Math 228
Chapter 1
Practice Test
Answer to the practice test questions
Note that the explanations are not complete.
1.
In each of the following sequences, list the next two terms and the nth term. Indicate if the
sequence is arithmetic, geometric, quadratic, or neither. Show your work to illustrate
how you generated the next two terms. Justify your answer in relation to the type of
sequence you indicated: Why it is arithmetic, geometric, quadratic, or neither?
a) 4, 7, 12, 19, 28,
39 , 52
An = n2 + 3
Type of Sequence
Quadratic
Reason: The second difference is constant
A6 = 28 + 11 = 39
A7 = 39 + 13 = 52
b) 22, 29, 36, 43, 50, _57 , 64
An = 22 + 7(n – 1)
Type of Sequence: Arithmetic sequence
Reason: There is a common difference of 7
between two consecutive terms.
A6 = 50 + 7 = 57
A7 = 57 + 7 = 64
c) 1, 4, 16, 64, 256, 1024, 4096
An = 4n - 1
Type of Sequence: Geometric sequence
Reason: There is a common ratio of 4
between two consecutive terms
A6 = 256(4) = 1024
A7 = 1024(4) = 4096
2. Find all real-value solutions to the following equations: Show your work.
a) x + 4 = 7 – (3 – x)
x+4=7–3+x
x–x=7–3–4
0 = 0 (True statement)
The solutions are all real numbers
b) 3(x – 2) = 3(x + 1)
x – 2 = x + 1 (Dividing both sides by 3)
x–x=1+2
0 = 3 (False statement)
No real number solution
c) (x – 3)(x + 3) = 7
x2 – 9 = 7
x2 – 16 = 0
(x – 4)(x + 4) = 0
x = 4 or x = - 4
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Math 228
Chapter 1
Practice Test
3. Find the sum of the following sequences. Show your work.
a) 9 + 16 + 23 + 30 +… + 156 + 163
Double the sequence and write the second sequence in reverse order.
9 + 16 + 23 + 30 +… + 156 + 163
163 + 156 + …
+ 16 + 9
The corresponding pairs in the two sequences are added together and the sum for each pair is
172. The number of pairs is determined by using the equation An = A1 + d(n – 1). By
substitution of the given values, 163 = 9 + 7(n – 1), or n = 23.
The sum is 23(172)/2 = 1978
b) 25 + 26 + 27 + 28 + 29 + … + 416 + 417
Double the sequence and write the second sequence in reverse order.
25 + 26 + 27 + 28 + 29 + … + 416 + 417
417 + 416 + …
+ 26 + 25
The sum of the corresponding pairs in the two sequences is 442. There were (417 – 24) or 393
sums of 442. Therefore the sum is 393(442)/2, or 86853
4. How many terms are in the following sequences? Show your solution
a) 6, 10, 14, 18, 22, …, 290
n – 1 = (290 – 6)/4
n = 72 terms
b) 3, 6, 12, 24, 48, …, 768
An = A1rn-1
768 = 3(2n – 1)
256 = 2n – 1
28 = 2n – 1
8=n–1
n = 9 terms
5. A conference room is set up in such a way that there are 32 seats in the first row and adds 6
additional seats in each consecutive row. The last row has 146 seats. How many rows are
there in the conference room? What is the total number of seats? SHOW YOUR WORK
AND BRIEFLY EXPLAIN.
An = A1 + d(n – 1)
146 = 32 + 6(n – 1)
n = 20, The conference room has 20 rows
Total number of seats = 20(32 + 146)/2 = 1780 seats in the conference room
6. A theater has 1358 seats. If the first row has 8 seats and each successive row has 3 additional
seats. How many rows are there in the theater? How many seats are there in the last row?
Let L be the number of seats in the last row
n be the number of rows
1358 = n(8 + L)/2, call this equation(1)
L = 8 + 3(n – 1), or L = 3n + 5, call this equation (2)
Substitute equation (2) to equation (1)
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Math 228
Chapter 1
Practice Test
We get, 1358 = n[8 + 3n + 5]/2
1358 = n(3n + 13)/2
2716 = n(3n + 13) (multiply both sides by 2)
3n2 + 13n – 2716 = 0
By Factoring, we get (n – 28)(3n + 97) = 0
Solving for n, we get n = 28 or n = -97/3
Use n = 28 and reject n = - 97/3
Therefore there are 28 rows of seats in the theater and the last row has
L = 3(28) + 5, or 89 seats.
7. At a bargain sale, the following advertisement was provided:
Total value: $802
Total value: $762
Total value: $1100
One Laptop (Computer)
Two Laptop (Computers)
Two Desktop (Computers)
One Desktop (Computer)
One Printer
One Laptop (Computer)
One Printer
What was the price for one printer? SHOW YOUR WORK AND EXPLAIN.
Let D represents the value of the Desktop
L represents the value of the Laptop
P represents the value of the printer
Combining the first and last options we get (L D P + D D + L) and the total value is 802 +
1100 = 1902. Taking away (L L P from (L D P D D L) we get (D D D) and it is now worth 1902
– 762 = 1140. That means that 3D = 1140 or D = 380.
Hence, one Desktop computer is worth $380.
From the third option, 2D + L = 1100
L = 1100 – 2D
= 1100 – 2(380)
= 340, one Laptop is worth $340
Using the second option, 2L + P = 762
P = 762 – 2L
P = 762 – 2(340)
= 82, one printer is worth $82
8. Fifteen people came to a party. If each person shake hands with every other person exactly
once. How many handshakes were made? Explain your work
The number of handshakes = 1 + 2 + 3 + 4 + … + 13 + 14
= 14(1 + 14)/2
= 14(15)/2
= 105 handshakes
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Math 228
Chapter 1
Practice Test
MATHEMATICS EDUCATION
9. Briefly describe three process standards exemplified by the National Council of Teachers of
Mathematics (NCTM, 2000).
10. Explain briefly three benefits of relational understanding.
11. Explain briefly three principles that governed the NCTM (2000) Standards
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