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IB Mathematics AI SL Arithmetic sequences and series Study Notes - New Syllabus

IB Mathematics AI SL Arithmetic sequences and series Study Notes

LEARNING OBJECTIVE

  • Use of the formulae for the nth term and the sum of the first n terms of the sequence. Use of sigma notation for sums of arithmetic sequences.

Key Concepts: 

  • Arithmetic sequences and series.

MAI HL and SL Notes – All topics

SEQUENCES IN GENERAL – SERIES

♦ SEQUENCE
A sequence is an ordered list of numbers (terms). For example:
$2, 5, 13, 5, -4, …$
(1st term, 2nd term, 3rd term, etc.)

Sequences often follow patterns:
Even numbers: $0, 2, 4, 6, 8, …$
Odd numbers: $1, 3, 5, 7, 9, …$
Powers of $2: 2, 4, 8, 16, 32, …$

Notation: \( u_n \) for the \( n \)-th term.

♦ SERIES
A series is the sum of terms:
\( S_n = u_1 + u_2 + \cdots + u_n \) (sum of first \( n \) terms)
\( S_\infty = u_1 + u_2 + \cdots \) (infinite series)

Odd numbers: 1, 3, 5, 7, …
\( u_1 = 1 \), \( u_6 = 11 \), \( u_{10} = 19 \)
 \( S_3 = 1 + 3 + 5 = 9 \)

♦ Sigma Notation (\( \sum \))
Compact representation of sums
 \( \sum_{n=1}^9 u_n = u_1 + u_2 + \cdots + u_9 \)

\( \sum_{n=1}^3 2^n = 2 + 4 + 8 = 14 \)
 \( \sum_{n=1}^4 \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{25}{12} \)
 Infinite series: \( \sum_{n=1}^\infty \frac{1}{2^n} = 1 \)

♦ Describing Sequences
1. General Formula: \( u_n = 2n \) $→ 2, 4, 6, 8, …$
2. Recursive Relation:
 \( u_1 = 10 \), \( u_{n+1} = u_n + 2 \) $→ 10, 12, 14, …$

 \( u_n = n^2 \) $→ 1, 4, 9, 16, …$
\( u_n = 2^n \) $→ 2, 4, 8, 16, …$

(Recursive)
\( u_1 = 3 \), \( u_{n+1} = 2u_n + 5 \) $→ 3, 11, 27, 59, …$

♦ Fibonacci Sequence
 \( u_1 = 1 \), \( u_2 = 1 \), \( u_{n+1} = u_n + u_{n-1} \) $→ 1, 1, 2, 3, 5, 8, …$

ARITHMETIC SEQUENCE (A.S.)

 

♦ Definition
A sequence where the difference (\( d \)) between consecutive terms is constant.
 Example: \( u_1 = 5 \), \( d = 3 \) $→ 5, 8, 11, 14, …$

♦ General Formula:
\( u_n = u_1 + (n-1)d \)

Example

(i)  \( u_1 = 1 \), \( d = 2 \) 
(ii)  \( u_1 = -10 \), \( d = 5 \)

▶️Answer/Explanation

Solution:

(i) $→ 1, 3, 5, 7, …$
(ii) $→ -10, -5, 0, 5, …$

Example

 \( u_1 = 3 \), \( d = 5 \)

▶️Answer/Explanation

Solution:

$→$ \( u_{100} = 3 + 99 \times 5 = 498 \)

♦ Sum of First \( n \) Terms (\( S_n \)):
1. \( S_n = \frac{n}{2}(u_1 + u_n) \)
2. \( S_n = \frac{n}{2}[2u_1 + (n-1)d] \)

Example

Sum of 1 to 100:

▶️Answer/Explanation

Solution:

 \( S_{100} = \frac{100 \times 101}{2} = 5050 \)

Example

Given \( u_3 = 0 \), \( S_{15} = -300 \):
 Solve for \( u_1 = 8 \), \( d = -4 \)

▶️Answer/Explanation

Solution:

 \( S_{10} = -100 \)

♦ Consecutive Terms:
For \( a, x, b \) in A.S., \( x = \frac{a + b}{2} \).

Example

Find \( x \) if \( x+1, 3x, 6x-5 \) are in A.S.

▶️Answer/Explanation

Solution:

\( x = 4 \) (terms: 5, 12, 19).

Real-Life Arithmetic Models: Overview

♦In real-world scenarios, data often does not follow a perfect arithmetic pattern, but it may approximate one. An arithmetic sequence is defined by a common difference between consecutive terms:

$
\text{Arithmetic sequence: } a,\, a + d,\, a + 2d,\, \dots
$

In real-life contexts, the data might not increase or decrease by the exact same amount each time, but close enough that an arithmetic model can be applied for prediction or analysis.

♦Why Real Data Is Not Perfectly Arithmetic

Real-world data can be affected by:
Rounding errors
 Measurement inaccuracies
External factors (e.g., weather, economics)
 Changing rates of growth/decay over time

As a result, the differences between terms may vary slightly.

♦How to Approximate the Common Difference

1. List consecutive values of the sequence.
2. Calculate differences between each pair of consecutive terms.
3. Find an average of the differences to estimate a “common difference”:

$d \approx \frac{\text{Sum of consecutive differences}}{\text{Number of differences}}$

Example

Scenario: A plant’s height is measured weekly.

 

▶️Answer/Explanation

Solution:

Differences
12.3 – 10.2 = 2.1
14.4 – 12.3 = 2.1
16.7 – 14.4 = 2.3
18.6 – 16.7 = 1.9

$
d \approx \frac{2.1 + 2.1 + 2.3 + 1.9}{4} = \frac{8.4}{4} = 2.1
$
$
\text{Height}_n = 10.2 + (n – 1)\cdot2.1
$
$
\text{Height}_n = 10.2 + (n – 1)\cdot2.1
$
$
\text{Height}_8 = 10.2 + (8 – 1)\cdot2.1 = 10.2 + 14.7 = \boxed{24.9\, \text{cm}}
$

♦Interpretation & Limitations

Model is valid only short-term: Over time, the growth rate might change.
Check the residuals (differences between predicted and actual values) to assess model accuracy.
Don’t extrapolate too far beyond the data range.

Graphic Display Calculator TI-84 Plus Codes

♦The TVM (Time-Value-of-Money) Solver in TI-84 Plus is a useful tool for calculating compound interest problems. 

Example

You invest $5000 in a savings account with an annual interest rate of 5%, compounded monthly, and want to know the value of your investment after 10 years.

▶️Answer/Explanation

Solution:

 Step 1:
Press the APPS button on the calculator, then select TVM Solver.

 Step 2:
Enter the following values:

  • N = 120 → (12 months × 10 years)

  • I/Y = 5 → (Annual interest rate as a whole percentage)

  • PV = -5000 → (Initial investment; entered as a negative value because it’s a cash outflow)

  • PMT = 0 → (No additional payments)

  • FV = ? → (We want to find this)

  • P/Y = 12, C/Y = 12 → (Monthly compounding)

Note: Ensure that P/Y and C/Y are both set to 12, since interest is compounded monthly.

 Step 3:
Move the cursor to the FV field, press ALPHA, then ENTER (SOLVE).
The calculator will compute:

FV = $\$8,235.05$

♦Code:

  • APPS → TVM Solver

  • Enter:

    • N = 120

    • I/Y = 5

    • PV = -5000

    • PMT = 0

    • FV = ?

    • P/Y = 12, C/Y = 12

  • Move to FV, press ALPHA → ENTER

  • Result: FV = $8,235.05

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