Finite Geometric Series Revision Notes for Grade 12 NSC Matric Mathematics

Finite Geometric Series

What is a finite geometric series?

A finite geometric series is created when you add up a specific number of terms from a geometric sequence. In a geometric sequence, each term is found by multiplying the previous term by a constant value called the common ratio.

Understanding geometric series is fundamental to many mathematical applications, from calculating compound interest to analysing population growth patterns.

infoNote

A geometric sequence follows a predictable pattern where each term is obtained by multiplying the previous term by the same constant value. This constant multiplier is what we call the common ratio.

General term of a geometric sequence

Before working with geometric series, you need to understand how geometric sequences work. The general formula for the nth term of a geometric sequence is:

$T_n = a \cdot r^{(n-1)}$

Where:

$n$ is the position of the term in the sequence
$T_n$ is the nth term of the sequence
$a$ is the first term
$r$ is the common ratio (constant multiplier)

infoNote

Understanding the Formula Components

The formula $T_n = a \cdot r^{(n-1)}$ tells us that to find any term in the sequence, we multiply the first term by the common ratio raised to the power of (position - 1). This is because the first term already represents $r^0 = 1$ .

Sum formula for finite geometric series

When you want to find the sum of the first n terms of a geometric series, you can use one of two equivalent formulas. Both give the same answer, but one may be easier to use depending on the value of r.

Formula 1: When r < 1

$S_n = \frac{a(1 - r^n)}{1 - r} \text{ where } r \neq 1$

This version is more convenient when the common ratio is less than 1 (like 1/2, 1/3, etc.).

Formula 2: When r > 1

$S_n = \frac{a(r^n - 1)}{r - 1} \text{ where } r \neq 1$

This version is more convenient when the common ratio is greater than 1 (like 2, 3, etc.).

chatImportant

Critical Condition: r ≠ 1

Both formulas require that $r \neq 1$ . If $r = 1$ , then all terms are the same, and the sum is simply $n \times a$ . This is because when the common ratio equals 1, you're just adding the same number n times.

Worked example 1: Basic sum calculation

lightbulbExample

Worked Example: Calculating a Basic Sum

Question: Calculate the sum of the first 6 terms of the series: $32(\frac{1}{2})^{(k-1)}$ where k goes from 1 to 6.

Solution:

Step 1: Write out the first few terms to identify the pattern

When $k = 1$ : $T_1 = 32(\frac{1}{2})^0 = 32$
When $k = 2$ : $T_2 = 32(\frac{1}{2})^1 = 16$
When $k = 3$ : $T_3 = 32(\frac{1}{2})^2 = 8$

The series becomes: $32 + 16 + 8 + ...$

Step 2: Identify the values of a and r

$a = 32$ (first term)
$r = \frac{T_2}{T_1} = \frac{16}{32} = \frac{1}{2}$

Step 3: Apply the sum formula

Since $r = \frac{1}{2} < 1$ , we use: $S_n = \frac{a(1 - r^n)}{1 - r}$

$S_6 = \frac{32(1 - (\frac{1}{2})^6)}{1 - \frac{1}{2}}$
$= \frac{32(1 - \frac{1}{64})}{\frac{1}{2}}$
$= \frac{32(\frac{63}{64})}{\frac{1}{2}}$
$= 2 \times 32 \times \frac{63}{64}$
$= 64 \times \frac{63}{64} = 63$

Answer: The sum equals 63.

Worked example 2: Finding unknown values

lightbulbExample

Worked Example: Finding Unknown Values

Question: A geometric series has $T_1 = -4$ and $T_4 = 32$ . If $S_n = 84$ , find the values of r and n.

Solution:

Step 1: Find the common ratio r

$a = T_1 = -4$
$T_4 = ar^3 = 32$
Therefore: $-4r^3 = 32$
$r^3 = -8$
$r = -2$

Step 2: Write the general term

The series is: $-4 + 8 - 16 + 32 - ...$

Notice the alternating signs because $r < 0$ .

General term: $T_n = -4(-2)^{(n-1)}$

Step 3: Use the sum formula to find n

Since $r = -2$ , we use: $S_n = \frac{a(1 - r^n)}{1 - r}$

$84 = \frac{-4(1 - (-2)^n)}{1 - (-2)}$
$84 = \frac{-4(1 - (-2)^n)}{3}$

Multiply both sides by $\frac{3}{-4}$ :

$-63 = 1 - (-2)^n$
$(-2)^n = 64$
$(-2)^n = (-2)^6$

Therefore: $n = 6$

Answer: r = -2 and n = 6.

Worked example 3: Working backwards from a sum

lightbulbExample

Worked Example: Working Backwards from a Known Sum

Question: Find the value of k if the sum $k(\frac{1}{2})^n$ from n=1 to 8 equals $\frac{255}{64}$ .

Solution:

Step 1: Write out the first few terms

$n = 1$ : $T_1 = k(\frac{1}{2})^1 = \frac{k}{2}$
$n = 2$ : $T_2 = k(\frac{1}{2})^2 = \frac{k}{4}$
$n = 3$ : $T_3 = k(\frac{1}{2})^3 = \frac{k}{8}$

Step 2: Factor out k

The series becomes: $k(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...)$

This means: $k \times [\text{sum of } (\frac{1}{2})^n \text{ from } n=1 \text{ to } 8] = \frac{255}{64}$

Step 3: Find the sum of the geometric series

$a = \frac{1}{2}$
$r = \frac{1}{2}$
$n = 8$
$S_8 = \frac{\frac{1}{2}(1 - (\frac{1}{2})^8)}{1 - \frac{1}{2}}$
$= \frac{\frac{1}{2}(1 - \frac{1}{256})}{\frac{1}{2}}$
$= 1 - \frac{1}{256} = \frac{255}{256}$

Step 4: Solve for k

$k \times \frac{255}{256} = \frac{255}{64}$
$k = \frac{255}{64} \times \frac{256}{255} = \frac{256}{64} = 4$

Answer: k = 4.

Exam tips

Here are essential strategies for success with finite geometric series problems:

Always identify whether r < 1 or r > 1 to choose the most convenient formula
Check that r ≠ 1 before applying the formulas
When r is negative, the series will have alternating positive and negative terms
Show all steps clearly, especially when finding unknown values
Double-check your arithmetic, particularly with fractions and powers

infoNote

Problem-Solving Strategy

When approaching geometric series problems, follow this systematic approach: First identify the first term (a), then find the common ratio (r), determine how many terms you need (n), and finally choose the appropriate sum formula based on whether r is greater than or less than 1.

bookmarkSummary

Key Points to Remember:

A finite geometric series sums a specific number of terms from a geometric sequence
The general term formula is $T_n = a \cdot r^{(n-1)}$
Use $S_n = \frac{a(1 - r^n)}{1 - r}$ when $r < 1$ , and $S_n = \frac{a(r^n - 1)}{r - 1}$ when $r > 1$
Both formulas require r ≠ 1
Always identify a, r, and n before applying the sum formula

Finite Geometric Series (Grade 12 NSC Matric Mathematics): Revision Notes