The ACT will probably contain two to three questions on each ACT test from a collection of additional topics. Even though none of the topics will not appear more than once on a test,collectively they could make a significant different to your ACT Math score.

AP AND GP SERIES

For the purpose of the ACT, there are only TWO important series:

1. Arithmetic Progression

Consider these two common sequences

• 1,3, 5, 7, . . .
•  0, 10, 20, 30, 40, . . . .

It is easy to see how these sequences are formed. They each start with a particular first term, and then to get successive terms we just add a fixed value to the previous term. In the first sequence we add 2 to get the next term, and in the second sequence we add 10.

So the difference between consecutive terms in each sequence is a constant. We could also subtract a constant instead, because that is just the same as adding a negative constant.

Any sequence with this property is called an ARITHMETIC PROGRESSION, or AP for short

The following are the commonly used notations in an AP series:

• 'a’ stand for the first term of the sequence
• ‘d’ stand for the common difference between successive terms.
• 'n' stands for the nth term in the series
• 'Sn' stands for the sum of the first n terms in the series

If we wanted to write down the n-th term, we would have

EXAMPLE: An arithmetic progression has 3 as its first term. Also,the sum of the first 8 terms is twice the sum of the first 5 terms. Find the common difference.We are given that a = 3. We are given information about\(S_8\) and \(S_5\), and we want to find the common difference.

\(S_8\) = (2a +(8-1)d)              --->         \(S_8\) = 4(2(3)+7d) and

\(S_5\) =(2a +(5-1)d)               --->         \(S_5\) = (2(3)+4d)

So, using the given fact that \(S_8=2(S_5\) , we see that

\(4(2(3)+7d)=2\cdot \frac{5}{2}(2(3)+4d)\)

\(8d=6\)

\(d=\frac{3}{4}\)

EXAMPLE 1:

Find the sum of all integers between 100 and 1000 which are divisible by 9.

A.55300

B.53500

C.55350

D.55530

Solution: The first integer greater than 100 and divisible by 9 is 108 and the integer just smaller than 1000 and divisible by 9 is 999. Thus, we have to find the sum of the series:

108 + 117 + 126 + …………. + 999.

Here,

         \( a= 108;\)          \(  d = 9; \)        \(a_n= 999 \)

Using the formulae\(a_n = a + (n-1)d\) ,we get

                                 \(999= 108 + (9 - 1) d \) 

                                    \(n=100\)

To find the sum, we use the sum formulae:

            \(S_n= (2a + (n-1)d) = (2(108) + (100-1)9) = 55350.\)

The correct answer is C.

2. Geometric Progression

A GEOMETRIC PROGRESSION, or GP, is a sequence where each new term after the first is obtained by multiplying the preceding term by a constant ‘r’, called the common ratio. If the first term of the sequence is ‘a’ then the geometric progression is

\(a, ar, ar^2 , ar^3 , . . .\)

The n-th term is: 

The sum of the terms of a geometric series is calculated by:

EXAMPLE: How many terms are there in the geometric progression 2, 4,8, . . ., 128?

In this sequence a = 2 and r = 2. We also know that the n-th term is 128. But the formula for the n-th term is \(ar^{n-1}\) .

\(128 = 2\cdot 2^{n-1}\) 

\(64 = 2^{n-1}\)

\(2^6=2^{n-1}\)

\(n = 7\)

EXAMPLE 2:

Since the beginning of 1990, the number of squirrels in a certain wooded area has tripled during every 3-year period of time. If there were 5,400 squirrels in the wooded area at the beginning of 1999, how many squirrels were in the wooded area at the beginning of 1990?

A.150

B.200

C.250

D.300

‍Solution: The number of squirrels triples every three years,so this is a geometric sequence. We first need to count how many times three years has passed between 1990 and 1999.Including the year 1990 and the year 1999, there are 4 terms for every 3 years between 1990 and 1999

                                                       1990,1993, 1996, 1999

This means that 1999 is our 4th term and 1990 is our 1st term. an = 5400, since there were 5400 squirrels in 1999.Now let’s plug in our values into our formula:

\(a_n=ar^{n-1}\)

\(a_n=a(3^{4-1})\)

\(5400 = a(27) \)

\(200 = a\)

The correct answer is B.

Matrices

A MATRIX is used to arrange number or data into rows and columns.

 Here is an EXAMPLE of a matrix

\(\begin{bmatrix}

5&2\\

4&-1 \\

-5&2

\end{bmatrix}\)

Matrices can be classified by listing first the number of rows then the number of columns. The matrix above is a 3x2 matrix because it has 3 rows and 2 columns.

The following matrix:

\(\begin{bmatrix}

3\\-8

\\ 7

\end{bmatrix}\)

is a 3x1 matrix because it has 3 rows and only 1 column.

Each number within a matrix is called an ENTRY.

Matrices are considered to be equal only if all the corresponding entries are equal. Thus

\(\begin{bmatrix}

2 & -1\\

6 & 5

\end{bmatrix}=\begin{bmatrix}

2 & -1\\

6 & 5

\end{bmatrix}\)

\(\begin{bmatrix}

7 & 0\\

-3 & 2

\end{bmatrix}=\begin{bmatrix}

7 & 0\\

-3 & 2

\end{bmatrix}\)

Adding & Subtracting Matrices

Matrices can only be added if they have the same number of rows and the same number of columns.

To add or subtract matrices....

• Make sure they have the same number of rows and columns (or you cannot add them)
• Add or subtract each set of corresponding entries

EXAMPLE:   

\(\begin{bmatrix}

6 & -5 & 6 \\

-3 & 4 & 7

\end{bmatrix} + \begin{bmatrix}

5 & 2 & 0 \\

-8 & -1 & 9

\end{bmatrix}=\begin{bmatrix}

(6+5) & (-5+2) & (6+0) \\

(-3-8) & (4-1) & (7+9)

\end{bmatrix}\)

SETS

Translation problems which involve two or more given sets of data that partially intersect with each other are termed overlapping sets.

For EXAMPLE: 30 people are in a room. 20 of them play golf. 15 of them play golf and tennis.If everyone plays at least one of the two sports, how many of the people play tennis only?

This problem involves two sets: people who play golf and people who play tennis.

The two sets overlap because some of the people who play golf also play tennis. Thus, these two sets can actually be divided into four categories:

1. People who only play golf

2. People who only play tennis

3. People who play golf and tennis

4. People who play neither sport

Solving double-set ACT problems, such as in the example above, involves finding values for these four categories

The Double-Set Matrix

For ACT problems involving only two categorizations or decisions, the most efficient tool is theDOUBLE-SET MATRIX, a table whose rows correspond to the options for one decision, and whose columns correspond to the options for the other decision.

The last row and the last column contain totals, so the bottom right corner contains the total number of everything or everyone in the problem.

Even if you are accustomed to using Venn diagrams for these problems, you should switch to the double-set matrix for problems with only two sets of options. The double-set matrix conveniently displays all possible combinations of options, including totals, whereas the Venn diagram only displays a few of them easily.

EXAMPLE: Of 30 integers, 15 are in set A, 22 are in set B, and 8 are in both set A and B. How many of the integers are in NEITHER set A nor set B?

Once the information given in the problem has been filled in, complete the chart, using the totals to guide you. (Each row and each column sum to a total value.)

The question asks for the number of integers that are in neither set. Look at the chart to find the number of integers that are NOT A and NOT B; you’ll find that the answer is 1.

‍WHEN YOU CONSTRUCT A DOUBLE-SET MATRIX, BE CAREFUL! As mentioned above, the rows should correspond to the mutually exclusive options for one decision. Likewise, the columns should correspond to the mutually exclusive options for the other.

For instance, if a problem deals with students getting either right or wrong answers on problems 1 and 2, the columns should not be “problem 1” and “problem 2,” and the rows should not be “right” and “wrong.” Instead, the columns should list options for one decision

• ‍problem 1 correct, problem 1 incorrect, total
• and the rows should list options for the other decision
• ‍problem 2 correct, problem 2 incorrect, total.

Overlapping Sets and Percents

Many overlapping-sets problems involve percents or fractions. The double-set matrix is still effective on these problems, especially if you pick a SMART NUMBER for the grand total. For problems involving percents, pick a total of 100. For problems involving fractions, pick a common denominator for the total.

For EXAMPLE, pick 15 or 30 if the problem mentions categories that are 1/3 & 2/5 of the total.

EXAMPLE, 70% of the guests at Company X's annual holiday party are employees of Company X.10% of the guests are women who are not employees of Company X. If half the guests at the party are men, what percent of the guests are female employees of Company X?

First, fill in 100 for the total number of guests at the party. Then, fill in the other information given in the problem: 70% of the guests are employees, and 10% are women who are not employees. You also know that half the guests are men.

Next, calculate the rest of the information in the matrix:

                                             100 - 70 = 30 guests who are not employees

                                             30 - 10 = 20 men who are not employees

                                             50 - 10 = 40 female employees

                                             50 + 20 = 30 male employees

Thus, 40% of the guests at the party are female employees of Company X. Note: the problem doesn't require you to complete the matrix with the number of male employees. However,completing the matrix is an excellent way to check your computation. The last box you fill in must work both vertically and horizontally.

As in other problems involving Smart Numbers, you can only assign a number to the total if it is undetermined to start with. If the problem contains only fractions and/or percents, but no actual numbers of items or people, then go ahead and pick a total of 100 (for percent problems) or a common denominator (for fraction problems). But if actual quantities appear anywhere in the problem, then all the totals are already determined. In that case, you cannot assign numbers,but must solve for them instead.

Overlapping Sets and Algebraic Expressions

When solving overlapping sets problems, you must pay close attention to the wording of the problem.

EXAMPLE: Santa estimates that 10% of the children in the world have been good this year but do not celebrate Christmas, and that 50% of the children who celebrate Christmas have been good this year. If 40% of the children in the world have been good, what percentage of children in the world are not good and do not celebrate Christmas?

It is tempting to fill in the number 50 to represent the percent of good children who celebrate Christmas. However, this approach is incorrect. Notice that you are told that 50% of the children who celebrate Christmas have been good.

This is different from being told that 50% of the children in the world have been good. In this problem, the information you have is a fraction of an unknown number. You do not yet know how many children celebrate Christmas.

Therefore, you cannot yet write a number for the good children who celebrate Christmas.Instead, you represent the unknown total number of children who celebrate Christmas with the variable x. Thus, you can represent the number of good children who celebrate Christmas with the expression 0.5x.

From the relationships in the table, you can set up an equation to solve for x:

0.5x + 10 = 40

x = 60

With this information, you can fill in the rest of the table:

Therefore, 30% of the children are not good and do not celebrate Christmas.

3-Set Problems: Venn Diagrams

Problems that involve three overlapping sets can be solved by using a Venn Diagram. The THREE OVERLAPPING SETS are usually three teams or clubs, and each person is either on or not on any given team or club.

For EXAMPLE: Workers are grouped by their areas of expertise and are placed on at least one team. There are 20 workers on the Marketing team, 30 on the Sales team, and 40 on the Vision team. 5 workers are on both the Marketing and Sales teams, 6 workers are on both the Sales and Vision teams, 9 workers are on both the Marketing and Vision teams, and 4 workers are on all three teams. How many workers are there in total?

In order to solve this problem, use a Venn Diagram. A Venn Diagram should be used ONLY for problems that involve three sets. Stick to the double-set matrix for two-set problems.

Begin your Venn Diagram by drawing three overlapping circles and labeling each one. Notice that there are seven different sections in a Venn Diagram. There is one innermost section (A) where all three circles overlap. This contains individuals who are on all three teams.

‍

There are three sections (B, C, and D)

where two circles overlap. These

contain individuals who are on two

teams. There are three non-overlapping

sections (E, F, and G) that contain

individuals who are on only one team.

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Venn Diagrams are easy to work with, if you remember one simple rule: WORK FROM THE INSIDE OUT.

That is, it is easiest to begin by filling in a number in the innermost section (A). Then, fill in numbers in the middle sections (B, C, and D). Fill in the outermost sections (E, F, and G) last.

1. Workers on all three teams. Fill in the innermost circle. This is given in the problem as 4.

2. Workers on two teams. Here you must remember to subtract those workers who are all three teams.

For EXAMPLE: the problem says that there

are 5 workers on the Marketing and Sales

teams. However, this includes the 4 workers

who are on all three teams. Therefore, in order

to determine the number of workers who are

on the Marketing and Sales teams exclusively,

you must subtract the 4 workers who are on

all three teams. You are left with 5 - 4 = 1 .

‍

The number of workers on the Marketing and Vision teams exclusively is 9 - 4 = 5. The number of workers on the Sales and Vision teams exclusively is 6 - 4 = 2.

‍3. Workers on one team only. Here you must remember to subtract those workers who are on two teams and those     workers who are on three teams. For EXAMPLE, the problem says that there are 20 workers on the Marketing     team. But this includes the 1 worker who is on the Marketing and Sales teams, the 5 workers who are on the     Marketing and Vision teams, and the 4 workers who are on all three teams. You must subtract all of these workers to find that there are 20 - 1 - 5 - 4 = 10 people who     are on the Marketing team exclusively. There are 30 - 1 - 2 - 4 = 23 people on the Sales team exclusively.

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‍

There are 40 - 2 - 5 - 4 = 29

people on the Vision team

exclusively. In order to determine

the total, just add all seven

numbers together = 74

total workers.

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Patterns

Some problem-types on the ACT are practically impossible to solve if you’ve never seen them before, but insanely easy to solve once you’ve done them once or twice. The following “sequence identification” problem-type is one such problem type.

In the context of the ACT, PATTERN RECOGNITION involves spotting a repeating cycle or other simple relationship underlying a series of numbers. If you can grasp the rule, you can predict numbers that appear later in the series. The series may be part of a defined sequence, or it may arise from a general list of possibilities.

Here's a simple EXAMPLE: If Today is Tuesday, what day will it be 300 days from now?

Uuuuuuhhhh…..how the heck could I ever solve this? Do they expect me to count till 300 days ago?

‍No, But you can answer this question in literally two seconds if I know how too. The answer is MONDAY. Here’s how I know that:

1. I figure out how many terms are in the sequence before it starts to repeat. In this case, there are 7 since there are seven days in a week.

‍2. I figure out the closest number to my target number divisible by the answer to #1. In other words, what’s the closest number to 300 that’s divisible by 7? That is 294.     That will be the last term in my sequence -> Monday in our example. Therefore the 294thterm will be Monday.

‍3. Count manually until you get to the one you’re looking for. Therefore, 295th term is Tuesday, the 296th term is Wednesday, 297th term is Thursday, 298th term is     Friday,299th term is Saturday and 300th Term is Monday.

The answer is Monday.

In the following sequence: A, B, C, D, E, F, A, B, C, D, E, F, what is the 602nd term?

A. A

B. B

C. D

D. F

Solution: Since there are 6 terms in this sequence, anything divisible by 6 will be F, the last term of the sequence!

600 is divisible by 6, which means that F is the 600th term. Now we just count manually. 600 is F, 601 is A, 602 is B - done! Therefore, the correct answer is B.

This trick ONLY works with the LAST term of the sequence.

You need to think of the entire pattern as one giant unit that can’t be broken up. Try these two problems on your own, and look at the answer explanations if you have any trouble:

‍EXAMPLE: In the following sequence: 1, 2, 2, 0, 4, 1, 2, 2, 0, 4, 1, 2, 2, 0, 4 (repeated indefinitely),

what is the sum of the first 613 terms of this sequence?

1. Figure out what the sequence looks like. “1, 2, 2, 0, 4” - that’s the unit that repeats. There are 5 terms in this sequence

2. Figure out the sum of that sequence. 1 + 2 + 2 + 0 + 4 = 9. So for every 5 terms, you have a sum of “9”

3. Divide 613 by 5. We get 122.6

4. Realize that this sequence repeats itself in full 122 times. So multiply 122 by the sum of the sequence, which is 9, and we get: 122 x 9 = 1,098

5. Get rid of the “.6” at the end of 122 and multiply that by 5. 122 x 5 = 610. That means that the sum of the first 610 terms is 1,098

6. Count manually from 610. We still have terms 611, 612, and 613 to go, which are 1, 2, and 2, respectively. So we add 1 + 2 + 2 =5 to our former total, 1,098, and we get     1,103.

EXAMPLE 5:

In the following sequence, A, B, C, D, E, F, G, H (repeated indefinitely), what’s the 456,791st term?

A. A

B. D

C. F

D. G

Solution:

1. Count the term numbers: There are 8

2. Divide 456,791 by 8. We get 57,098.875

3. Get rid of the “.875”

4. Multiply the remainder by 8. 57,098 x 8 = 456,784.

5. Count manually. 456,784 = H, 456,785 = A, 456,786 = B, 456,787 = C, 456,788 = D, 456,789 = E,

    456,790 = F, 456,791 = G. The correct answer is C