Hey there, fellow math explorers! Ever wondered how we can predict the future, or at least model growth and change, with numbers? Today, we’re diving deep into one of the most practical and fascinating topics in IB Math: Sequences and Series. It’s not just abstract theory; it’s a powerful toolkit for understanding everything from compound interest to, well, the relentless expansion of a bacterial colony!
To kick things off, let’s confront a gripping real-world scenario: The Bacterial Colony Conundrum. Imagine a scientific lab observing a unique strain of bacteria. On the very first day, they count 150 individual bacteria. Here’s the twist: each subsequent day, the colony robustly grows by 8% of its population from the day before. Pretty intense, right?
We’ve got a couple of pressing questions to tackle:
- What will the bacterial population be on the 10th day?
- The colony’s environment can only sustain a grand total of 6000 bacteria cumulatively before the entire system collapses. Will this happen within the first 18 days of observation?
This isn’t just a biology puzzle; it’s a perfect playground for sequences and series. Ready to unlock the mathematical magic required to solve it?
Table of Contents
- Unraveling the Basics: Sequences and Series
- The General Term: Your Sequence Map
- When to Stop? Finite vs. Infinite
- Summing It All Up: What is a Series?
- Solving the Bacterial Colony Conundrum
- Step 1: Identify and Classify
- Part A: What is the Population on Day 10?
- Step 2: Apply the General Term Formula
- Part B: Will the Colony Crash by Day 18?
- Step 3: Use the Sum Formula for a Geometric Series
- Step 4: The Final Verdict
- Wrapping It Up
- Additional Practice
Unraveling the Basics: Sequences and Series
Before we dive into bacterial chaos, let’s get our foundational definitions solid. Trust me, these are the bread and butter of understanding this topic!
Definition: Sequence
A sequence is simply an ordered list of numbers, often called terms. Think of it like a neatly arranged line of numbers, where each one has a specific spot.
Example: A Simple Sequence
Consider the sequence: $$5, 10, 15, 20, \dots$$
Here, 5 is the first term, 10 is the second, and so on. The ellipsis ($$\dots$$) tells us it continues.
Each number in a sequence is called a term, and we denote it using $u_n$, where $n$ represents its position in the sequence. So, for our example, $u_1 = 5$, $u_2 = 10$, $u_3 = 15$, and so forth.
The General Term: Your Sequence Map
The real power comes from the general term, also written as $u_n$. This is a formula that allows you to find *any* term in the sequence just by knowing its position $n$. It’s like having a GPS for your sequence!
Example: Finding the General Term
For the sequence $$5, 10, 15, 20, \dots$$ you might quickly spot that each term is simply 5 multiplied by its position. So, the general term is $$u_n = 5n$$.
Let’s check:
- For $n=1$, $u_1 = 5(1) = 5$
- For $n=2$, $u_2 = 5(2) = 10$
- For $n=3$, $u_3 = 5(3) = 15$
It works!
IB Exam Hint: Master the General Term!
Finding the general term for various types of sequences (especially arithmetic and geometric) is a fundamental skill often tested in IB Math exams (both Paper 1 and Paper 2). While the formulas for specific types are in your booklet, understanding how to derive $u_n$ from a pattern is crucial!
When to Stop? Finite vs. Infinite
Sequences can either have an end or continue indefinitely:
Finite vs. Infinite Sequences
- A finite sequence has a limited, countable number of terms. Example: $$1, 3, 5, 7$$
- An infinite sequence continues without end, indicated by an ellipsis ($$\dots$$). Example: $$1, 3, 5, 7, \dots$$
Summing It All Up: What is a Series?
Definition: Series
While a sequence is a list, a series is the sum of the terms in a sequence. We often use summation notation (the Greek capital letter sigma, $$\sum$$) to represent a series.
Example: Series Sum
For the sequence $$1, 3, 5$$, the series would be $$1 + 3 + 5 = 9$$.
In summation notation, this is $$\sum_{k=1}^{3} u_k = u_1 + u_2 + u_3$$.
More generally, the sum of the first $n$ terms is $$\sum_{k=1}^{n} u_k = u_1 + u_2 + \dots + u_n$$.
Key Distinction & IB Link
It’s absolutely crucial to distinguish between a sequence (a list) and a series (a sum). The IB Math AA Formula Booklet (Section 1.2) provides explicit formulas for the sum of arithmetic and geometric series ($S_n$), which will be your best friend in exam situations.
IB Internal Assessment (IA) Connection: Sequences and series are fantastic for modelling! Think population dynamics, financial investments (compound interest), radioactive decay, or even the spread of information online. They allow you to investigate growth patterns or cumulative totals in a quantitative way.
Solving the Bacterial Colony Conundrum
Alright, foundations laid! Let’s get back to our rapidly expanding bacterial friends. Remember our setup:
- Initial population (Day 1): 150 bacteria.
- Daily increase: 8% of the previous day’s size.
Step 1: Identify and Classify
The first step in any sequences and series problem is to figure out what type of sequence you’re dealing with. Is it arithmetic, geometric, or something else?
Since the colony increases by 8% of its previous day’s size, this means we’re multiplying the previous day’s count by a constant factor. An 8% increase means we’re retaining 100% and adding 8%, so we multiply by $1.08$. This constant multiplier is the hallmark of a Geometric Sequence.
From this, we can extract our key parameters:
- First term ($u_1$): The initial population on Day 1, which is $$u_1 = 150$$.
- Common ratio ($r$): The factor by which the population grows each day, which is $$r = 1.08$$.
Part A: What is the Population on Day 10?
This question asks for the population on a specific day, not a cumulative total. So, we’re looking for a specific term in the sequence.
Step 2: Apply the General Term Formula for a Geometric Sequence
Your trusty IB Formula Booklet (Section 1.2) reminds us that for a geometric sequence, the general term is:
$$u_n = u_1 r^{n-1}$$
We want to find the population on the 10th day, so $n=10$. Let’s plug in our values:
$$u_{10} = 150 (1.08)^{10-1}$$
$$u_{10} = 150 (1.08)^9$$
Now, let’s calculate $$(1.08)^9$$ first:
$$(1.08)^9 \approx 1.9990046$$
Then multiply by $150$:
$$u_{10} = 150 \times 1.9990046 \approx 299.85069$$
Important Note on Rounding!
Since we’re talking about individual bacteria, a fractional number doesn’t make sense in this context. We should round to the nearest whole number. So, $u_{10} \approx 300$ bacteria.
Always consider the real-world context of your answer. Would a half-bacterium make sense? Probably not!
So, on the 10th day, the colony will have approximately 300 bacteria.
Part B: Will the Colony Crash by Day 18?
This question is all about the cumulative total. We need to sum up all the bacteria from Day 1 to Day 18. This means we’re dealing with a series.
Step 3: Use the Sum Formula for a Geometric Series
Again, consult your IB Formula Booklet (Section 1.2). The sum of the first $n$ terms of a geometric series is:
$$S_n = \frac{u_1(r^n – 1)}{r-1}$$
We need the sum for the first 18 days, so $n=18$. Let’s substitute our values ($u_1=150$, $r=1.08$):
$$S_{18} = \frac{150((1.08)^{18} – 1)}{1.08 – 1}$$
Let’s break down the calculation:
- Calculate $$(1.08)^{18}$$: $$ (1.08)^{18} \approx 3.996019$$
- Subtract 1: $$3.996019 – 1 = 2.996019$$
- Multiply by $u_1$ (150): $$150 \times 2.996019 = 449.40285$$
- Calculate the denominator $r-1$: $$1.08 – 1 = 0.08$$
- Divide the numerator by the denominator: $$S_{18} = \frac{449.40285}{0.08} = 5617.5356$$
Rounding to the nearest whole bacterium for the cumulative total:
$$S_{18} \approx 5618 \text{ bacteria}$$
Step 4: Compare with the crash limit
The colony crashes if the cumulative total exceeds 6000 bacteria.
Our calculated cumulative total after 18 days is approximately $S_{18} = 5618$ bacteria.
Since $$5618 < 6000$$, the cumulative total has not exceeded the crash limit within the first 18 days.
Conclusion: Bacterial Colony Status
The colony will NOT crash within the first 18 days.
Wrapping It Up
And there you have it! From a simple population count to a critical survival prediction, sequences and series are incredibly versatile tools. The key takeaways? First, accurately identify your sequence type (arithmetic or geometric) and whether you need to find a specific term (sequence, $u_n$) or a cumulative total (series, $S_n$). Second, lean on your IB Formula Booklet – it’s there to help!
We’ve focused on geometric sequences and series today, but remember, arithmetic sequences (where you add/subtract a common difference) follow a similar logic with their own set of formulas.
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
Ready to flex those math muscles? Try these practice problems to solidify your understanding of sequences and series. Remember to identify the type of sequence first!
Question 1: For the sequence $$2, 7, 12, 17, \dots$$, find the 15th term ($u_{15}$).
Show Answer
Solution:
This is an arithmetic sequence with $u_1 = 2$ and common difference $d = 5$.
The general term for an arithmetic sequence is $u_n = u_1 + (n-1)d$.
$u_{15} = 2 + (15-1)5 = 2 + (14)5 = 2 + 70 = 72$.
The 15th term is 72.
Question 2: A ball is dropped from a height of 8 meters. After each bounce, it reaches 75% of its previous height. How high does it reach after the 5th bounce?
Show Answer
Solution:
This is a geometric sequence. Let $u_n$ represent the height after the $n$-th bounce.
Height after 1st bounce ($u_1$): $8 \times 0.75 = 6$ m.
Common ratio ($r$) = 0.75.
We need the height after the 5th bounce, which is $u_5$.
$u_n = u_1 r^{n-1}$
$u_5 = 6 \times (0.75)^{5-1} = 6 \times (0.75)^4 = 6 \times 0.31640625 = 1.8984375$ m.
The height after the 5th bounce is approximately 1.90 meters (3 s.f.).
Question 3: Find the sum of the first 20 terms of the arithmetic series $$3 + 7 + 11 + \dots$$
Show Answer
Solution:
This is an arithmetic series with $u_1 = 3$ and common difference $d = 4$. We need $S_{20}$.
The sum formula for an arithmetic series is $S_n = \frac{n}{2}(2u_1 + (n-1)d)$.
$S_{20} = \frac{20}{2}(2(3) + (20-1)4) = 10(6 + (19)4) = 10(6 + 76) = 10(82) = 820$.
The sum of the first 20 terms is 820.
Question 4: What is the general term ($u_n$) for the sequence $$1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dots$$?
Show Answer
Solution:
This is a geometric sequence with $u_1 = 1$ and common ratio $r = \frac{1}{2}$.
The general term is $u_n = u_1 r^{n-1}$.
$u_n = 1 \left(\frac{1}{2}\right)^{n-1} = \left(\frac{1}{2}\right)^{n-1}$.
Question 5: A theater has 15 rows. The first row has 12 seats, and each subsequent row has 2 more seats than the row in front of it. How many seats are in the theater in total?
Show Answer
Solution:
This is an arithmetic series. $u_1 = 12$ (seats in the first row), $d = 2$ (common difference), $n = 15$ (number of rows).
We need the sum of all seats, $S_{15}$.
$S_n = \frac{n}{2}(2u_1 + (n-1)d)$
$S_{15} = \frac{15}{2}(2(12) + (15-1)2) = \frac{15}{2}(24 + (14)2) = \frac{15}{2}(24 + 28) = \frac{15}{2}(52) = 15 \times 26 = 390$.
There are 390 seats in the theater.
Question 6: The first term of a geometric sequence is 5 and the common ratio is 2. Which term is 320?
Show Answer
Solution:
We have $u_1 = 5$, $r = 2$, and $u_n = 320$. We need to find $n$.
$u_n = u_1 r^{n-1}$
$320 = 5 \times 2^{n-1}$
Divide by 5: $64 = 2^{n-1}$
We know that $2^6 = 64$.
So, $n-1 = 6 \implies n = 7$.
The 7th term is 320.
Question 7: An investment starts with $1000 and increases by 4% compounded annually. What is the value of the investment after 7 years?
Show Answer
Solution:
This is a geometric sequence. $u_1 = 1000$ (initial investment). The increase of 4% means $r = 1.04$.
If $u_1$ represents the initial amount, then after 1 year, the amount is $u_2$, after 2 years it’s $u_3$, and so on. So, after 7 years, we are looking for the 8th term, $u_8$.
$u_n = u_1 r^{n-1}$
$u_8 = 1000 \times (1.04)^{8-1} = 1000 \times (1.04)^7$
$u_8 = 1000 \times 1.315931779 \approx 1315.93$
The value of the investment after 7 years will be approximately $1315.93.
Question 8: The first term of an arithmetic sequence is 10, and the 6th term is 35. Find the common difference ($d$).
Show Answer
Solution:
We have $u_1 = 10$ and $u_6 = 35$.
For an arithmetic sequence, $u_n = u_1 + (n-1)d$.
$u_6 = u_1 + (6-1)d$
$35 = 10 + 5d$
$25 = 5d$
$d = 5$.
The common difference is 5.
Question 9: Calculate the sum of the first 8 terms of the geometric series where $u_1 = 6$ and $r = \frac{1}{3}$.
Show Answer
Solution:
We have $u_1 = 6$, $r = \frac{1}{3}$, and $n = 8$.
$S_n = \frac{u_1(1-r^n)}{1-r}$ (This form is often preferred when $r < 1$)
$S_8 = \frac{6(1 – (1/3)^8)}{1 – 1/3} = \frac{6(1 – 1/6561)}{2/3}$
$S_8 = \frac{6(6560/6561)}{2/3} = 6 \times \frac{6560}{6561} \times \frac{3}{2}$
$S_8 = 3 \times \frac{6560}{6561} \times 3 = \frac{59040}{6561} \approx 8.9986$
The sum of the first 8 terms is approximately 8.9986.
IB Exam Style Question
Question 10: A new social media platform gains users rapidly. In its first month, it has 1200 active users. Each subsequent month, the number of new users joining is 200 less than the number of new users who joined the previous month. After the 6th month, the platform gains 200 new users.
- Show that the number of new users joining in the first month was 1200.
- Find the number of new users who joined in the 3rd month.
- Calculate the total number of active users on the platform at the end of the 8th month.
Reveal Solution
Solution:
This problem involves an arithmetic sequence for the *new users joining each month*, and then a sum for the total active users.
Let $N_n$ be the number of new users joining in month $n$.
We are given that $N_6 = 200$.
The number of new users joining each month decreases by 200, so this is an arithmetic sequence with a common difference $d = -200$.
Part a: Show that the number of new users joining in the first month was 1200.
We use the arithmetic general term formula: $N_n = N_1 + (n-1)d$.
For $n=6$, $N_6 = N_1 + (6-1)(-200)$.
$200 = N_1 + 5(-200)$
$200 = N_1 – 1000$
$N_1 = 1200$.
Therefore, the number of new users joining in the first month was 1200.
Part b: Find the number of new users who joined in the 3rd month.
Using $N_1 = 1200$ and $d = -200$ for $n=3$:
$N_3 = N_1 + (3-1)d = 1200 + 2(-200) = 1200 – 400 = 800$.
The number of new users joining in the 3rd month was 800.
Part c: Calculate the total number of active users on the platform at the end of the 8th month.
This requires the sum of the new users for the first 8 months, $S_8$.
We use the sum of an arithmetic series formula: $S_n = \frac{n}{2}(2N_1 + (n-1)d)$.
$S_8 = \frac{8}{2}(2(1200) + (8-1)(-200))$
$S_8 = 4(2400 + 7(-200))$
$S_8 = 4(2400 – 1400)$
$S_8 = 4(1000)$
$S_8 = 4000$.
The total number of active users on the platform at the end of the 8th month is 4000.