Ever feel like math is just a bunch of abstract rules? Well, today we’re going to shatter that myth by diving into one of the most practical tools in your IB Math toolkit: Arithmetic Series. Imagine a new startup owner who makes a profit of \$2,000 in her first month. Pumped by her success, she sets a goal to increase her monthly profit by a steady \$500 every month. How much profit will she make in her 12th month? And more importantly, what will her total profit be after two full years? This isn’t just a random puzzle; it’s a real-world scenario that you can solve with ease using the power of arithmetic series. Let’s break it down and turn this seemingly complex problem into a piece of cake.
Table of Contents
What is an Arithmetic Series, Anyway?
Before we can run, we’ve got to walk. The journey into arithmetic series starts with its sibling concept: the arithmetic sequence.
Definition
An arithmetic sequence is a list of numbers where the difference between any two consecutive terms is always the same. We call this consistent difference the common difference, or just ‘$d$’.
For example, think about a simple fitness plan: you do 5 push-ups on day 1, 8 on day 2, 11 on day 3, and so on. The sequence is: $5, 8, 11, 14, \dots$ Here, the common difference, $d$, is 3.
An arithmetic series is what you get when you add up all the terms of an arithmetic sequence. So, for our fitness plan, the series would be the total push-ups done over time: $5 + 8 + 11 + 14 + \dots$
Decoding Sigma Notation (Σ)
Writing out long sums like $5 + 8 + 11 + \dots + 62$ is tedious. Mathematicians love efficiency, which is where the super-cool Sigma notation comes in. It’s the Greek capital letter Sigma, $\Sigma$, and it’s your best friend for writing sums concisely.
Rule: Sigma Notation
The general form looks like this: $$\sum_{k=1}^{n} u_k$$
- $\Sigma$: This is the summation symbol. It means “add everything up.”
- $k=1$: This is the lower limit. It tells you which term to start with (in this case, the first term).
- $n$: This is the upper limit. It tells you which term to end with.
- $u_k$: This is the general term or formula for the terms in the sequence.
Example: Let’s say we have the expression $$\sum_{k=1}^{4} (5k – 2)$$ To evaluate this, we just plug in the values of $k$ from 1 to 4 and add them up:
$$= (5(1)-2) + (5(2)-2) + (5(3)-2) + (5(4)-2)$$
$$= 3 + 8 + 13 + 18 = 42$$
Your Secret Weapons: The IB Formulas
Here’s the best news you’ll hear all day: you don’t have to memorize the main formulas for arithmetic series! They are right there in your IB Formula Booklet. Your job is to understand what they mean and when to use them.
IB Exam Tip
Always have your Formula Booklet handy when you practice. The more you use it, the faster you’ll be at locating the right formula under pressure during an exam. Familiarity is key!
Finding Any Term: The $n^{th}$ Term Formula
Need to find the value of a specific term way down the line, like the 50th term? There’s a formula for that.
Rule: The $n^{th}$ Term ($u_n$)
The formula to find any term ($u_n$) in an arithmetic sequence is:
$$u_n = u_1 + (n-1)d$$
- $u_n$: The term you want to find.
- $u_1$: The very first term of the sequence.
- $n$: The position of the term in the sequence (e.g., for the 5th term, $n=5$).
- $d$: The common difference.
Summing It All Up: The $S_n$ Formulas
When you need the total sum of a certain number of terms, you’ll use one of the two $S_n$ formulas.
Rule: Sum of the First $n$ Terms ($S_n$)
You have two options for finding the sum ($S_n$):
Formula 1: $$S_n = \frac{n}{2}(2u_1 + (n-1)d)$$
Use this one when you know the first term ($u_1$), the number of terms ($n$), and the common difference ($d$).
Formula 2: $$S_n = \frac{n}{2}(u_1 + u_n)$$
This one is a fantastic shortcut. Use it when you know the first term ($u_1$), the last term ($u_n$), and the number of terms ($n$).
Putting It All Together: A Worked Example
Let’s tackle a classic IB-style problem to see these formulas in action.
IB Exam Question
The Dedicated Coder’s Challenge
A software developer writes 50 lines of code on her first day of a new project. To build momentum, she decides to increase her output by exactly 15 lines of code each subsequent day.
Part A: How many lines of code will she write on the 12th day?
Part B: What is the total number of lines of code she will have written after 30 days?
Solution Walkthrough
First, let’s identify our variables from the problem description:
- The first term ($u_1$): 50 lines
- The common difference ($d$): 15 lines
Solving Part A:
We need to find the number of lines written on the 12th day, so we’re looking for $u_{12}$. This means $n=12$. We’ll use the $n^{th}$ term formula:
$$u_n = u_1 + (n-1)d$$
Substitute the values:
$$u_{12} = 50 + (12-1)(15)$$
$$u_{12} = 50 + (11)(15)$$
$$u_{12} = 50 + 165 = 215$$
Answer A: She will write 215 lines of code on the 12th day.
Solving Part B:
Now we need the total lines written after 30 days, so we’re looking for $S_{30}$. Here, $n=30$. Since we have $u_1$, $d$, and $n$, the first sum formula is our best bet.
$$S_n = \frac{n}{2}(2u_1 + (n-1)d)$$
Substitute the values:
$$S_{30} = \frac{30}{2}(2(50) + (30-1)(15))$$
$$S_{30} = 15(100 + (29)(15))$$
$$S_{30} = 15(100 + 435)$$
$$S_{30} = 15(535) = 8025$$
Answer B: She will have written a total of 8,025 lines of code after 30 days.
Key Takeaways
And that’s the core of it! You’ve seen how a real-world scenario with steady growth can be perfectly modeled by an arithmetic series. Remember these key points:
- An arithmetic sequence has a common difference ($d$). A series is the sum of that sequence.
- Your IB Formula Booklet is your ally. Know where to find the formulas for $u_n$ and $S_n$.
- Identify your variables first! Before you can solve anything, you need to know your $u_1$, $d$, and $n$.
- Choose the right tool for the job. Pick the $S_n$ formula that best fits the information you have.
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: $2, 6, 10, 14, \dots$
Show Answer
Solution: Here, $u_1=2$, $d=4$, and $n=25$. Using $u_n = u_1 + (n-1)d$, we get $u_{25} = 2 + (25-1)(4) = 2 + 24(4) = 2 + 96 = 98$.
Question 2: An arithmetic sequence has a first term of -5 and a common difference of 3. What is the sum of the first 20 terms?
Show Answer
Solution: We have $u_1=-5$, $d=3$, and $n=20$. Using $S_n = \frac{n}{2}(2u_1 + (n-1)d)$, we get $S_{20} = \frac{20}{2}(2(-5) + (20-1)(3)) = 10(-10 + 19(3)) = 10(-10 + 57) = 10(47) = 470$.
Question 3: The first term of an arithmetic sequence is 100 and the 10th term is 28. Find the common difference.
Show Answer
Solution: We know $u_1=100$, $n=10$, and $u_{10}=28$. Using $u_n = u_1 + (n-1)d$, we have $28 = 100 + (10-1)d$. This simplifies to $28 = 100 + 9d$, so $-72 = 9d$, which gives $d = -8$.
Question 4: Find the sum of the series: $7 + 12 + 17 + \dots + 102$.
Show Answer
Solution: First, find the number of terms, $n$. We have $u_1=7, d=5, u_n=102$. Use $u_n = u_1 + (n-1)d$: $102 = 7 + (n-1)5$. So, $95 = 5(n-1)$, which means $19 = n-1$, and $n=20$. Now use the sum formula $S_n = \frac{n}{2}(u_1 + u_n)$: $S_{20} = \frac{20}{2}(7 + 102) = 10(109) = 1090$.
Question 5: Evaluate the sum: $$\sum_{k=1}^{15} (4k+1)$$
Show Answer
Solution: This is an arithmetic series with $n=15$. The first term is $u_1 = 4(1)+1 = 5$. The last term is $u_{15} = 4(15)+1 = 61$. Using $S_n = \frac{n}{2}(u_1 + u_n)$: $S_{15} = \frac{15}{2}(5 + 61) = 7.5(66) = 495$.
Question 6: The sum of the first 14 terms of an arithmetic series is 735. If the common difference is 5, what is the first term?
Show Answer
Solution: We have $S_{14}=735$, $n=14$, $d=5$. Use $S_n = \frac{n}{2}(2u_1 + (n-1)d)$. So, $735 = \frac{14}{2}(2u_1 + (14-1)5)$. This becomes $735 = 7(2u_1 + 13(5))$, so $105 = 2u_1 + 65$. Then $40 = 2u_1$, so $u_1 = 20$.
Question 7: How many terms are there in the sequence $18, 24, 30, \dots, 192$?
Show Answer
Solution: We have $u_1=18, d=6, u_n=192$. We need to find $n$. Use $u_n = u_1 + (n-1)d$: $192 = 18 + (n-1)6$. So, $174 = 6(n-1)$, which means $29 = n-1$, so $n=30$. There are 30 terms.
Question 8: An arithmetic series has $u_5 = 22$ and $u_{11} = 52$. Find the first term, $u_1$, and the common difference, $d$.
Show Answer
Solution: Set up a system of equations: $u_5 = u_1 + 4d = 22$ and $u_{11} = u_1 + 10d = 52$. Subtract the first equation from the second: $(u_1 + 10d) – (u_1 + 4d) = 52 – 22$. This gives $6d=30$, so $d=5$. Substitute $d=5$ back into the first equation: $u_1 + 4(5) = 22$, so $u_1 + 20 = 22$, which means $u_1 = 2$.
Question 9: The sum of an arithmetic series is 1200. Its first term is 10 and its last term is 150. How many terms are in the series?
Show Answer
Solution: We have $S_n=1200, u_1=10, u_n=150$. Use the shortcut formula $S_n = \frac{n}{2}(u_1 + u_n)$: $1200 = \frac{n}{2}(10 + 150)$. This is $1200 = \frac{n}{2}(160)$, or $1200 = 80n$. Dividing gives $n=15$.
Question 10 (IB Exam Style Question): In an arithmetic sequence, the sum of the first eight terms is 248 ($S_8 = 248$) and the fifth term is 25 ($u_5 = 25$).
(a) Find the first term and the common difference.
(b) Find the sum of the first 20 terms.
Show Answer
Solution:
(a) We need to set up a system of two equations with two variables ($u_1$ and $d$).
From $u_5 = 25$, we get the equation: $u_1 + (5-1)d = 25 \implies u_1 + 4d = 25$. (Equation 1)
From $S_8 = 248$, we get the equation: $\frac{8}{2}(2u_1 + (8-1)d) = 248 \implies 4(2u_1 + 7d) = 248 \implies 2u_1 + 7d = 62$. (Equation 2)
From Equation 1, we can write $u_1 = 25 – 4d$. Substitute this into Equation 2:
$2(25 – 4d) + 7d = 62$
$50 – 8d + 7d = 62$
$50 – d = 62 \implies d = -12$.
Now find $u_1$ using $u_1 = 25 – 4d$: $u_1 = 25 – 4(-12) = 25 + 48 = 73$.
Answer (a): The first term is 73 and the common difference is -12.
(b) Now we need to find $S_{20}$ using $u_1=73$, $d=-12$, and $n=20$.
$S_{20} = \frac{20}{2}(2(73) + (20-1)(-12))$
$S_{20} = 10(146 + 19(-12))$
$S_{20} = 10(146 – 228)$
$S_{20} = 10(-82) = -820$.
Answer (b): The sum of the first 20 terms is -820.