Ever wonder how a silly online challenge can suddenly explode, captivating millions overnight? Or how a small startup’s user base can skyrocket in a matter of weeks? It might seem like magic, but it’s pure mathematics in action. This is the world of exponential growth, and its engine is something we call a Geometric Sequence. Today, we’re not just going to learn the formulas; we’re going to crack the code behind this rapid growth and give you the tools to predict it, analyze it, and absolutely own it on your next IB exam.
Table of Contents
What’s a Geometric Sequence?
Let’s get straight to it. Imagine a list of numbers where you get from one term to the next by always multiplying by the same number. That’s it. That’s a geometric sequence. This special multiplier has a name: the common ratio, which we denote with the letter $r$.
Definition: Geometric Sequence
A sequence of numbers where each term after the first ($u_1$) is found by multiplying the previous term by a fixed, non-zero number called the common ratio ($r$).
For example, in the sequence $5, 15, 45, 135, …$, the first term $u_1 = 5$ and the common ratio $r = 3$.
Pinpointing a Specific Term ($u_n$)
What if you wanted to find the 20th term in that sequence? You wouldn’t want to multiply by 3 nineteen times! Luckily, there’s a formula for that. This formula is your best friend for finding the value of any term ($u_n$) down the line.
Rule: The General Term Formula
The formula to find the $n^{th}$ term of a geometric sequence is:
$$u_n = u_1 r^{n-1}$$
Where:
- $u_n$ is the term you want to find.
- $u_1$ is the very first term of the sequence.
- $r$ is the common ratio.
- $n$ is the position of the term in the sequence (e.g., 5th, 20th, etc.).
Formula Booklet Alert!
Good news! You don’t have to memorize this formula. It’s right there in your IB Math AA/AI SL & HL formula booklet. Your job is to know what each variable means and when to use it.
From Sequence to Series
So, what happens when you stop just listing the terms and start adding them up? You get a Geometric Series. Instead of $5, 15, 45, …$, you’re looking at $5 + 15 + 45 + …$ This is crucial for answering questions like, “What’s the *total* number of participants after one week?”
Summing a Finite Number of Terms ($S_n$)
Just like with the general term, we have a handy formula (actually, two versions of the same formula) to find the sum of the first $n$ terms, which we call $S_n$.
Rule: Sum of the First n Terms
$$S_n = \frac{u_1(r^n – 1)}{r-1} \quad \text{or} \quad S_n = \frac{u_1(1 – r^n)}{1-r}$$
Both formulas are identical and also in your formula booklet. A good pro-tip is to use the first version when $|r| > 1$ and the second when $|r| < 1$ to keep the denominator positive and avoid sign errors!
The Magic of an Infinite Sum ($S_\infty$)
This is where things get really cool. What if you added up the terms of a sequence… forever? Sometimes, this sum will just grow to infinity. But other times, if the terms are getting smaller and smaller, the sum will actually approach a single, finite number. This is called a convergent series.
Rule: Sum to Infinity
The sum to infinity exists only if the convergence condition is met: $|r| < 1$. This means $r$ must be between -1 and 1.
If the condition is met, the formula is:
$$S_\infty = \frac{u_1}{1-r}$$
IB Exam Power-Ups: Pro Tips
Alright, let’s talk strategy. Knowing the formulas is one thing, but applying them under pressure is another. Here are some key tips to keep in mind.
Exam Hints
- Nail $u_1$ and $r$ First: Before you even think about which formula to use, read the question carefully and identify the first term ($u_1$) and the common ratio ($r$). Getting these wrong means everything that follows will be wrong.
- $n-1$ vs. $n$: This is a classic slip-up. The power in the general term formula ($u_n$) is $n-1$. The power in the sum formula ($S_n$) is $n$. Double-check your exponents!
- State the Condition: When you use the sum to infinity formula, you MUST state why you’re allowed to use it. Explicitly write down that $|r| < 1$. If it’s not, state that the series diverges and the sum to infinity does not exist. This shows the examiner you understand the theory.
- Logarithms are Your Friend: If a question asks “on which day…” or “how many terms…”, it’s asking you to find $n$. Since $n$ is often in the exponent, you’ll almost certainly need to use logarithms to solve for it. Make sure your log rules are sharp!
Worked Example: The FitApp Frenzy
Time to put this all together with a fresh example. Let’s see how theory becomes practice.
IB Style Question
A new fitness app, “FitApp,” launches. On its first day, 500 users sign up. The marketing team finds that each subsequent day, the number of new sign-ups is 20% greater than the previous day.
(a) How many new users sign up on the 10th day?
(b) What is the total number of users who have signed up after two weeks (14 days)?
(c) The company’s server capacity can handle 50,000 users. On which day will the total number of users first exceed this capacity?
Solution Walkthrough
Step 1: Identify $u_1$ and $r$.
The number of users on day 1 is 500, so $u_1 = 500$.
The number of new users grows by 20% each day. This means we multiply by 120%, or 1.2. So, $r = 1.2$.
Part (a): New users on day 10.
This is asking for a specific term, $u_{10}$. We use the general term formula: $u_n = u_1 r^{n-1}$.
$u_{10} = 500 (1.2)^{10-1} = 500 (1.2)^9$
$u_{10} \approx 500 (5.15978) \approx 2579.89$
Since we can’t have a fraction of a user, we round to the nearest whole number. Approximately 2,580 new users sign up on the 10th day.
Part (b): Total users after 14 days.
“Total” is our keyword for a sum. We need to find $S_{14}$ using the sum formula. Since $r > 1$, let’s use the version with $(r^n – 1)$.
$S_n = \frac{u_1(r^n – 1)}{r-1}$
$S_{14} = \frac{500((1.2)^{14} – 1)}{1.2-1}$
$S_{14} = \frac{500(12.8391… – 1)}{0.2}$
$_{14} = \frac{500(11.8391…)}{0.2} \approx 29597.9$
Approximately 29,598 total users have signed up after 14 days.
Part (c): Day total exceeds 50,000.
Here we need to find the value of $n$ for which $S_n$ first goes over 50,000. We set up an inequality.
$S_n > 50000$
$\frac{500((1.2)^n – 1)}{1.2-1} > 50000$
$\frac{500((1.2)^n – 1)}{0.2} > 50000$
$2500((1.2)^n – 1) > 50000$
$(1.2)^n – 1 > \frac{50000}{2500}$
$(1.2)^n – 1 > 20$
$(1.2)^n > 21$
Now, we unleash the logarithms to solve for $n$. Let’s use the natural log ($\ln$).
$\ln((1.2)^n) > \ln(21)$
$n \ln(1.2) > \ln(21)$
$n > \frac{\ln(21)}{\ln(1.2)}$
$n > \frac{3.0445…}{0.1823…} \approx 16.7$
Since $n$ must be an integer representing the day, it must be the next whole number up. Therefore, on the 17th day, the total number of users will first exceed 50,000.
Wrapping It Up
And there you have it! Geometric sequences and series are the mathematical backbone of exponential growth. By mastering the core formulas for $u_n$, $S_n$, and $S_\infty$, and by knowing when and how to apply them (especially with logarithms!), you can tackle some of the most challenging and interesting problems in the IB curriculum. From viral trends to financial investments, the applications are everywhere.
Have questions or want to discuss a problem? Share your thoughts in the comments below! Engaging with the material and your peers is a fantastic way to deepen your understanding and analytical skills in mathematics.
Additional Practice
Question 1: Find the 8th term of the geometric sequence 3, 12, 48, …
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Solution: First, identify $u_1$ and $r$. Here, $u_1 = 3$. To find $r$, divide the second term by the first: $r = \frac{12}{3} = 4$.
Use the formula $u_n = u_1 r^{n-1}$ with $n=8$.
$u_8 = 3 \times (4)^{8-1} = 3 \times 4^7 = 3 \times 16384 = 49152$.
Question 2: A geometric sequence has a first term of 100 and a common ratio of 0.5. Find the sum of its first 6 terms.
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Solution: We have $u_1 = 100$, $r = 0.5$, and $n = 6$. Since $|r| < 1$, it’s convenient to use $S_n = \frac{u_1(1 – r^n)}{1-r}$.
$S_6 = \frac{100(1 – (0.5)^6)}{1 – 0.5} = \frac{100(1 – 0.015625)}{0.5} = \frac{100(0.984375)}{0.5} = 196.875$.
Question 3: Find the sum to infinity of the geometric series 24 + 12 + 6 + 3 + …
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Solution: Here, $u_1 = 24$. The common ratio is $r = \frac{12}{24} = 0.5$.
First, check the convergence condition: Is $|r| < 1$? Yes, $|0.5| < 1$, so the sum to infinity exists.
Use the formula $S_\infty = \frac{u_1}{1-r}$.
$S_\infty = \frac{24}{1 – 0.5} = \frac{24}{0.5} = 48$.
Question 4: The 3rd term of a geometric sequence is 18 and the 6th term is 486. Find the first term ($u_1$) and the common ratio ($r$).
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Solution: We have two equations: $u_3 = u_1 r^{3-1} = u_1 r^2 = 18$ and $u_6 = u_1 r^{6-1} = u_1 r^5 = 486$.
Divide the second equation by the first: $\frac{u_1 r^5}{u_1 r^2} = \frac{486}{18}$.
$r^3 = 27$, so $r = 3$.
Now substitute $r=3$ back into the first equation: $u_1 (3)^2 = 18 \Rightarrow u_1 \times 9 = 18 \Rightarrow u_1 = 2$.
Question 5: For the sequence 4, 6, 9, …, which term is the first to exceed 100?
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Solution: We have $u_1 = 4$ and $r = \frac{6}{4} = 1.5$. We want to find the smallest integer $n$ such that $u_n > 100$.
$u_1 r^{n-1} > 100 \Rightarrow 4 (1.5)^{n-1} > 100 \Rightarrow (1.5)^{n-1} > 25$.
Take logarithms: $(n-1)\ln(1.5) > \ln(25)$.
$n-1 > \frac{\ln(25)}{\ln(1.5)} \approx \frac{3.2188}{0.4054} \approx 7.93$.
$n > 8.93$. Since $n$ must be an integer, the first term to exceed 100 is the 9th term.
Question 6: How many terms of the series 2 + 8 + 32 + … are needed for the sum to exceed 1,000,000?
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Solution: Here, $u_1 = 2$ and $r = 4$. We want to find $n$ where $S_n > 1,000,000$.
$\frac{2(4^n – 1)}{4-1} > 1000000 \Rightarrow \frac{2(4^n – 1)}{3} > 1000000$.
$4^n – 1 > 1500000 \Rightarrow 4^n > 1500001$.
$n \ln(4) > \ln(1500001) \Rightarrow n > \frac{\ln(1500001)}{\ln(4)} \approx \frac{14.22}{1.386} \approx 10.24$.
Since n must be an integer, 11 terms are needed.
Question 7: A bouncy ball is dropped from a height of 10 meters. After each bounce, it reaches 80% of its previous height. Find the total vertical distance traveled by the ball before it comes to rest.
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Solution: This is a tricky one! The total distance is the initial drop PLUS the sum of all the ups and downs.
Initial drop: 10 m.
Upward distances: The first upward journey is $10 \times 0.8 = 8$. The next is $8 \times 0.8$, and so on. This is a geometric series with $u_1 = 8$ and $r=0.8$. The total upward distance is $S_\infty = \frac{8}{1-0.8} = \frac{8}{0.2} = 40$ m.
Downward distances (after the first drop): The first downward journey is also 8 m, the next is $8 \times 0.8$, etc. This is the exact same series, so the total downward distance (after the first drop) is also 40 m.
Total distance = Initial Drop + Total Up + Total Down = $10 + 40 + 40 = 90$ meters.
Question 8: A geometric series has a sum to infinity of 20 and a first term of 4. What is the common ratio?
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Solution: Use the formula $S_\infty = \frac{u_1}{1-r}$.
$20 = \frac{4}{1-r}$.
$20(1-r) = 4 \Rightarrow 20 – 20r = 4 \Rightarrow 16 = 20r \Rightarrow r = \frac{16}{20} = 0.8$.
Question 9: The sum of the first two terms of a geometric series is 15, and the sum to infinity is 27. Find the possible values for the first term and common ratio.
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Solution: We have two equations:
1) $u_1 + u_2 = u_1 + u_1r = u_1(1+r) = 15$
2) $S_\infty = \frac{u_1}{1-r} = 27$
From (2), $u_1 = 27(1-r)$. Substitute this into (1):
$27(1-r)(1+r) = 15 \Rightarrow 27(1-r^2) = 15$
$1-r^2 = \frac{15}{27} = \frac{5}{9} \Rightarrow r^2 = 1 – \frac{5}{9} = \frac{4}{9}$
So, $r = \frac{2}{3}$$ or $$r = \frac{-2}{3}$. (Both satisfy $|r|<1$).
If $r = \frac{2}{3}$, then $u_1 = 27(1 – \frac{2}{3}) = 27(\frac{1}{3}) = 9$.
If $r = \frac{-2}{3}$, then $u_1 = 27(1 – (\frac{-2}{3})) = 27(\frac{5}{3}) = 45$.
The two possible pairs are ($u_1=9, r=\frac{2}{3}$) and ($u_1=45, r=\frac{-2}{3}$).
Question 10: IB Exam Style Question
In a geometric sequence, the fourth term is 135 and the seventh term is 3645.
(a) Find the common ratio.
(b) Find the first term.
(c) Find the sum of the first 10 terms of the sequence.
Show Answer
Solution:
(a) Find the common ratio.
We are given $u_4 = u_1 r^3 = 135$ and $u_7 = u_1 r^6 = 3645$.
To find $r$, we can divide the equation for $u_7$ by the equation for $u_4$:
$\frac{u_1 r^6}{u_1 r^3} = \frac{3645}{135}$
$r^3 = 27$
$r = 3$
(b) Find the first term.
Substitute $r=3$ into the equation for $u_4$:
$u_1 (3)^3 = 135$
$u_1 \times 27 = 135$
$u_1 = \frac{135}{27} = 5$
(c) Find the sum of the first 10 terms.
Now we have $u_1 = 5$, $r = 3$, and we need to find $S_{10}$.
Using the formula $S_n = \frac{u_1(r^n – 1)}{r-1}$:
$S_{10} = \frac{5(3^{10} – 1)}{3-1}$
$S_{10} = \frac{5(59049 – 1)}{2}$
$S_{10} = \frac{5(59048)}{2} = 5 \times 29524 = 147620$