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

IB Mathematics AI SL Geometric 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 geometric sequences.

Key Concepts: 

  • geometric sequences and series.

MAI HL and SL Notes – All topics

GEOMETRIC SEQUENCE (G.S.)

 GEOMETRIC SEQUENCE (G.S.)

♦ Definition

Sequence where the ratio (\( r \)) between consecutive terms is constant.
Example: \( u_1 = 5 \), \( r = 2 \) $→ 5, 10, 20, 40, …$

♦ General Formula:

\( u_n = u_1 r^{n-1} \)

♦ Consecutive Terms:

For \( a, x, b \) in G.S., \( x^2 = ab \).

♦ Sum of First \( n \) Terms (\( S_n \)):

\( S_n = \frac{u_1(r^n – 1)}{r – 1} \quad (r \neq 1) \)

♦ Sum to infinity (when \( |r| < 1 \)):

\( S_\infty = \frac{a}{1 – r} \)

Example

Find the Terms

(i) \( u_1 = 1 \), \( r = -2 \)

(ii) \( u_1 = 1 \), \( r = \frac{1}{2} \)

▶️Answer/Explanation

Solution:

(i) $→ 1, -2, 4, -8, …$

(ii) $→ 1,$ \( \frac{1}{2} \), \( \frac{1}{4} \), …

Example

\( u_1 = 3 \), \( r = 2 \) find $u_{100}$

▶️Answer/Explanation

Solution:

$→$ \( u_{100} = 3 \times 2^{99} \)

Example

Sum of \( 2 + 4 + 8 + \cdots + 2^{10} \)

▶️Answer/Explanation

Solution:

\( S_{10} = 2046 \)

Example

For 10, \( x \), 90 in G.S.

▶️Answer/Explanation

Solution:

\( x = \pm 30 \)

Sigma Notation (Geometric Sequences)

♦ Definition

Compact representation of geometric sums

♦ A geometric sequence has the form:

$u_n = ar^{n-1}$

where:

$a$ is the first term
$r$ is the common ratio
$n$ is the term number

♦ In Sigma notation,

$\sum_{n=1}^{k} ar^{n-1}$

♦ This represents:

$ar^0 + ar^1 + ar^2 + \cdots + ar^{k-1}$

Example

Evaluate:

$
\sum_{n=1}^{4} 3 \cdot 2^{n-1}
$

Show all the Steps.

▶️Answer/Explanation

Solution:

This is a geometric sequence with first term $a = 3$ and common ratio $r = 2$:

$
3 \cdot 2^{0} + 3 \cdot 2^{1} + 3 \cdot 2^{2} + 3 \cdot 2^{3} = 3 + 6 + 12 + 24 = \boxed{45}
$

Example

Evaluate:

$
\sum_{n=0}^{5} 5 \cdot \left(\frac{1}{3}\right)^n
$

Show all the Steps.

▶️Answer/Explanation

Solution:

This is a geometric sequence with $a = 5$, $r = \frac{1}{3}$:

$
5 + \frac{5}{3} + \frac{5}{9} + \frac{5}{27} + \frac{5}{81} + \frac{5}{243}
$

Convert to a single fraction (optional) or use geometric sum formula:

$
S = a \cdot \frac{1 – r^{n+1}}{1 – r} = 5 \cdot \frac{1 – \left(\frac{1}{3}\right)^6}{1 – \frac{1}{3}} = 5 \cdot \frac{1 – \frac{1}{729}}{\frac{2}{3}} = \boxed{\frac{3640}{81}} \approx \boxed{44.94}
$

APPLICATIONS

♦ Real-Life Applications of Geometric Sequences

ScenarioDescription
Population growthPopulations growing by a fixed percentage each year follow geometric growth.
Spread of diseaseInfection rates that double or triple over time show geometric increase.
Salary changesAnnual raises or deductions by a fixed percentage are modeled geometrically.
DepreciationValue of a car or asset decreasing by a fixed % yearly.

Disease Spread

A virus spreads to twice as many people each day. On day 1, 5 people are infected.

\( a = 5 \), \( r = 2 \)

Day 4 infected: ?

Total infected after 5 days: ?

▶️ Answer/Explanation

Day 4 infected:
\( u_4 = 5 \cdot 2^3 = 40 \)

Total after 5 days:
\( S_5 = 5 \cdot \frac{1 – 2^5}{1 – 2} = 5 \cdot 31 = 155 \)

Salary Increase

A worker earns \$30,000 and receives a 5% raise annually.

\( a = 30{,}000 \), \( r = 1.05 \)

After 3 years: ?

Total earned over 5 years: ?

▶️ Answer/Explanation

After 3 years:
\( u_3 = 30{,}000 \cdot (1.05)^2 = 33{,}075 \)

Total over 5 years:
\( S_5 = 30{,}000 \cdot \frac{1 – (1.05)^5}{1 – 1.05} \approx 165{,}765 \)

Depreciation of Car

A car worth \$20,000 depreciates by 20% per year.

\( a = 20{,}000 \), \( r = 0.8 \)

Value after 4 years: ?

▶️ Answer/Explanation

Value after 4 years:
\( u_5 = 20{,}000 \cdot (0.8)^4 = 8{,}192 \)

Interpretation:

\( r > 1 \) → exponential growth
\( 0 < r < 1 \) → exponential decay

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