Hey there, fellow IB math explorers! Ever felt like math problems are just codes waiting to be cracked? Well, today we’re tackling one of the most satisfying codes out there: arithmetic sequences. These guys are all about predictable, linear patterns, but the IB loves to throw us curveballs. How do you find a term that’s a hundred steps down the line without writing everything out? Or, if you know where a sequence ends, how can you figure out how long the journey was? Stick around, because we’re about to turn these tricky questions into total cakewalks.
Table of Contents
What’s the Deal with Arithmetic Sequences?
Before we jump into the heavy lifting, let’s get our fundamentals straight. Think of an arithmetic sequence as a list of numbers where you’re always taking the same size step to get from one number to the next. It’s like walking up a staircase where every single step is the exact same height.
Definition: Arithmetic Sequence
An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant. This constant difference is called the common difference, and we denote it with the variable $d$.
For example, in the sequence $4, 10, 16, 22, \dots$, the common difference $d$ is $6$, because you add 6 each time.
The Secret Weapon: The General Term Formula
Now, you could find the 50th term by just adding the common difference 49 times… but who has time for that? Especially not in an exam! This is where our secret weapon comes in: the general term formula. This little beauty lets us find any term we want, instantly.
Rule: The General Term Formula
The formula for the $n$-th term of an arithmetic sequence is:
$$U_n = U_1 + (n-1)d$$
Where:
- $U_n$ is the term you are looking for (the ‘nth’ term).
- $U_1$ is the very first term of the sequence.
- $n$ is the term number (e.g., 5th, 20th, 100th).
- $d$ is the common difference.
IB Pro Tip
Don’t panic about memorizing this! This formula is your best friend, and it’s located in Topic 1: Number and Algebra of your IB Math Formula Booklet. Always have it handy.
Mission 1: Finding a Term in the Future
Let’s put the formula into action. This is the most common type of question you’ll see: given the start and the pattern, find a specific term later on.
Example: The Growing Plant
A botanist is tracking a special bamboo plant. On the first day, it measures 5 cm tall. She observes that it grows by a constant 2.5 cm every single day. What will be the height of the plant on the 30th day?
Step 1: Identify your variables.
- The first term ($U_1$) is the starting height: $U_1 = 5$.
- The common difference ($d$) is the daily growth: $d = 2.5$.
- The term number ($n$) is the day we’re interested in: $n = 30$.
- We need to find $U_{30}$.
Step 2: Substitute into the formula.
$$U_n = U_1 + (n-1)d$$$$U_{30} = 5 + (30-1)(2.5)$$
Step 3: Solve it!
$$U_{30} = 5 + (29)(2.5)$$$$U_{30} = 5 + 72.5$$$$U_{30} = 77.5$$
So, on the 30th day, the bamboo plant will be 77.5 cm tall. See? No need to add 2.5 over and over again!
Mission 2: Counting the Steps (How Many Terms?)
This is the reverse problem. You know where you started, you know your step size, and you know your final destination. The question is: how many steps did it take to get there?
Example: Saving Up
You’re saving for a new laptop. You start with \$50 in your piggy bank. You commit to adding exactly \$15 every week. If the final amount you need is $425, how many weeks will it take to reach your goal?
Step 1: Identify your variables.
- The first term ($U_1$) is your starting amount: $U_1 = 50$.
- The common difference ($d$) is your weekly saving: $d = 15$.
- The final term ($U_n$) is your goal: $U_n = 425$.
- This time, we need to find $n$, the number of weeks.
Step 2: Substitute into the formula.
$$U_n = U_1 + (n-1)d$$$$425 = 50 + (n-1)(15)$$
Step 3: Use algebra to solve for $n$.
Our goal is to isolate $n$. Let’s peel back the layers.
$$425 – 50 = (n-1)(15)$$$$375 = (n-1)(15)$$$$\frac{375}{15} = n-1$$$$25 = n-1$$$$n = 25 + 1$$$$n = 26$$
It will take 26 weeks to save up for the laptop. That’s exactly half a year!
Watch Out for This Common Pitfall!
When solving for $n$, a common mistake is to mess up the order of operations. In the example above ($375 = (n-1)(15)$), remember to divide by 15 before you add the 1. Don’t distribute the 15 unless you absolutely have to—it often creates more steps and more room for error. Isolate the bracket first!
Final Debrief: Key Takeaways
Mastering arithmetic sequence problems really boils down to a solid, repeatable strategy:
- Identify Your Knowns: Read the problem carefully and figure out which of the four key variables ($U_1, d, n, U_n$) you have.
- Identify Your Unknown: Determine which one of the four variables you need to find.
- Formula Time: Grab the general term formula, $U_n = U_1 + (n-1)d$, from your formula booklet.
- Substitute and Solve: Plug in your knowns and carefully use your algebra skills to find the unknown.
That’s it! Whether you’re finding a future term or the length of the sequence, the process is the same. It’s all about identifying what you have and what you need.
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 25th term of the arithmetic sequence: $3, 7, 11, 15, \dots$
Show Answer
Solution:
Here, $U_1 = 3$, $d = 7 – 3 = 4$, and $n = 25$.
$U_{25} = 3 + (25-1)(4)$
$U_{25} = 3 + (24)(4)$
$U_{25} = 3 + 96 = 99$.
Question 2: An arithmetic sequence has a first term of -8 and a common difference of 5. Find the 40th term.
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Solution:
Here, $U_1 = -8$, $d = 5$, and $n = 40$.
$U_{40} = -8 + (40-1)(5)$
$U_{40} = -8 + (39)(5)$
$U_{40} = -8 + 195 = 187$.
Question 3: How many terms are in the arithmetic sequence: $10, 13, 16, \dots, 70$?
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Solution:
Here, $U_1 = 10$, $d = 3$, and $U_n = 70$. We need to find $n$.
$70 = 10 + (n-1)(3)$
$60 = (n-1)(3)$
$20 = n-1$
$n = 21$. There are 21 terms.
Question 4: The first term of a sequence is 100, and the common difference is -6. Which term in the sequence is equal to 16?
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Solution:
Here, $U_1 = 100$, $d = -6$, and $U_n = 16$. We need to find $n$.
$16 = 100 + (n-1)(-6)$
$16 – 100 = (n-1)(-6)$
$-84 = (n-1)(-6)$
$\frac{-84}{-6} = n-1$
$14 = n-1$
$n = 15$. The 15th term is 16.
Question 5: A theatre has 20 seats in the first row. Each subsequent row has 2 more seats than the row in front of it. If the last row has 54 seats, how many rows are in the theatre?
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Solution:
This is an arithmetic sequence where $U_1 = 20$, $d = 2$, and $U_n = 54$.
$54 = 20 + (n-1)(2)$
$34 = (n-1)(2)$
$17 = n-1$
$n = 18$. There are 18 rows in the theatre.
Question 6: Find the 101st term of the sequence defined by $U_n = 8n – 15$.
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Solution:
For this, we can just substitute $n=101$ directly into the formula.
$U_{101} = 8(101) – 15$
$U_{101} = 808 – 15 = 793$.
Question 7: What is the first term of an arithmetic sequence if the 20th term is 99 and the common difference is 4?
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Solution:
Here, $n=20$, $U_{20} = 99$, and $d=4$. We need to find $U_1$.
$99 = U_1 + (20-1)(4)$
$99 = U_1 + (19)(4)$
$99 = U_1 + 76$
$U_1 = 99 – 76 = 23$.
Question 8: How many multiples of 7 are there between 100 and 300?
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Solution:
First, find the first multiple of 7 after 100. $100 \div 7 \approx 14.28$, so the first multiple is $7 \times 15 = 105$. So, $U_1 = 105$.
Next, find the last multiple of 7 before 300. $300 \div 7 \approx 42.85$, so the last multiple is $7 \times 42 = 294$. So, $U_n = 294$.
The common difference is $d=7$.
$294 = 105 + (n-1)(7)$
$189 = (n-1)(7)$
$27 = n-1$
$n = 28$. There are 28 multiples of 7.
Question 9: An arithmetic sequence has $U_1 = 50$ and $d = -3.5$. What is the first term in the sequence to be negative?
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Solution:
We want to find the smallest integer $n$ for which $U_n < 0$.
$U_1 + (n-1)d < 0$
$50 + (n-1)(-3.5) < 0$
$50 < 3.5(n-1)$
$\frac{50}{3.5} < n-1$
$14.285… < n-1$
$15.285… < n$
Since $n$ must be an integer, the smallest integer value for $n$ is 16. The 16th term is the first negative term.
IB Exam Style Question
The terms of an arithmetic sequence are given by the expression $U_n = 52 – 3n$.
(a) Find the first term of the sequence.
(b) Find the common difference of the sequence.
(c) The last term of the sequence is -20. Find the number of terms in the sequence.
Reveal Solution
Solution:
(a) Find the first term ($U_1$).
To find the first term, we set $n=1$.
$U_1 = 52 – 3(1)$
$U_1 = 52 – 3 = 49$.
(b) Find the common difference ($d$).
To find the common difference, we can find the second term ($U_2$) and then calculate $d = U_2 – U_1$.
$U_2 = 52 – 3(2) = 52 – 6 = 46$.
$d = U_2 – U_1 = 46 – 49 = -3$.
Alternatively, the coefficient of $n$ in the general expression is always the common difference, so $d=-3$.
(c) Find the number of terms ($n$).
We are given the last term $U_n = -20$. We use the formula with our known values $U_1=49$ and $d=-3$.
$-20 = 49 + (n-1)(-3)$
$-20 – 49 = (n-1)(-3)$
$-69 = (n-1)(-3)$
$\frac{-69}{-3} = n-1$
$23 = n-1$
$n = 24$.
There are 24 terms in the sequence.