# Sequences and Series

> IB Mathematics: Analysis and Approaches HL · Number & Algebra
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u1-sequences-and-series/

This subtopic covers key properties of arithmetic and geometric sequences and series, including calculating nth terms, partial sums, and infinite convergent geometric series, with applications to common IB exam problems.

**Prerequisites:** Basic algebraic manipulation; [Exponent rules](https://www.owlsprep.com/study/ib-math-aa-hl-u1-exponent-rules/); [Absolute value inequalities](https://www.owlsprep.com/study/ib-math-aa-hl-algebra-absolute-value-inequalities/)

## Learning objectives

- Distinguish between arithmetic and geometric sequences and series
- Calculate the nth term and sum of the first n terms for both progressions
- Determine the sum of an infinite convergent geometric series
- Apply geometric sequences to financial contexts: compound interest, depreciation and real value under inflation
- Solve applied and exam-style problems involving sequences and series

## Arithmetic Sequences and Series

**Arithmetic Progression (AP)** — An arithmetic progression is a sequence where each term after the first is obtained by adding a constant common difference to the previous term.

*Notation:* First term $u_1$, common difference $d$

*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.

## Geometric Sequences and Series

**Geometric Progression (GP)** — A geometric progression is a sequence where each term after the first is obtained by multiplying the previous term by a constant common ratio.

*Notation:* First term $u_1$, common ratio $r$

*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. $$S_4 = (-2) \frac{1 - (-3)^4}{1 - (-3)} = (-2) \frac{1 - 81}{4} = 40$$

## Infinite Geometric Series and Convergence

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.

**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. $$u_4 = u_1 r^3 = 12 \left( \frac{1}{3} \right)^3 = 12 \times \frac{1}{27} = \frac{4}{9}$$

## Financial Applications: Compound Interest and Depreciation

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 $k$.

**Compound Interest** — $PV$ is the present value (initial amount), $FV$ 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.

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

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

The value forms a geometric sequence with common ratio $\left(1 + \frac{r}{100k}\right)$ 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.

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

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

*Example:* A car worth $18000 depreciating at 15% per year is worth $18000(0.85)^n$ after $n$ 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 $FV$ received in $n$ years has real value $\dfrac{FV}{\left(1 + \frac{i}{100}\right)^{n}}$ in today's money.

**Worked example:** A savings account grows $8000 to $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 today's money it is worth about $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 $kn$, not $n$. A common slip is to use the annual rate over quarterly periods.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using $n$ instead of $n-1$ in the nth term formula for AP/GP
  - Why it fails: 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: 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:** Discarding negative solutions for the common ratio $r$ in GPs
  - Why it fails: IB exam questions regularly use negative common ratios, but many students incorrectly discard negative solutions assuming $r$ is positive.
  - Correct: Keep negative values of $r$ unless the problem explicitly states all terms of the sequence are positive.
- **Wrong:** Using the infinite sum formula for $|r| \geq 1$
  - Why it fails: The infinite sum formula only applies to convergent series. If $|r| \geq 1$, terms grow in magnitude and the series diverges.
  - Correct: 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:** Incorrectly counting the number of terms for a sum between two positions
  - Why it fails: For example, the sum from the 3rd to 7th term has 5 terms, not $7-3=4$ terms.
  - Correct: Calculate the number of terms as $(\text{last position} - \text{first position}) + 1$, or use $S_{\text{last}} - S_{\text{first}-1}$ to get the correct sum automatically.

## 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 $\|r\| \geq 1$ |

## 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.

- [Binomial Theorem](https://www.owlsprep.com/study/ib-math-aa-hl-u1-binomial-theorem/)
- [Exponents and logarithms](https://www.owlsprep.com/study/ib-math-aa-hl-u1-exponents-and-logarithms/)
- [Counting Principles](https://www.owlsprep.com/study/ib-math-aa-hl-u1-counting-principles/)

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