Home / Edexcel A Level / Study notes

Edexcel IAL - Pure Maths 2- 4.4 Geometric Sequences and Series; Sum to Infinity- Study notes  - New syllabus

Edexcel IAL – Pure Maths 2- 4.4 Geometric Sequences and Series; Sum to Infinity -Study notes- New syllabus

Edexcel IAL – Pure Maths 2- 4.4 Geometric Sequences and Series; Sum to Infinity -Study notes -Edexcel A level Physics – per latest Syllabus.

Key Concepts:

  • 4.4 Geometric Sequences and Series; Sum to Infinity

Edexcel IAL Maths-Study Notes- All Topics

Geometric Sequences and Series

A geometric sequence is a sequence in which each term is obtained by multiplying the previous term by a constant value.

This constant multiplier is called the common ratio, denoted by \( r \).

Geometric Sequence

A geometric sequence has the form:

  

\( a,\ ar,\ ar^2,\ ar^3,\ \ldots \)

where:

  • \( a \) is the first term
  • \( r \) is the common ratio

Formula for the nth Term

The nth term of a geometric sequence is given by:

\( a_n = ar^{\,n-1} \)

This formula allows any term to be found directly.

Geometric Series

A geometric series is the sum of the terms of a geometric sequence.

The sum of the first \( n \) terms is denoted by \( S_n \).

Sum of a Finite Geometric Series

If \( r \ne 1 \), the sum of the first \( n \) terms is:

This formula is valid for any real value of \( r \), except \( r = 1 \).

Proof of the Finite Sum Formula

Let:

\( S_n = a + ar + ar^2 + \cdots + ar^{n-1} \)

Multiply both sides by \( r \):

\( rS_n = ar + ar^2 + \cdots + ar^n \)

Subtract:

\( S_n – rS_n = a – ar^n \)

Factorise:

\( S_n(1 – r) = a(1 – r^n) \)

Divide by \( (1 – r) \):

\( S_n = \dfrac{a(1 – r^n)}{1 – r} \)

Hence proved.

Sum to Infinity of a Geometric Series

If the common ratio satisfies:

\( |r| < 1 \)

then the geometric series is convergent.

As \( n \to \infty \), we have:

\( r^n \to 0 \)

The sum to infinity is written as \( S_\infty \) and is given by:

When Does the Sum to Infinity Exist?

  • If \( |r| < 1 \): sum to infinity exists
  • If \( |r| \ge 1 \): series diverges

Using Logarithms to Find \( n \)

When the sum \( S_n \) or a term is given and \( n \) is required, logarithms may be used.

From:

\( a_n = ar^{n-1} \)

take logarithms:

\( \log a_n = \log a + (n-1)\log r \)

This allows \( n \) to be found.

Example 

Find the 8th term of the geometric sequence:

\( 3,\ 6,\ 12,\ \ldots \)

▶️ Answer / Explanation

\( a = 3,\ r = 2 \)

\( a_8 = 3 \times 2^7 = 3 \times 128 = 384 \)

Answer: 384

Example 

Find the sum to infinity of the geometric series:

\( 5 + 2.5 + 1.25 + \cdots \)

▶️ Answer / Explanation

\( a = 5,\ r = \dfrac{1}{2} \)

Since \( |r| < 1 \), the sum to infinity exists.

\( S_\infty = \dfrac{5}{1 – \frac{1}{2}} = 10 \)

Answer: 10

Example 

The sum of the first \( n \) terms of a geometric series is:

\( S_n = 62 \)

The first term is 2 and the common ratio is 3. Find the value of \( n \).

▶️ Answer / Explanation

Use the sum formula:

\( 62 = \dfrac{2(1 – 3^n)}{1 – 3} \)

\( 62 = \dfrac{2(1 – 3^n)}{-2} \)

\( 62 = 3^n – 1 \)

\( 3^n = 63 \)

Take logarithms:

\( n\log 3 = \log 63 \)

\( n = \dfrac{\log 63}{\log 3} \approx 3.76 \)

Answer: \( n \approx 3.76 \)

Scroll to Top