If you’ve ever felt like some math problems are more like puzzles or mysteries, you’re in the right place. Today, we’re going deep into one of my favorite topics: arithmetic sequences and series. These aren’t just rows of numbers; they’re patterns with a hidden code. Given just a few clues—like a sum, a product, or a specific term—can we uncover the entire sequence? The answer is a resounding yes! We’re going to move beyond the basic formulas and tackle four classic scenarios that will sharpen your problem-solving skills and get you ready for whatever the IB throws your way. Let’s get started!
Table of Contents
The Arithmetic Toolkit: Key Formulas & Definitions
Before we jump into the puzzles, let’s make sure our toolkit is ready. An arithmetic sequence is a list of numbers where the difference between any two consecutive terms is constant. This constant is our superstar, the common difference ($d$).
IB Formula Booklet Essentials
Your IB Math Formula Booklet is your best friend here. These are the two key formulas for arithmetic sequences and series:
- The nth Term: $u_n = u_1 + (n-1)d$
This helps you find any term in the sequence if you know the first term ($u_1$) and the common difference ($d$). - The Sum of n Terms: You get two versions!
- $S_n = \frac{n}{2}(2u_1 + (n-1)d)$
- $S_n = \frac{n}{2}(u_1 + u_n)$
The second one is a slick shortcut when you know the first and last terms!
Puzzle 1: The Case of the Three Numbers
This is a classic opener. You’re given clues about the sum and product of three consecutive terms and asked to play detective and find the terms themselves.
Example: Sum & Product
The sum of three consecutive terms in an arithmetic sequence is 21, and their product is 231. Find the three terms.
Step 1: The Strategic Setup
Instead of calling the terms $x, x+d, x+2d$, we can be much cleverer. Let’s define the middle term as ‘$a$’. Then the terms before and after it are $a-d$ and $a+d$. Our sequence is: $a-d, a, a+d$.
Step 2: Use the Sum Clue
The sum is 21. Look what happens when we add our terms:
$$(a-d) + a + (a+d) = 21$$
$$3a = 21$$
$$a = 7$$
Just like that, we’ve found the middle term! It’s 7.
Step 3: Use the Product Clue
The product is 231. Let’s substitute $a=7$ into our product equation:
$$(7-d)(7)(7+d) = 231$$
Divide both sides by 7:
$$(7-d)(7+d) = 33$$
Recognize that? It’s a difference of squares! $(x-y)(x+y) = x^2 – y^2$.
$$7^2 – d^2 = 33$$
$$49 – d^2 = 33$$
$$d^2 = 16$$
$$d = \pm 4$$
Step 4: Find the Two Possible Sequences
We have two valid values for $d$, which means we have two possible sequences. They’re just mirror images of each other!
- Case 1 (d = 4): The terms are $7-4, 7, 7+4$, which gives us 3, 7, 11.
- Case 2 (d = -4): The terms are $7-(-4), 7, 7+(-4)$, which gives us 11, 7, 3.
Both sequences fit the clues perfectly!
Pro Tip: Variable Choice is Key
The choice to represent the terms as $a-d, a, a+d$ is a game-changer. It makes the common difference $d$ cancel out immediately when you sum them, giving you a direct path to one of your variables. If you’re ever asked about 5 consecutive terms, use the same logic: $a-2d, a-d, a, a+d, a+2d$. The symmetry is your friend!
Puzzle 2: From Formula to Sum
Sometimes you’re not given the sequence itself, but the general formula for the nth term, $u_n$. The first task is to confirm it’s actually arithmetic.
Example: General Term
A sequence is defined by the general term $u_n = 4n – 1$.
a) Show that the sequence is arithmetic.
b) Find the sum of the first 25 terms.
Part a: The Proof
To prove a sequence is arithmetic, we need to show there’s a constant common difference. The easiest way is to calculate the first few terms and check the gaps.
- $u_1 = 4(1) – 1 = 3$
- $u_2 = 4(2) – 1 = 7$
- $u_3 = 4(3) – 1 = 11$
Now, let’s check the differences:
$u_2 – u_1 = 7 – 3 = 4$
$u_3 – u_2 = 11 – 7 = 4$
Since the difference between consecutive terms is a constant value of 4, the sequence is indeed arithmetic with $d=4$.
Part b: The Sum
We need to find $S_{25}$. We have $n=25$, $u_1=3$, and $d=4$. We could use the first sum formula, but let’s use the shortcut! We just need to find the 25th term, $u_{25}$.
$$u_{25} = 4(25) – 1 = 100 – 1 = 99$$
Now, plug it into the slick formula $S_n = \frac{n}{2}(u_1 + u_n)$:
$$S_{25} = \frac{25}{2}(3 + 99)$$
$$S_{25} = \frac{25}{2}(102)$$
$$S_{25} = 25 \times 51 = 1275$$
The sum of the first 25 terms is 1275.
Hint: Spotting the Common Difference
When the general term $u_n$ is a linear expression like $u_n = an+b$, the sequence will always be arithmetic! Even better, the common difference $d$ is simply the coefficient of $n$. In our example $u_n = 4n – 1$, we can see immediately that $d=4$ without even calculating the first few terms. This is a fantastic time-saver in an exam.
Puzzle 3: Finding the Missing Count
This time, we know the beginning, the end, and the total sum. Our job is to figure out how many terms it took to get there.
Example: Finding n
The sum of an arithmetic series is 440. The first term is 5 and the last term is 75. How many terms are in the series?
Step 1: Identify Your Tools
We are given $S_n = 440$, $u_1 = 5$, and $u_n = 75$. We need to find $n$. This is a perfect setup for the shortcut sum formula!
$$S_n = \frac{n}{2}(u_1 + u_n)$$
Step 2: Substitute and Solve
Let’s plug in the numbers we know and solve for the one we don’t.
$$440 = \frac{n}{2}(5 + 75)$$
$$440 = \frac{n}{2}(80)$$
$$440 = 40n$$
Now, just divide both sides by 40:
$$n = \frac{440}{40} = 11$$
So, there are 11 terms in the sequence.
Puzzle 4: The Simultaneous Equation Showdown
Welcome to the main event! This is a very common IB-style problem where you’re given two separate pieces of information, and you have to combine them to find your unknowns. It’s time to put on your algebra hat.
IB Style Example
In an arithmetic sequence, the 4th term is 18 and the sum of the first 9 terms is 198. Find the first term ($u_1$) and the common difference ($d$).
Step 1: Translate Clues into Equations
We have two unknowns ($u_1$ and $d$), so we need two equations. Let’s use our formula booklet to create them.
Clue 1: The 4th term is 18.
Using the nth term formula, $u_n = u_1 + (n-1)d$:
$$u_4 = u_1 + (4-1)d = 18$$
$$\bf u_1 + 3d = 18 \quad \text{(Equation 1)}$$
Clue 2: The sum of the first 9 terms is 198.
Using the sum formula, $S_n = \frac{n}{2}(2u_1 + (n-1)d)$:
$$S_9 = \frac{9}{2}(2u_1 + (9-1)d) = 198$$
$$\frac{9}{2}(2u_1 + 8d) = 198$$
Let’s simplify this. Multiply both sides by 2/9:
$$2u_1 + 8d = 198 \times \frac{2}{9}$$
$$2u_1 + 8d = 22 \times 2 = 44$$
We can simplify further by dividing the entire equation by 2:
$$\bf u_1 + 4d = 22 \quad \text{(Equation 2)}$$
Step 2: Solve the System of Equations
Now we have a neat system to solve:
1) $u_1 + 3d = 18$
2) $u_1 + 4d = 22$
Substitution is great, but elimination looks even faster here! Let’s subtract Equation 1 from Equation 2:
$$(u_1 + 4d) – (u_1 + 3d) = 22 – 18$$
$$d = 4$$
Brilliant! We’ve found the common difference.
Step 3: Back-substitute to Find the Final Piece
Now, substitute $d=4$ back into either equation to find $u_1$. Let’s use Equation 1:
$$u_1 + 3(4) = 18$$
$$u_1 + 12 = 18$$
$$u_1 = 6$$
And we’re done! The first term is 6 and the common difference is 4.
Your Mission, Should You Choose to Accept It
As you can see, solving these problems is all about strategy. It’s about recognizing the clues, picking the right tool (formula) for the job, and executing the algebra cleanly. From the clever $a-d, a, a+d$ setup to solving simultaneous equations, these techniques form the backbone of your work with sequences. Master them, and you’ll be able to dismantle any arithmetic sequence problem with confidence.
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: An arithmetic sequence has a first term of -5 and a common difference of 3. What is the 15th term?
Show Answer
Solution: Use the formula $u_n = u_1 + (n-1)d$.
$u_{15} = -5 + (15-1)(3)$
$u_{15} = -5 + (14)(3) = -5 + 42 = 37$.
Question 2: The sum of three consecutive terms of an arithmetic sequence is 45 and their product is 3240. Find the terms.
Show Answer
Solution: Let the terms be $a-d, a, a+d$.
Sum: $(a-d) + a + (a+d) = 3a = 45 \implies a = 15$.
Product: $(15-d)(15)(15+d) = 3240 \implies (15-d)(15+d) = 216$.
$225 – d^2 = 216 \implies d^2 = 9 \implies d = \pm 3$.
The terms are either 12, 15, 18 or 18, 15, 12.
Question 3: Find the sum of the first 50 terms of the arithmetic sequence 4, 7, 10, …
Show Answer
Solution: Here, $u_1=4$, $d=3$, and $n=50$.
Use the formula $S_n = \frac{n}{2}(2u_1 + (n-1)d)$.
$S_{50} = \frac{50}{2}(2(4) + (50-1)(3)) = 25(8 + 49 \times 3) = 25(8 + 147) = 25(155) = 3875$.
Question 4: The sum of an arithmetic series is 180, its first term is 2, and its last term is 34. How many terms are there in the series?
Show Answer
Solution: Use $S_n = \frac{n}{2}(u_1 + u_n)$.
$180 = \frac{n}{2}(2 + 34) = \frac{n}{2}(36) = 18n$.
$n = \frac{180}{18} = 10$. There are 10 terms.
Question 5: A sequence is given by $u_n = 7 – 2n$. Find the sum of the first 22 terms.
Show Answer
Solution: First, find $u_1$ and $u_{22}$.
$u_1 = 7 – 2(1) = 5$.
$u_{22} = 7 – 2(22) = 7 – 44 = -37$.
Now use $S_n = \frac{n}{2}(u_1 + u_n)$.
$S_{22} = \frac{22}{2}(5 + (-37)) = 11(-32) = -352$.
Question 6: In an arithmetic sequence, the 3rd term is 14 and the 10th term is 49. Find the first term and common difference.
Show Answer
Solution: Set up a system of equations.
Eq 1: $u_3 = u_1 + 2d = 14$.
Eq 2: $u_{10} = u_1 + 9d = 49$.
Subtracting Eq 1 from Eq 2: $(u_1+9d) – (u_1+2d) = 49-14 \implies 7d = 35 \implies d=5$.
Substitute $d=5$ into Eq 1: $u_1 + 2(5) = 14 \implies u_1 + 10 = 14 \implies u_1=4$.
The first term is 4 and the common difference is 5.
Question 7: How many terms of the sequence 5, 9, 13, … are needed to make a sum of 465?
Show Answer
Solution: We have $u_1=5$, $d=4$, $S_n=465$. We need to find $n$.
Use $S_n = \frac{n}{2}(2u_1 + (n-1)d)$.
$465 = \frac{n}{2}(2(5) + (n-1)4) = \frac{n}{2}(10 + 4n – 4) = \frac{n}{2}(4n + 6) = n(2n+3)$.
$2n^2 + 3n = 465 \implies 2n^2 + 3n – 465 = 0$.
Using the quadratic formula or a solver: $n = \frac{-3 \pm \sqrt{3^2 – 4(2)(-465)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 3720}}{4} = \frac{-3 \pm \sqrt{3729}}{4} = \frac{-3 \pm 61}{4}$.
Since $n$ must be positive, $n = \frac{58}{4}$ is not an integer. Let me recheck my math. Ah, $n(2n+3)$ is not correct. It’s $n(2n+3)$ from $ rac{n}{2}(4n+6)$. $465 = 2n^2+3n$. Let’s try factoring differently. Let’s see… $2n^2 + 3n – 465 = 0$. $n=15$ gives $2(225) + 3(15) – 465 = 450 + 45 – 495
eq 0$. Oh wait. $450+45=495$. No, $450+45=495$. I meant $450+45-465 = 30$. Wait. Let’s re-calculate. Ah, $2n^2+3n-465=0$. Let’s try $n=15$. $2(15)^2 + 3(15) – 465 = 2(225) + 45 – 465 = 450 + 45 – 465 = 495 – 465 = 30$. My quadratic is correct. Let me re-check the quadratic formula. $n = rac{-3 + 61}{4} = rac{58}{4}$, not an integer. Let me re-check the question setup… Ah, perhaps I made an error in the question design. Let’s change the sum. Let the sum be 440. No, that was a previous example. Let’s try to make the factors work. If $n=15$, sum is $S_{15} = rac{15}{2}(2(5) + 14(4)) = rac{15}{2}(10+56) = rac{15}{2}(66) = 15 imes 33 = 495$. OK, let’s change the question sum to 495. Re-writing solution.
Corrected Question: How many terms of the sequence 5, 9, 13, … are needed to make a sum of 495?
Solution: We have $u_1=5$, $d=4$, $S_n=495$. We need to find $n$.
Use $S_n = \frac{n}{2}(2u_1 + (n-1)d)$.
$495 = \frac{n}{2}(2(5) + (n-1)4) = \frac{n}{2}(10 + 4n – 4) = \frac{n}{2}(4n + 6) = n(2n+3)$.
$2n^2 + 3n – 495 = 0$.
Solving the quadratic (e.g., using a GDC or the quadratic formula), we get $n = 15$ or $n = -16.5$. Since $n$ must be a positive integer, $n=15$.
Question 8: The first term of an arithmetic sequence is 20 and the last term is -10. If the sum of the terms is 75, find the common difference.
Show Answer
Solution: First find $n$. $S_n=75, u_1=20, u_n=-10$.
$75 = \frac{n}{2}(20 – 10) = \frac{n}{2}(10) = 5n \implies n=15$.
Now use the nth term formula to find $d$. $u_{15} = u_1 + 14d = -10$.
$20 + 14d = -10 \implies 14d = -30 \implies d = -\frac{30}{14} = -\frac{15}{7}$.
Question 9: The sum of the first 8 terms of an arithmetic series is 100. The sum of the first 15 terms is 375. Find the first term.
Show Answer
Solution: Set up a system of equations using $S_n = \frac{n}{2}(2u_1 + (n-1)d)$.
Eq 1 ($S_8=100$): $100 = \frac{8}{2}(2u_1 + 7d) = 4(2u_1+7d) \implies 25 = 2u_1 + 7d$.
Eq 2 ($S_{15}=375$): $375 = \frac{15}{2}(2u_1 + 14d) = 15(u_1+7d) \implies 25 = u_1+7d$.
We have: $2u_1+7d=25$ and $u_1+7d=25$.
Subtracting the second from the first: $(2u_1+7d) – (u_1+7d) = 25-25 \implies u_1 = 0$.
Question 10 (IB Exam Style Question): An arithmetic sequence has first term $u_1$ and common difference $d$. The sum of the first 10 terms is 175. The fifth term is twice the second term.
- (a) Write down two equations in $u_1$ and $d$.
(b) Find the value of $u_1$ and the value of $d$.
(c) Find the value of the 20th term of the sequence.
Reveal Solution
Solution:
(a) Setting up the equations:
Clue 1: $S_{10} = 175$. Using the sum formula: $175 = \frac{10}{2}(2u_1 + (10-1)d) = 5(2u_1 + 9d)$. Dividing by 5 gives our first equation: $35 = 2u_1 + 9d$.
Clue 2: $u_5 = 2u_2$. Using the term formula: $u_1 + (5-1)d = 2(u_1 + (2-1)d) \implies u_1 + 4d = 2(u_1 + d) \implies u_1 + 4d = 2u_1 + 2d$. Rearranging this gives our second equation: $u_1 = 2d$.
So the two equations are: $2u_1 + 9d = 35$ and $u_1 = 2d$.
(b) Solving for $u_1$ and $d$:
This is a perfect setup for substitution. Substitute $u_1 = 2d$ into the first equation:
$2(2d) + 9d = 35 \implies 4d + 9d = 35 \implies 13d = 35 \implies d = \frac{35}{13}$.
Now find $u_1$: $u_1 = 2d = 2(\frac{35}{13}) = \frac{70}{13}$.
So, $u_1 = \frac{70}{13}$ and $d = \frac{35}{13}$.
(c) Finding the 20th term:
Use the term formula $u_n = u_1 + (n-1)d$ with our new values:
$u_{20} = \frac{70}{13} + (20-1)(\frac{35}{13}) = \frac{70}{13} + 19(\frac{35}{13}) = \frac{70 + 665}{13} = \frac{735}{13}$.
The 20th term is $\frac{735}{13}$.