Ever looked at a grand structure like an arched bridge or a stadium and wondered, “How on earth did they get every single piece to line up so perfectly?” It’s not magic; it’s math. Specifically, it’s the elegant, predictable power of patterns. Today, we’re pulling back the curtain on one of the most fundamental patterns in mathematics: the arithmetic sequence.
We’ll start with a puzzle. Imagine you’re an architect designing a stunning archway supported by pillars. The first, tallest pillar is 10 meters. Because of the curve, each pillar that follows must be exactly 1.5 meters shorter than the one before it. If your design specifies the shortest pillar can be no less than 1 meter, how many pillars can you actually build? This isn’t just a textbook question; it’s a real-world design constraint. By the end of this post, you’ll not only be able to solve this puzzle but also tackle similar problems with confidence. Let’s get building!
Table of Contents
What Exactly is an Arithmetic Sequence?
Let’s start with the basics. At its heart, an arithmetic sequence is just a list of numbers with a very specific, predictable rule: you get from one term to the next by adding (or subtracting) the exact same number every single time. It’s this constant step that makes it ‘arithmetic’.
Definition: Arithmetic Sequence
An arithmetic sequence (or arithmetic progression) is a sequence of numbers where the difference between any two consecutive terms is constant. This constant value is called the common difference.
Consider the sequence: $4, 9, 14, 19, 24, \dots$
To get from 4 to 9, you add 5. From 9 to 14, you add 5. From 14 to 19, you add 5. This consistent step of ‘add 5’ is the defining feature. It’s simple, but powerful!
The Key Ingredients: First Term ($u_1$) and Common Difference ($d$)
To work with any arithmetic sequence, you only need two pieces of information: where it starts and what the constant step is. In IB math, we have specific notation for these.
Rule: Key Terminology
- The First Term ($u_1$): This is simply the number that kicks off the sequence. In our example $4, 9, 14, \dots$, the first term is $u_1 = 4$. (Sometimes you might see it written as ‘a’).
- The Common Difference ($d$): This is that constant step between terms. For $4, 9, 14, \dots$, the common difference is $d = 5$. You can always find it by taking any term and subtracting the one before it: $d = u_n – u_{n-1}$. For instance, $u_2 – u_1 = 9 – 4 = 5$.
With just $u_1$ and $d$, we can find any term in the sequence without having to write them all out. This is where the superstar formula comes in.
Your Best Friend: The IB Formula Booklet
You don’t need to memorize this! The formula for the n-th term of an arithmetic sequence is right there in your IB Math Formula Booklet (Topic 1: Number and Algebra). Your job is to know what it means and how to use it.
$u_n = u_1 + (n-1)d$
Here, $u_n$ is the term you’re looking for, $u_1$ is your starting point, $n$ is the position of the term in the sequence (e.g., 5th, 20th, 100th), and $d$ is the common difference.
Going Up or Down? Increasing vs. Decreasing Sequences
The common difference, $d$, does more than just define the step; it tells you the story of the sequence. Is it climbing towards infinity, or is it shrinking towards zero or negative numbers? The sign of $d$ gives it all away.
- If $d$ is positive ($d > 0$), you’re always adding a positive number, so the terms get bigger. This is an increasing sequence. Example: $10, 13, 16, 19, \dots$ where $d = 3$.
- If $d$ is negative ($d < 0$), you’re always adding a negative number (which is the same as subtracting), so the terms get smaller. This is a decreasing sequence. Example: $50, 48, 46, 44, \dots$ where $d = -2$.
Exam Tip: Decode the Language
In word problems, the sign of ‘d’ is often hidden in the language. Be a detective! Words like “shorter,” “decreases,” “loses,” or “reduces” are huge clues that your common difference ‘d’ will be negative. The archway problem is a perfect example of this!
Putting It All Together: Solving The Diver’s Descent
Okay, theory is great, but let’s see how it works in practice. We’re going to tackle a new problem from scratch.
Example: The Diver’s Descent
Problem: A scuba diver is exploring a reef. She starts her controlled descent at a depth of 5 meters below the surface. Every minute, she descends a further 3 meters. Her dive computer has a maximum safe depth alarm set at 32 meters.
Part A: What is her depth after 7 minutes?
Part B: How many full minutes can she descend before the alarm goes off?
Step 1: Identify Your Variables
First, let’s translate the words into math. The sequence represents her depth at each minute.
- Her starting depth is 5 meters. So, the first term is $u_1 = 5$.
- She descends (gets deeper) by 3 meters each minute. This means her depth value is increasing. So, the common difference is $d = 3$.
Part A: Depth After 7 Minutes
We want to find her depth after 7 minutes. This is the 7th term in our sequence, so we are looking for $u_7$. We’ll use our formula: $u_n = u_1 + (n-1)d$.
Substitute the values we know: $n=7$, $u_1=5$, and $d=3$.
$$u_7 = 5 + (7-1)(3)$$
$$u_7 = 5 + (6)(3)$$
$$u_7 = 5 + 18$$
$$u_7 = 23$$
Answer A: After 7 minutes, the diver is at a depth of 23 meters.
Part B: Time to Max Depth
Here, we know the final depth (a term in the sequence), but we don’t know its position (the time ‘n’). We are given that the final depth, $u_n$, is 32 meters. We need to find ‘n’.
Let’s use the formula again: $u_n = u_1 + (n-1)d$.
Substitute the knowns: $u_n=32$, $u_1=5$, and $d=3$.
$$32 = 5 + (n-1)(3)$$
Now, we just solve this equation for ‘n’.
$$32 – 5 = (n-1)(3)$$
$$27 = 3(n-1)$$
$$\frac{27}{3} = n-1$$
$$9 = n-1$$
$$n = 10$$
Answer B: She can descend for 10 full minutes before the alarm goes off.
Key Takeaways & What’s Next
And there you have it! The logic of arithmetic sequences allowed us to model a real-world scenario and make precise predictions. The key is always to identify your starting point ($u_1$), your constant change ($d$), and then use the n-th term formula to find whatever piece of the puzzle is missing.
This is just the beginning of our journey with sequences. What if you wanted to know the total distance the diver descended? Or the total length of steel needed for all the pillars in the archway? For that, we need to learn how to add up the terms in a sequence. But first, in the next lesson, we’ll tackle more examples about Arithmetic Sequences.
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 Problems
Question 1: Find the 20th term of the arithmetic sequence $2, 6, 10, 14, \dots$
Show Answer
Solution:
First, identify $u_1$ and $d$.
$u_1 = 2$
$d = u_2 – u_1 = 6 – 2 = 4$
We want the 20th term, so $n = 20$.
Using the formula $u_n = u_1 + (n-1)d$:
$u_{20} = 2 + (20-1)(4)$
$u_{20} = 2 + (19)(4) = 2 + 76 = 78$
The 20th term is 78.
Question 2: An arithmetic sequence has a first term of 15 and a common difference of -3. What is the 12th term?
Show Answer
Solution:
We are given:
$u_1 = 15$
$d = -3$
We need to find the 12th term, so $n = 12$.
Using the formula $u_n = u_1 + (n-1)d$:
$u_{12} = 15 + (12-1)(-3)$
$u_{12} = 15 + (11)(-3) = 15 – 33 = -18$
The 12th term is -18.
Question 3: The fourth term of an arithmetic sequence is 25 and the ninth term is 45. Find the common difference.
Show Answer
Solution:
We have $u_4 = 25$ and $u_9 = 45$.
We can think of the jump from the 4th term to the 9th term as $9-4=5$ steps (or 5 additions of ‘d’).
So, $u_9 = u_4 + 5d$.
$45 = 25 + 5d$
$45 – 25 = 5d$
$20 = 5d$
$d = 4$
The common difference is 4.
Question 4: In the sequence $100, 97, 94, \dots$, which term is the first to be negative?
Show Answer
Solution:
First, identify $u_1=100$ and $d = 97 – 100 = -3$.
We want to find the first ‘n’ for which $u_n < 0$.
Set up the inequality: $u_1 + (n-1)d < 0$
$100 + (n-1)(-3) < 0$
$100 – 3n + 3 < 0$
$103 – 3n < 0$
$103 < 3n$
$\frac{103}{3} < n$
$34.333… < n$
Since ‘n’ must be an integer, the first integer value of ‘n’ greater than 34.333… is 35. So, the 35th term is the first negative term.
Question 5: A baker sells 40 loaves of bread on his first day. He plans to increase his sales by 6 loaves each day. On which day will he first sell over 100 loaves?
Show Answer
Solution:
This is an arithmetic sequence where $u_1 = 40$ and $d = 6$.
We want to find the first day ‘n’ where sales $u_n > 100$.
$40 + (n-1)(6) > 100$
$40 + 6n – 6 > 100$
$34 + 6n > 100$
$6n > 66$
$n > 11$
The first integer value of ‘n’ greater than 11 is 12. So, on the 12th day, he will first sell over 100 loaves.
Question 6: Find the first term of an arithmetic sequence if the 5th term is 30 and the common difference is 4.
Show Answer
Solution:
We know $u_5 = 30$, $d=4$, and $n=5$. We need to find $u_1$.
Using the formula $u_n = u_1 + (n-1)d$:
$30 = u_1 + (5-1)(4)$
$30 = u_1 + (4)(4)$
$30 = u_1 + 16$
$u_1 = 30 – 16 = 14$
The first term is 14.
Question 7: Which term in the sequence $5, 8, 11, \dots$ is equal to 62?
Show Answer
Solution:
Here, $u_1=5$ and $d=3$. We are given a term value, $u_n = 62$, and we need to find its position, ‘n’.
$62 = 5 + (n-1)(3)$
$57 = (n-1)(3)$
$19 = n-1$
$n = 20$
62 is the 20th term in the sequence.
Question 8: The 3rd term of an arithmetic sequence is 10 and the 8th term is -5. Find the first term and the common difference.
Show Answer
Solution:
We have a system of two equations:
$u_3 = u_1 + (3-1)d = u_1 + 2d = 10$ (Equation 1)
$u_8 = u_1 + (8-1)d = u_1 + 7d = -5$ (Equation 2)
Subtract Equation 1 from Equation 2:
$(u_1 + 7d) – (u_1 + 2d) = -5 – 10$
$5d = -15$
$d = -3$
Now substitute $d=-3$ back into Equation 1:
$u_1 + 2(-3) = 10$
$u_1 – 6 = 10$
$u_1 = 16$
The first term is 16 and the common difference is -3.
Question 9: A phone’s battery is at 95% and loses 4% every hour of use. After how many hours will the battery level be at 15%?
Show Answer
Solution:
This is a decreasing arithmetic sequence. The initial state (at 0 hours) would be 95%. After 1 hour ($u_1$) it will be $95-4=91%$. So, $u_1=91$ and $d=-4$. We want to find ‘n’ (number of hours of use) when the battery level is 15%. However, it’s easier to think of the sequence as starting with $u_0=95$. Let’s stick to the $u_1$ convention. The battery level at the end of hour ‘n’ is $u_n$.
$u_1=91, d=-4, u_n=15$.
$15 = 91 + (n-1)(-4)$
$15 – 91 = -4(n-1)$
$-76 = -4(n-1)$
$19 = n-1$
$n=20$
After 20 hours, the battery will be at 15%.
IB Exam Style Question
Question 10: A concert hall has rows of seats forming an arithmetic sequence. The first row (Row 1) has 20 seats. Each subsequent row has 2 more seats than the row in front of it.
- (a) Find the number of seats in the 15th row.
(b) A particular row has 54 seats. Find which row number this is.
(c) The concert hall has a total of 30 rows. Find the number of seats in the last row.
Show Answer
Solution:
First, identify the parameters of the arithmetic sequence.
First term (seats in Row 1): $u_1 = 20$
Common difference (increase per row): $d = 2$
(a) Find the number of seats in the 15th row.
We need to find $u_{15}$. Using the formula $u_n = u_1 + (n-1)d$:
$u_{15} = 20 + (15-1)(2)$
$u_{15} = 20 + (14)(2) = 20 + 28 = 48$
There are 48 seats in the 15th row.
(b) A particular row has 54 seats. Find which row number this is.
Here, we are given $u_n = 54$ and we need to find ‘n’.
$54 = 20 + (n-1)(2)$
$54 – 20 = 2(n-1)$
$34 = 2(n-1)$
$17 = n-1$
$n = 18$
It is the 18th row that has 54 seats.
(c) The concert hall has a total of 30 rows. Find the number of seats in the last row.
The last row is the 30th row, so we need to find $u_{30}$.
$u_{30} = 20 + (30-1)(2)$
$u_{30} = 20 + (29)(2) = 20 + 58 = 78$
There are 78 seats in the last row.