Always use $u_{n} = u_{1} + (n - 1) d$ for AP and $u_{n} = u_{1} r^{n - 1}$ for GP, unless the problem explicitly defines the sequence starting at index 0.

Wrong move:

Discarding negative solutions for the common ratio $r$ in GPs

Why:

IB exam questions regularly use negative common ratios, but many students incorrectly discard negative solutions assuming $r$ is positive.

Correct move:

Keep negative values of $r$ unless the problem explicitly states all terms of the sequence are positive.

Wrong move:

Using the infinite sum formula for $∣ r ∣ \geq 1$

Why:

The infinite sum formula only applies to convergent series. If $∣ r ∣ \geq 1$ , terms grow in magnitude and the series diverges.

Correct move:

Always confirm $∣ r ∣ < 1$ before using $S_{\infty} = \frac{u _{1}}{1 - r}$ , and state that the infinite sum does not exist if $∣ r ∣ \geq 1$ .

Wrong move:

Incorrectly counting the number of terms for a sum between two positions

Why:

For example, the sum from the 3rd to 7th term has 5 terms, not $7 - 3 = 4$ terms.

Correct move:

Calculate the number of terms as $(last position - first position) + 1$ , or use $S_{last} - S_{first - 1}$ to get the correct sum automatically.

6. Quick Reference Cheatsheet

Concept	nth term	Sum of first n terms	Key rule/formula
Arithmetic Progression	$u_{n} = u_{1} + (n - 1) d$	$S_{n} = \frac{n}{2} (u_{1} + u_{n})$	Constant common difference $d$
Geometric Progression ( $r \neq = 1$ )	$u_{n} = u_{1} r^{n - 1}$	$S_{n} = u_{1} \frac{1 - r ^{n}}{1 - r}$	Constant common ratio $r$
Geometric Progression ( $r = 1$ )	$u_{n} = u_{1}$	$S_{n} = n u_{1}$	All terms equal to $u_{1}$
Infinite Geometric Series	Same as finite GP	N/A	$S_{\infty} = \frac{u _{1}}{1 - r}, ∣ r ∣ < 1$ ; diverges if $∣ r ∣ \geq 1$

7. Frequently Asked

When is an infinite geometric series convergent?

An infinite geometric series converges if and only if the absolute value of the common ratio is less than 1: $∣ r ∣ < 1$ . If $∣ r ∣ \geq 1$ , the series diverges and has no finite sum.

What is the difference between a sequence and a series?

A sequence is an ordered list of individual terms, while a series is the sum of the terms of a sequence.

Going deeper

syllabusIB AA HL Official Syllabus Topic 1Check for latest syllabus updates

What's Next

Sequences and series are foundational for almost all further topics in IB AA HL, including the binomial theorem, Maclaurin and Taylor series in calculus, and discrete probability models. This topic regularly appears in both Paper 1 and Paper 2 exams, often combined with logarithms, exponentials or financial math to create extended response questions. Mastery of AP/GP formulas and convergence conditions will make it much easier to tackle more advanced series topics later in the syllabus.