Sum of a GP - Formula, Derivation & Examples
Sum of a GP - Formula, Derivation & Examples
TL;DR
The sum of the first n terms of a geometric progression is
$$ S_n = \frac{a(1 - r^n)}{1 - r} $$
when the common ratio $ r \neq 1 $, and $ S_n = na $ when $ r = 1 $. Here $ a $ is the first term and $ r $ is the common ratio. This article derives the formula by the multiply-and-subtract trick, explains why the $ r=1 $ case is separate, and works six examples from clean integers to a folding-paper word problem.
What Is The Sum Of A GP?
The sum of a GP is the result of adding the terms of a geometric progression - a sequence where each term is the previous term multiplied by a fixed number called the common ratio. If the progression is $ a, ar, ar^2, ar^3, \dots $, then the sum of its first $ n $ terms, written $ S_n $, is the total of those $ n $ terms added together.
What Is The Formula For The Sum Of n Terms Of A GP?
There are two cases, and choosing the right one is the whole game.
When $ r \neq 1 $:
$$ S_n = \frac{a(1 - r^n)}{1 - r} $$
An equivalent form, handy when $ r > 1 $ so you work with positive numbers:
$$ S_n = \frac{a(r^n - 1)}{r - 1} $$
When $ r = 1 $:
$$ S_n = na $$
Variable glossary:
| Symbol | Meaning |
|---|---|
| $ a $ | the first term of the progression |
| $ r $ | the common ratio (each term ÷ the one before) |
| $ n $ | the number of terms being added |
| $ S_n $ | the sum of those first $ n $ terms |
Where Does The Formula Come From?
Write the sum out in full:
$$ S_n = a + ar + ar^2 + \dots + ar^{n-1} $$
Now multiply every term by $ r $:
$$ rS_n = ar + ar^2 + ar^3 + \dots + ar^n $$
Line the two equations up and subtract the second from the first:
$$ S_n - rS_n = a - ar^n $$
Factor both sides:
$$ S_n(1 - r) = a(1 - r^n) $$
Divide by $(1 - r)$:
$$ S_n = \frac{a(1 - r^n)}{1 - r} $$
The $ r=1 $ case is straightforward since there is no division by zero. If $ r=1 $, adding $ n $ terms equals $ n \cdot a $.
Examples Of The Sum Of A GP
Example 1
Find the sum of the first 5 terms of 1, 2, 4, 8, 16.
Here $ a=1 $, $ r=2 $, $ n=5 $. Since $ r>1 $, use the $ r^n - 1 $ form:
$$ S_5 = \frac{1(2^5 - 1)}{2 - 1} = 31 $$
Example 2
Find the sum of the first 6 terms of 4, 8, 16,…
Adjusting for $ r=2 $, we find:
$$ S_6 = \frac{4(2^6 - 1)}{2 - 1} = 252 $$
Example 3
Find the sum of the first 8 terms of 1, 1/2, 1/4,…
With $ r=1/2 $, we calculate:
$$ S_8 = \frac{1(1 - (1/2)^8)}{1 - 1/2} = \frac{255}{128} $$
Example 4
A GP has first term 5 and common ratio 3. Find the sum of the first 4 terms.
Thus:
$$ S_4 = \frac{5(3^4 - 1)}{3 - 1} = 200 $$
Example 5
A GP with negative ratio: find the sum of the first 5 terms of 2, −6, 18, −54…
Calculating:
$$ S_5 = \frac{2((-3)^5 - 1)}{-3 - 1} = 122 $$
Example 6
A folding-paper problem. Each fold doubles the thickness. Building on:
$$ S_7 = 0.2(2^7 - 1) = 25.4 \text{ mm} $$
Why The Multiply-And-Subtract Trick Matters Beyond GPs
Geometric growth is multiplicative and appears in various real-world applications.
Common Errors That Cost Marks
Mistake 1: Forgetting to check whether $ r = 1 $.
Mistake 2: Using the wrong power on $ r $.
Mistake 3: Sign slips when $ r $ is negative.
Practice Questions On The Sum Of A GP
- Find the sum of the first 6 terms of 2, 4, 8….
- A GP has first term 3 and common ratio 1. Find the sum of its first 5 terms.
- Find the sum of the first 4 terms of 1, 1/3, 1/9….
Answers:
- $ S_6 = 126 $
- $ S_5 = 15 $
- $ S_4 = \frac{40}{27} $
The Short Version
The sum of a GP for the first $ n $ terms is $$ S_n = \frac{a(1 - r^n)}{1 - r} $$ when $ r \neq 1 $.
When $ r = 1 $, every term is equal, so $$ S_n = na $.
Identify $ a $, $ r $, and $ n $ before choosing a formula.