VMHS Math Circle
VII. The Grand Algebra Review (Part II: Data Sets, Sequences, and Series)
Data sets, sequences, and series all have this common little theme: lists of numbers. Whether
you’re trying to interpret a list of numbers you have or find out whatever term is in there, whatever the
AMC asks you concerning lists of numbers mostly go along the lines of data sets, sequences, and series.
To start off, let’s talk about some stuff that’s really good to know concerning data sets. Since what I’m
going to talk about is pretty straightforward, there’s not necessarily much background knowledge you
should have under your belt. If you know this stuff already and feel like it’s review, just check the
practice problems out. I put them in a different font partly because of that. Let’s begin!
Mean, Median, and Mode
I’m pretty much assuming that you know what mean, median, and mode mean (Ha ha ha.). But
for a quick recap, the mean is the average of all terms, the median is the middle term (or the average of
the two middle terms if we have an even number of terms), and the mode is the most abundant, or most
occurring term. Let’s start with an easy practice problem:
Given the set S of numbers {1, 1, 1, 2, 2, 5, 6, 8, 12, 12}, find the mean, median, and mode of set S.
The mode seems easiest to approach. We see that the number 1 occurs the most times, so our mode is
1. Our median is the middle term, and since we have an even number of terms, we just need to add our
middle terms and divide by 2. Hence, our median is (2 + 5)/2 = 3.5. Our mean is the average of the
terms. To find this, we need to find the sum of the terms and divide by how many terms there are. This
is (1 + 1 + 1 + 2 + 2 + 5 + 6 + 8 + 12 + 12)/10 = 50/10 = 5.
Pretty straightforward. With this out of the way, I’m going to mention some pointers:
A set can have more than one mode.
The median has to be in the middle. You can’t simply add the lowest and highest numbers in the
set, divide it by two, and call it your median. It doesn’t work like that. Even if you had the set {1,
2, 100}, the number 2 would still be the median, not 50.
 The mean, median, and mode in a set can be the same. Take this set, for example: {2, 2, 2, 2, 2}.
It’s obvious that the average, the middle value, and the most occurring number is 2.
Again, how does the AMC make this harder? Bending some concepts of what a mean or median
is, combining that with some algebra, and tying in little tricks or things you must watch out for in order to
not miss questions and lose points. Here are some problems to show what I’m talking about.


A set of positive integers has 5 terms as follows: 1, 5, 3, 9 and x. For what values of x is the median of
this set equal to the mean?
Okay, I’m going to throw out some pointers, but let’s order these terms first. Now, our set of integers is
as follows: 1, 3, 5, 9, and x. We have multiple possible medians in 3, 5, and x, not just one! We know
that neither 7 nor 1 can be a possible median because 1 is either the lowest or second lowest number in
the set, and 7 is either the highest or second highest number in the set, depending on where we put x.
Also, since we want to find suitable means, we have to find the sum of the numbers we have and divide
by 5. Let’s call our mean “m.” It follows that we have the equation
(16 + x)/5 = m  16 + x = 5m.
The problem tells us that our median must be the same as the mean, and we know that such values for
our mean can be 3, 5, or x. Thus, we make three separate equations, substituting these values for “m.”
Thus, we have:
16 + x = 5x  16 = 4x  x = 4
16 + x = 5(3) = 16 + x = 15  x = -1
16 + x = 5(5)  16 + x = 25  x = 9
Our problem says that we must have positive integers. Hence, the answer is x = 4 or 9.
If we have a set S of 8 positive integers with a mean of 10, a median of 8, and a unique mode of 9, how
many times must 9 be included for S to contain the largest integer it can possibly have?
Since we have a kind-of big set of numbers, it would help to just experiment right away, rather than do
something complicated. Let’s say the mode 9 occurs twice. Remember that our median has to be equal
to 8. Since we have an even number of terms, the two middle terms must average to 8. Moreover, the
mean is 10, so all 8 numbers must add up to 80. Experimenting to find our maximum value, we have a
set that looks good:
{1, 2, 3, 7, 9, 9, 10, 39}.
However, we do not want to assume that our maximum is 39 just yet! Mean, median, and mode
questions are those types of questions that always seem to trick people who aren’t careful. We move
on to a case where 9 occurs three times. This following set is optimal:
{1, 1, 2, 7, 9, 9, 9, 42}.
Now we see a number higher than 39. Again, we don’t want to be too careful. We move on to the next
case, when 9 appears four times.
{1, 1, 3, 7, 9, 9, 9, 9}.
Uh-oh. Our median has to be 8, and when 9 appears four times, we no longer get our numbers summing
to 80.  Hence, judging from the cases we already have, the number 9 must be included three times for
S to have its maximum value.
Now that we’ve seen some practice problems, there are a couple of things to consider. When you
see the word “mean,” think sum! Don’t think that the AMC is being mean to you (although it actually
might be in reality), but relate the sum of the numbers in a data set to its mean! Also, when you have
median questions, do assume that there are multiple possibilities for the median! Find all possibilities,
and work from there. For example, let’s consider our sample problem with the mean being the same as the
median. Don’t think that just because you’ve solved for one value of x in that problem, you’ve solved for
all possible values of x. Remember to be as careful as possible, read and understand the question, and
take a little more time to get through those mean, median, and mode questions.
That’s essentially all there is for mean, median, and mode! Let’s move on.
Sequences and Series
Here’s a little twist on having a list of numbers. You know that there’s a sequence in front of you
if you notice a certain pattern that occurs throughout a given set of numbers. In other words, if you see a
common change from one term to the next in a set of numbers, rest assured that set is probably a
sequence.
Let’s deal with the easier of sequences to comprehend: arithmetic sequences. This happens when
you’re adding the same number to each term to get to the next term until one point in the sequence (or the
sequence may never end). With this being said, there are formulas that you should know concerning
arithmetic sequences.
nth term:
an = a1 + d(n – 1)
Where an is the nth term, and d is the common difference (whatever you keep adding). Note that
subtracting is the same as adding by a negative number.
Sum of a Finite Arithmetic Series:
x
∑[a
1
+ d(m – 1)
]
= (x – n + 1)(an + ax)/2
m=n
Basically, you take the number of terms in the sequence, divide that by 2, and multiply that quantity by
what you get when you add the first and last terms.
Obviously, the AMC isn’t going to test this as a mere skill; there’s a problem solving context to
it. However, one good thing to consider is that you can use this in the context of counting how many even
integers there are, starting with 1 or 2 or whatever. It’s good to know that the sequence for odd counting
integers is:
1 + 2(n – 1) = -1 + 2n.
Similarly, it’s good to know that the sequence for even counting integers is:
2n.
This way, when the AMC asks you about the first n odd or even counting integers, you know
what the last term is before you can do some more complex math. Moving onto geometric series!
nth term:
an = a1(rn – 1)
Where an is the nth term, and r denotes the common ratio (what you keep on multiplying). Remember that
dividing means multiplying by a fraction with numerator 1.
Sum of a Finite Geometric Series:
x
∑[a (r
1
m–1
)
]
= an(1 – rx)/(1 – r)
m=n
Here, you take the first term of your sequence, multiply it by (1 – rx)/(1 – r). Now, the AMC doesn’t touch
on sums of finite geometric series very much because these are pretty calculation-heavy. Despite this, be
careful for one.
Sum of an Infinite Geometric Series:
∞
∑[a (r
1
]
) = an/(1 – r)
m–1
m=n
This is only possible if r is between 0 and 1, not inclusive. You’re going to see infinite geometric series
more often because these don’t require as much calculation.
Now that we’ve gotten these formulas out of the way, let’s try some practice problems. With
questions about series, the main thing to do is to rearrange the equations you get in any way possible that
helps you get closer to your answer. Let’s make these ones a little hard .
Let a1, a2, … , ak be a finite arithmetic sequence with
a4 + a7 + a10 = 17
a4 + a5 + a6 + a7 + a8 + a9 + a10 + a11 + a12 + a13 + a14 = 77
and ak = 13.
What is k?
Aieee. One thing to notice is that a4, a7, and a10 and a4 + a5 + a6 + a7 + a8 + a9 + a10 + a11 + a12 + a13 + a14 are
arithmetic sequences themselves. If any grouping of terms in an arithmetic sequence is equally spaced
out, that sequence itself is arithmetic. Why? In an arithmetic sequence, you’re adding the same thing
each time, right? If you’re adding the same thing each time, it doesn’t matter where you start in the
sequence, as long as the terms are equally spaced out. To those who haven’t figured out why said terms
should be equally spaced, it’s so that you can still add the same thing each time. For example, if I have
the sequence 1, 2, 3, 4, 5, 6, …, the 2, 4, 6 is still an arithmetic sequence because the terms have equal
spacing between them. Anyway, let’s get back to the problem.
Let’s treat a4 as our a1 without actually changing that to a1 (since we don’t want to make dumb
mistakes). Then, we have the sums:
a4, a4 + d, a4 + 2d, a4 + 3d, a4 + 4d, a4 + 5d, a4 + 6d, a4 + 7d, a4 + 8d, a4 + 9d, a4 + 10d = 77
a4, a4 + 3d, a4 + 6d = 17
Hey! That’s really cool because we have two variables and two equations. We can make a system of
equations here. We then have:
11a4 + d(1 + 2 + 3 + … + 10) = 77
3a4 + 9d = 17.
Notice that 1, 2, 3, … , 10 is an arithmetic sequence already, so we basically compute the sum of that,
which is (10/2)(1 + 10) = 5(11) = 55. Now, our first equation reduces to
11a4 + 55d = 77. Although you may freak out a bit because it doesn’t seem like we have integer solutions
for this system, you have to move on. You may get caught up trying to check your work, but if you can’t
find any mistakes, all you can do is chug. Let’s solve by elimination by multiplying our second equation
by -11/3. Our system is thus:
11a4 + 55d = 77
-11a4 – 33d = -187/3.
Adding, we have 22d = 44/3. Dividing, we get 44/(3 x 22) = 2/3. Now that’s not the worst solution for d.
Solve for a4, we have 3a4 + 9(2/3) = 17  3a4 = 11  a4 = 11/3. Not too bad, either.
Now, we try to solve for k. We know that the kth (say that out loud and tell me how good that feels) term
of this sequence is 13, so let’s use our formula. I’ll be using k here, but you must remember that we’re
starting with a4!
11/3 + 2(k – 1)/3 = 13  11/3 + 2k/3 – 2/3 = 13  2k/3 + 3 = 13  2k/3 = 10
Our k is equal to 15, and because we’re starting with a4, that’s three terms after a1. Hence, three terms
after 15 gets us our 18th term. If you don’t support this reasoning, just count fifteen fingers starting with
four and see where you land.
Let a, ar1, ar12, ar13, … and a, ar2, ar22, ar23, … be two different infinite geometric series of positive
numbers with the same first term “a.” The sum of the first series is r1, and the sum of the second series
is r2. What is r1 + r2?
We’re given the words “sum,” “infinite,” and “geometric series.” What do you get when you put all
those three words together in a sentence and add the words “of an” to it? This right here:
a1/(1 – r).
Since we’re given that the sum of our first sequence is r1, and the sum of the second is r2, simply make
two different equations. Note that a is the first term.
a/(1 – r1) = r1
a/(1 – r2) = r2
Rearranging to get a by itself and substituting, we have:
a = r1(1 – r1) = r2(1 – r2).
Now, considering the r1’s and r2’s, let’s distribute and see whether something good happens.
r1 – r12 = r2 – r22
Huh. Two minus signs and two squares, I’m thinking “difference of squares” here. Let’s rearrange some
more.
2
r2 – r12 + r1 – r2 = 0
(r2 + r1)(r2 – r1) = r2 – r1
r2 + r1 = 1.
Am I the only one who’s bugged by the number of r’s that I typed? Well, it gets worse. I once was solving
an Olympiad-level problem. As soon as I reached the solution, I counted how many r’s I wrote. Curious
about what that number is? FIVE HUNDRED AND FORTY SIX.
There is one last thing that I’ll cover before wrapping this up: recursions. The AMC might give
you a problem that gives you a sequence a1, a2, a3, blah blah blah, and tell you that the next term bears
some relation to the terms that come before it. Even though there’s no straight-up formula to calculate the
sum of a recursive sequence or find the nth term of that sequence from the very get-go, the most
important thing to have is an open mind. Make sure you recognize some patterns or possible
manipulations that you can utilize to attain your solution. Let’s throw in a problem in here.
Define a sequence of real numbers a1, a2, a3, . . . by a1 = 1 and an+13 = 99an3 for all n ≥ 1.
Then a100 equals:
One thing that really helps is to simplify that recursive formula. Doing so, we have:
an+1/an = √(99), or 991/3. If we start experimenting with numbers, we’ll see that the numerator of the
power we are raising 99 with goes up by one. With that figured out, we now know that a100 = 9999/3 (not
99100/3 because our first term was 1, and our second term, 991/3, had the numerator of 1). Simplifying
gets us 9933.
3
TIPS:
 I will reiterate: whenever the AMC combines a subscript and a power, it likes to line up the two
to make what you’re looking at foreign and confusing. Don’t fall for that. Just treat (incoming
screen cap)



as an raised to the third power.
Remember to consciously check whether you’ve counted your terms correctly. Being off by one
is something very frustrating to deal with.
I will also reiterate: remember to take some time with mean, median, and mode questions.
They’re easier questions to make dumb mistakes on.
Don’t forget that the number of terms you have is the number of the last term minus the number
of the first term plus one!