Summing a Geometric Series Revision Notes for HSC SSCE Mathematics Advanced

Summing a Geometric Series

Introduction

A useful formula exists for calculating the sum of the first $n$ terms of a geometric progression (GP). This method differs significantly from the approach used for arithmetic progressions.

Understanding how to sum a geometric series is essential for solving many mathematical problems, from compound interest calculations to analysing exponential growth patterns.

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The ability to sum geometric series efficiently is a powerful tool in mathematics. While individual terms in a GP can be calculated directly, finding the sum of many terms without a formula would be time-consuming and prone to errors.

Deriving the sum formula

To understand where the formula comes from, let's work through the derivation using a general GP with first term $a$ and common ratio $r$ .

Consider the sum of the first $n$ terms: $a + ar + ar^2 + \cdots + ar^{n-2} + ar^{n-1}$

Let's call this sum $S_n$ . We can write:

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

Now, if we multiply both sides by $r$ :

$rS_n = ar + ar^2 + ar^3 + \cdots + ar^{n-1} + ar^n \text{ ... (equation 2)}$

Notice that most terms in these two equations are the same. If we subtract equation 1 from equation 2:

$(r - 1)S_n = ar^n - a$

Provided that $r \neq 1$ , we can divide both sides by $(r - 1)$ :

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

When $r < 1$ , we can obtain a more convenient form by taking opposites of both the numerator and denominator:

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

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Key Insight: The clever trick here is multiplying the sum by $r$ and then subtracting. This eliminates most terms, leaving only the first and last terms in a simple expression. This technique is unique to geometric progressions and relies on the constant ratio between consecutive terms.

Two formulae for summing a GP

When you know the first term $a$ , the common ratio $r$ , and the number of terms $n$ , you can find the sum using one of these two formulae:

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When $r > 1$ : Use the formula $S_n = \frac{a(r^n - 1)}{r - 1}$

When $r < 1$ : Use the formula $S_n = \frac{a(1 - r^n)}{1 - r}$

The choice of formula depends on whether the common ratio is greater than or less than 1. Both formulae give the same result, but one will be easier to work with depending on the value of $r$ .

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Exam Tip: Choose the formula that avoids negative numbers in the denominator. This will make your calculations cleaner and reduce the chance of sign errors.

Worked examples

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Worked Example 1: Finding the sum of powers

Question: Find the sum of all the powers of 5 from $5^0$ to $5^7$ .

Solution:

The sum $5^0 + 5^1 + \cdots + 5^7$ is a geometric series with $a = 1$ and $r = 5$ .

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

Note that there are 8 terms in total (from $5^0$ to $5^7$ ).

S_8 = \frac{a(r^8 - 1)}{r - 1} S_8 = \frac{1 \times (5^8 - 1)}{5 - 1} S_8 = \frac{390625 - 1}{4} S_8 = \frac{390624}{4} S_8 = 97656

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Worked Example 2: Finding the sum with a negative ratio

Question: Find the sum of the first six terms of the geometric series $2 - 6 + 18 - \cdots$

Solution:

The series $2 - 6 + 18 - \cdots$ is a GP with $a = 2$ and $r = -3$ .

Since r < 1 (remember that $-3 < 1$ ), we use $S_n = \frac{a(1 - r^n)}{1 - r}$

S_6 = \frac{a(1 - r^6)}{1 - r} S_6 = \frac{2 \times (1 - (-3)^6)}{1 - (-3)} S_6 = \frac{2 \times (1 - 729)}{1 + 3} S_6 = \frac{2 \times (-728)}{4} S_6 = \frac{-1456}{4} S_6 = -364

Solving problems involving sums of GPs

When solving problems about geometric series, carefully read the question and write down all given information in symbolic form. This helps you identify which values you know and which you need to find.

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Worked Example 3: Finding the terms given the sum

Question: The sum of the first four terms of a GP with ratio 3 is 200. Find the four terms.

Solution:

We know that $S_4 = 200$ , $r = 3$ , and $n = 4$ .

Using the formula $S_n = \frac{a(r^n - 1)}{r - 1}$ :

\frac{a(3^4 - 1)}{3 - 1} = 200 \frac{a(81 - 1)}{2} = 200 \frac{80a}{2} = 200 40a = 200 a = 5

Therefore, the series is 5 + 15 + 45 + 135 + ...

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Worked Example 4: Using trial-and-error or logarithms

Question:

a) Find a formula for the sum of the first $n$ terms of the GP $2 + 6 + 18 + \cdots$

b) How many terms of this GP must be taken for the sum to exceed one billion?

Solution:

Part a) The sequence is a GP with $a = 2$ and $r = 3$ .

Since $r > 1$ , we use:

S_n = \frac{a(r^n - 1)}{r - 1} S_n = \frac{2(3^n - 1)}{3 - 1} S_n = \frac{2(3^n - 1)}{2} S_n = 3^n - 1

Part b) We need to find $n$ such that $S_n > 1000000000$ .

Using our formula: $3^n - 1 > 1000000000$

$3^n > 1000000001$

Method 1: Trial-and-error

Using a calculator to test values:

$3^{18} = 387420489$ (too small)
$3^{19} = 1162261467$ (exceeds one billion)

Therefore, S₁₉ is the first sum to exceed one billion.

Method 2: Using logarithms

3^n > 1000000001

Taking logarithms of both sides:

n \log_{10} 3 > \log_{10} 1000000001 n > \frac{\log_{10} 1000000001}{\log_{10} 3} n > 18.86\cdots

Since $n$ must be a whole number, we need n = 19.

Therefore, S₁₉ is the first sum to exceed one billion.

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Exam Tip: For simpler problems, trial-and-error is often quicker than using logarithms. However, logarithms provide a more systematic approach for complex problems and are particularly useful when you need to find exact values.

Special case: when the ratio equals 1

When the common ratio of a GP is 1, the standard formula doesn't work because we would be dividing by zero (since the denominator $r - 1$ would equal 0).

However, this case is actually straightforward. If $r = 1$ , then all terms in the series equal the first term $a$ :

$a + a + a + \cdots + a \text{ ($n$ terms)}$

Therefore, the sum is simply:

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When $r = 1$ : $S_n = an$

This makes sense because you're adding the same value $a$ exactly $n$ times.

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This series is also an arithmetic progression with first term $a$ and common difference $d = 0$ . Using the AP formula $S_n = \frac{n}{2}(a + \ell)$ where $\ell = a$ :

$S_n = \frac{n}{2}(a + a) = \frac{n}{2}(2a) = an$

Both approaches give the same result, which provides a nice connection between geometric and arithmetic progressions.

Summary

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Key Points to Remember:

There are two formulae for summing a geometric series: use $S_n = \frac{a(r^n - 1)}{r - 1}$ when $r > 1$ , and $S_n = \frac{a(1 - r^n)}{1 - r}$ when $r < 1$ .
The choice of formula depends on whether the common ratio is greater than or less than 1. Choose the formula that keeps calculations simple and avoids negative denominators.
When solving problems, always write down the known values ( $a$ , $r$ , $n$ , or $S_n$ ) in symbolic form before substituting into the formula.
For problems requiring you to find $n$ , you can use either trial-and-error (testing values systematically) or logarithms (for a more formal solution).
When the ratio equals 1, the formula doesn't apply. Instead, use $S_n = an$ because you're simply adding the same term $n$ times.

Summing a Geometric Series (HSC SSCE Mathematics Advanced): Revision Notes