Hey everyone! Ever get that feeling that math problems in textbooks are just a little too… neat? The lines are always straight, the patterns are always perfect. But what happens when we take our math skills out into the real world, where things are messy, unpredictable, and definitely not always linear? That’s exactly what we’re diving into today: using the tools of arithmetic sequences to model real-world growth, even when it doesn’t perfectly fit the mold. This is a huge skill for the IB Math exam and, honestly, for just thinking critically about the world around you.
Table of Contents
The Setup: A Coffee Shop’s Loyalty Program
Let’s imagine a local coffee shop called “The Daily Grind.” They just launched a new loyalty app to attract more customers. They’re tracking the number of new sign-ups each day for the first week to see how it’s going. Here’s their data:
- Day 1: 40 sign-ups
- Day 2: 52 sign-ups
- Day 3: 64 sign-ups
- Day 4: 74 sign-ups
- Day 5: 82 sign-ups
At first glance, it looks like steady growth. But is it the kind of predictable, constant growth we can model with a simple arithmetic sequence? Let’s play detective and find out.
The First Test: Is The Growth Truly Arithmetic?
First things first, we need to determine if this pattern of sign-ups is a perfect arithmetic sequence. Time for a quick refresher.
Definition: Arithmetic Sequence
An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant. This constant value is called the common difference, denoted by $d$.
To check if our coffee shop data is arithmetic, we just need to calculate the difference between each consecutive day’s sign-ups. Let’s call the sign-ups on day $n$ as $U_n$.
- $d_1 = U_2 – U_1 = 52 – 40 = 12$
- $d_2 = U_3 – U_2 = 64 – 52 = 12$
- $d_3 = U_4 – U_3 = 74 – 64 = 10$
- $d_4 = U_5 – U_4 = 82 – 74 = 8$
The verdict? The differences are 12, 12, 10, and 8. Since they are not all the same, the growth in app sign-ups is not a perfect arithmetic sequence. The growth is slowing down!
Building a Model: Finding the ‘Best Fit’
Okay, so it’s not perfect. But what if we wanted to make a quick prediction? In mathematical modeling, we often simplify a situation to get a useful approximation. The problem asks us to create an arithmetic model based on the growth from the first three days, which was more consistent.
For days 1, 2, and 3, the common difference was consistently 12. So, let’s build our model with these values:
- The first term, $U_1 = 40$.
- The common difference, $d = 12$.
The general formula for the $n^{th}$ term of an arithmetic sequence is:
$$ U_n = U_1 + (n-1)d $$
Plugging in our values for The Daily Grind, our model becomes:
$$ U_n = 40 + (n-1)12 $$
IB Pro Tip: The Power of Assumptions
In IB Math, especially for your Internal Assessment (IA), you’ll often need to make and justify assumptions. Stating “we assume the initial growth rate was constant” is a perfect example of this. It shows you understand that models are simplifications and allows you to apply a mathematical technique even when the data isn’t perfect. Always remember to state your assumptions clearly!
Crystal Ball Math: Making a Prediction
Now for the fun part! Let’s use our model, $U_n = 40 + (n-1)12$, to predict the number of sign-ups on Day 8. We just need to substitute $n=8$ into our formula:
$U_8 = 40 + (8-1) \times 12$
$U_8 = 40 + (7) \times 12$
$U_8 = 40 + 84$
$U_8 = 124$
So, our arithmetic model predicts that The Daily Grind will get 124 sign-ups on Day 8. But… how much faith should we have in this number?
The Reality Check: Why Our Model Isn’t Perfect
This is where your critical thinking skills, so vital for IB, come into play. A number from a model is just that—a number. We need to interpret it in the context of the real world. Why might the actual number of sign-ups on Day 8 be different from 124?
Real-World Factors
Here are a couple of reasons the actual growth might deviate from our linear model:
- Promotion Fatigue: The strong initial growth was likely driven by a launch promotion or the initial buzz. As this novelty wears off, the rate of new sign-ups naturally slows down, which is exactly what we saw in the data for days 4 and 5. Our model completely ignores this slowdown.
- External Factors: Real life is complex! Maybe a competing coffee shop opened nearby, or maybe Day 8 is a public holiday when the shop is less busy. Our simple model assumes conditions remain identical every single day, which is never true. Weekend vs. weekday traffic alone could throw the numbers off.
Key Takeaways
The biggest lesson here is that while arithmetic sequences provide a fantastic tool for understanding linear growth, real-world data is rarely that clean. The true skill of a mathematician isn’t just plugging numbers into formulas, but also understanding the context, making justified assumptions to build a model, and critically evaluating the model’s predictions and limitations. This process of analyzing, modeling, and critiquing is at the heart of what the IB DP Math course wants you to learn.
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: Is the sequence 5, 11, 17, 23, 29, … an arithmetic sequence? If so, what is the common difference?
Show Answer
Solution: Yes. The common difference is $d = 6$ (since $11-5=6$, $17-11=6$, etc.).
Question 2: An arithmetic sequence has a first term $U_1 = -4$ and a common difference $d = 3$. What is the 10th term, $U_{10}$?
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Solution: Using $U_n = U_1 + (n-1)d$, we get $U_{10} = -4 + (10-1) \times 3 = -4 + 9 \times 3 = -4 + 27 = 23$.
Question 3: The third term of an arithmetic sequence is 15 and the seventh term is 39. Find the first term ($U_1$) and the common difference ($d$).
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Solution:
We have two equations: $U_3 = U_1 + 2d = 15$ and $U_7 = U_1 + 6d = 39$.
Subtracting the first equation from the second gives: $(U_1 + 6d) – (U_1 + 2d) = 39 – 15$,
which simplifies to $4d = 24$,
so $d=6$.
Substituting $d=6$ into the first equation: $U_1 + 2(6) = 15$, so $U_1 + 12 = 15$, which means $U_1 = 3$.
Question 4: Find the number of terms in the arithmetic sequence 7, 11, 15, …, 143.
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Solution:
Here $U_1=7$, $d=4$, and the last term $U_n=143$.
We use the formula $U_n = U_1 + (n-1)d$.
So, $143 = 7 + (n-1)4$.
Subtract 7: $136 = (n-1)4$.
Divide by 4: $34 = n-1$.
Therefore, $n=35$. There are 35 terms in the sequence.
Question 5: A theater has 20 seats in the first row, 23 in the second, 26 in the third, and so on. How many seats are in the 15th row?
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Solution:
This is an arithmetic sequence with $U_1 = 20$ and $d = 3$. We want to find $U_{15}$.
$U_{15} = 20 + (15-1) \times 3 = 20 + 14 \times 3 = 20 + 42 = 62$.
There are 62 seats in the 15th row.
Question 6: Is the sequence 100, 90, 81, 73, … an arithmetic sequence?
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Solution:
No. The differences are $90-100 = -10$, $81-90 = -9$, and $73-81 = -8$. Since the difference is not constant, it is not an arithmetic sequence.
Question 7: What is the formula for the $n^{th}$ term of the sequence 8, 5, 2, -1, …?
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Solution:
Here, $U_1=8$ and $d = 5 – 8 = -3$. The formula is $U_n = U_1 + (n-1)d$, which becomes $U_n = 8 + (n-1)(-3) = 8 – 3n + 3 = 11 – 3n$.
Question 8: An arithmetic sequence has $U_1=50$ and $d=-5$. Which term in the sequence is equal to 0?
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Solution:
We want to find $n$ such that $U_n=0$.
Using the formula $U_n = 50 + (n-1)(-5)$.
Set it to 0: $0 = 50 – 5(n-1)$.
So, $5(n-1) = 50$, which means $n-1 = 10$, and $n=11$.
The 11th term is 0.
Question 9: A plant is 15 cm tall. It grows 2.5 cm each week. How many weeks will it take to be taller than 50 cm?
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Solution:
This is an arithmetic sequence where $U_1=15$ (Week 0 height) or we can set Week 1 height as $U_1=17.5$.
Let’s use $U_n = 15 + 2.5n$ where $n$ is the number of weeks.
We want to solve $15 + 2.5n > 50$.
$2.5n > 35$.
$n > 35/2.5$, so $n > 14$.
It will take 15 weeks to be taller than 50 cm.
IB Exam Style Question
A new online streaming service tracks its subscriber growth. In the first month, it gained 5000 subscribers. In the second month, it gained 5700 subscribers, and in the third month, it gained 6400 subscribers.
- a) Show that the monthly subscriber gain follows an arithmetic sequence and state the common difference, $d$.
- b) Assuming this growth rate continues, write down a formula for the number of new subscribers gained in the $n^{th}$ month, $U_n$.
- c) Calculate the number of new subscribers gained in the 12th month.
- d) The service needs a total of 150,000 subscribers to become profitable. Calculate the minimum number of months required to reach this goal. (Hint: You will need the sum formula for an arithmetic sequence, $S_n = \frac{n}{2}(2U_1 + (n-1)d)$).
Reveal Solution
Solution:
a) To show it’s an arithmetic sequence, we check the differences:
$d_1 = U_2 – U_1 = 5700 – 5000 = 700$.
$d_2 = U_3 – U_2 = 6400 – 5700 = 700$.
Since the difference is constant, the sequence is arithmetic with a common difference $d = 700$.
b) Here, $U_1=5000$ and $d=700$. The formula is $U_n = U_1 + (n-1)d$.
$U_n = 5000 + (n-1)700$.
c) For the 12th month, $n=12$:
$U_{12} = 5000 + (12-1)700 = 5000 + 11 \times 700 = 5000 + 7700 = 12,700$.
They are predicted to gain 12,700 new subscribers in the 12th month.
d) We need to find the smallest $n$ for which the total sum of subscribers, $S_n$, is at least 150,000. We use the sum formula: $S_n = \frac{n}{2}(2U_1 + (n-1)d) \geq 150000$.
$\frac{n}{2}(2(5000) + (n-1)700) \geq 150000$
$\frac{n}{2}(10000 + 700n – 700) \geq 150000$
$\frac{n}{2}(9300 + 700n) \geq 150000$
$n(4650 + 350n) \geq 150000$
$350n^2 + 4650n – 150000 \geq 0$.
We can solve the quadratic equation $350n^2 + 4650n – 150000 = 0$ using a calculator’s polynomial solver or the quadratic formula. The positive solution is approximately $n \approx 15.4$. Since the number of months must be an integer, we must round up. Therefore, it will take 16 months to reach the profitability goal.