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Chapter 1 Sequences and Series
Section 1.1 Arithmetic Sequences
Section 1.1 Page 16 Question 1
a) 32 − 16 = 16, 48 − 32 = 16, 64 − 48 = 16, … Since successive differences are constant, the sequence is arithmetic. t 1 = 16; d = 16. The next three terms are: 80 + 16 = 96 96 + 16 = 112 112 + 16 = 128
b) Not arithmetic, because the differences of consecutive terms are not constant: 4 − 2 = 2, 8 − 4 = 4,…
c) − 7 − (−4) = −3, − 10 − (−7) = −3, − 13 − (−10) = −3, … Since successive differences are constant, the sequence is arithmetic. t 1 = −4; d = −3. The next three terms are: −16 + (−3) = − 19 −19 + (−3) = − 22 −22 + (−3) = − 25
d) 0 − 3 = −3, − 3 − 0 = −3, − 6 − (−3) = −3, … Since successive differences are constant, the sequence is arithmetic. t 1 = 3; d = −3. The next three terms are: −9 + (−3) = − 12 −12 + (−3) = − 15 −15 + (−3) = − 18
Section 1.1 Page 16 Question 2
a) t 1 = 5, t 2 = 5 + 3 or 8, t 3 = 8 + 3 or 11, t 4 = 11 + 3 = 14 The first four terms are 5, 8, 11, 14.
b) t 1 = −1, t 2 = −1 + (−4) or −5, t 3 = −5 + (−4) or −9, t 4 = −9 + (−4) or − 13 The first four terms are −1, −5, −9, −13.
c) t 1 = 4, t 2 = 4 +
or
, t 3 =
, t 4 =
The first four terms are 4,
d) t 1 = 1.25, t 2 = 1.25 + (−0.25) or 1, t 3 = 1 + (−0.25) or 0.75, t 4 = 0.75 + (−0.25) = 0. The first four terms are 1.25, 1, 0.75, 0.5.
Section 1.1 Page 16 Question 3
a) t (^) n = 3 n + 8 t 1 = 3(1) + 8 t 1 = 11
b) t 7 = 3(7) + 8 t 7 = 29
c) t 14 = 3(14) + 8 t 14 = 50
Section 1.1 Page 16 Question 4
a) t 4 = 19, t 5 = 23 t 5 − t 4 = 23 − 19 or 4 d = 4 t (^) n = t 1 + ( n − 1) d 19 = t 1 + (4 − 1) 19 = t 1 + 12 t 1 = 7 Then, t 2 = 7 + 4 or 11, and t 3 = 11 + 4 or 15. The missing terms are 7, 11, 15; t 1 = 7 and d = 4.
b) t 3 = 3, t 4 =
t 4 − t 3 =
− 3 or
d =
t (^) n = t 1 + ( n − 1) d
3 = t 1 + (3 − 1)
3 = t 1 + (−3) t 1 = 6
Then, t 2 = 6 +
or
The missing terms are 6,
; t 1 = 6 and d =
c) t 2 = 4, t 5 = 10 t 5 = t 2 + 3 d 10 = 4 + 3 d 6 = 3 d d = 2 The missing terms are 2 and 6, 8; t 1 = 2 and d = 2.
n
n
So, −10 is the 17th term of the sequence.
Section 1.1 Page 16 Question 6
a) t 1 = 6, t 4 = 33 Substitute for t 4 in t (^) n = t 1 + ( n − 1) d. 33 = 6 + 3 d 33 − 6 = 3 d 27 = 3 d d = 9 Then, the second term is 6 + 9 or 15, and the third term is 15 + 9 or 24.
b) t 1 = 8, t 4 = 41 Substitute for t 4 in t (^) n = t 1 + ( n − 1) d. 41 = 8 + 3 d 41 − 8 = 3 d 33 = 3 d d = 11 Then, the second term is 8 + 11 or 19, and the third term is 19 + 11 or 30.
c) t 1 = 42, t 4 = 27 Substitute for t 4 in t (^) n = t 1 + ( n − 1) d. 27 = 42 + 3 d 27 − 42 = 3 d −15 = 3 d d = − 5 Then, the second term is 42 + (−5) or 37, and the third term is 37 + (−5) or 32.
Section 1.1 Page 17 Question 7
a) The y -values of each point are the terms. It appears that the first five terms are: 5, 8, 11, 14, 17.
b) From the pattern of the sequence, t 1 = 5 and d = 3. t (^) n = 5 + ( n − 1) t (^) n = 2 + 3 n
c) t 50 = 2 + 3(50) t 200 = 2 + 3(200) t 50 = 152 t 200 = 602
d) Answers may vary. Example: Use the points (1, 5) and (2, 8).
slope 2 1 slope 3
The slope is the same as the common difference. It is the coefficient of the variable term in the general term, 2 + 3 n.
e) If a line were drawn through the points, the y -intercept would be 2. This is the same as the constant value in the general term, 2 + 3 n.
Section 1.1 Page 17 Question 8
Consider A : t (^) n = 6 + ( n − 1) 34 = 6 + 4 n − 4 34 − 2 = 4 n 32 = 4 n n = 8 The sequence defined by tn = 6 + ( n − 1)4 has 34 as its 8th term.
Consider B : t (^) n = 3 n − 1 34 = 3 n − 1 35 = 3 n This does not give a natural number for n , the term number, so 34 cannot be a term of this sequence.
Consider C : t 1 = 12, d = 5. Then, t (^) n = t 1 + ( n − 1) d t (^) n = 12 + ( n − 1)5. 34 = 12 + 5.5 n − 5. 34 − 6.5 = 5.5 n 27.5 = 5.5 n
5
n
n
The sequence for which t 1 = 12 and d = 5.5 has 34 as its fifth term.
Consider D : From the pattern of the sequence, t 1 = 3 and d = 4. Then, t (^) n = 3 + ( n − 1) 34 = 3 + 4 n − 4 34 + 1 = 4 n This does not give a natural number for n , the term number, so 34 cannot be a term of this sequence.
Section 1.1 Page 17 Question 13
a) Perimeters of the four figures shown: 10, 16, 22, 28. The perimeters are an arithmetic sequence with t 1 = 10 and d = 6. So, an equation for the perimeter of figure n is Pn = 10 + ( n − 1)6 or Pn = 6 n + 4.
b) For the perimeter of Figure 9, substitute n = 9. P 9 = 10 + (9 − 1) P 9 = 58 The perimeter of Figure 9 is 58 units.
c) Determine the value of n when Pn = 76. 76 = 10 + ( n − 1) 76 = 10 + 6 n − 6 72 = 6 n n = 12 Figure 12 has a perimeter of 76 units.
Section 1.1 Page 17 Question 14
a) t 1 = 0, d = 8 The tee-off times are: 8:00, 8:08, 8:16, 8:24, … Considering 8:00 to be time 0, the sequence is 0, 8, 16, 24.
b) Extend the sequence to 60 min. 0, 8, 16, 24, 32, 40, 48, 56, … So within the first hour, 8 groups of four will have teed-off. This means 32 players will be on the course after 1 h.
c) t (^) n = 0 + ( n − 1) t (^) n = 8 n − 8
d) For 132 players, there will need to be 132 ÷ 4 or 33 groups teeing-off. Substitute n = 33 into t (^) n. t (^) n = 8(33) − 8 t (^) n = 256
This means that the last group will tee-off 256 min after the first group. 256 min = 4 h 16 min, so the last group will tee-off at 8:00 + 4:16 or 12:16.
e) Answers may vary. Examples: Rain may interrupt the tee-off times or players in a group might not be quite ready at their tee-off time.
Section 1.1 Page 18 Question 15
Area of the wall hanging = 22 × 27 = 594 Substitute t 1 = 48, n = 27, and t (^) n = 594 into tn = t 1 + ( n − 1) d. 594 = 48 + (27 − 1) d 594 = 48 + 26 d 594 − 48 = 26 d 546 = 26 d d = 21 Lucy completed 21 square inches of the wall hanging on each subsequent day.
Section 1.1 Page 18 Question 16
a) t 6 = 11, t 15 = 29 Substitute into t (^) n = t 1 + ( n − 1) d. For n = 6: 11 = t 1 + 5 d c For n = 15: 29 = t 1 + 14 d d Subtract the first equation from the second. 18 = 9 d d = 2 Substitute d = 2 into c to determine t 1. 11 = t 1 + 5(2) t 1 = 1 The general term that relates the number of sit-ups to the number of days is t (^) n = 1 + ( n − 1)2 or tn = 2 n − 1.
b) Substitute t (^) n = 100 into the general term and solve for n. 100 = 1 + ( n − 1) 101 = 2 n n = 50. Susan will be able to do 100 sit-ups on the 51st day of her program.
c) Answers may vary. Example: Assume that she is physically able to continue increasing the number of sit-ups by 2 each day.
Section 1.1 Page 18 Question 17
a) Carbon Atoms 1 2 3 4 Hydrogen Atoms 4 6 8 10
b) From the pattern in the table, t 1 = 4 and d = 2. t (^) n = t 1 + ( n − 1) d t (^) n = 4 + ( n − 1) t (^) n = 4 + 2 n − 2 t (^) n = 2 + 2 n or H = 2 + 2C
Section 1.1 Page 19 Question 19
a) Depth (ft) 0 30 60 90 Water Pressure (psi) 14.7^ 29.4^ 44.1^ 58.
The first four terms of the sequence of water pressure with depth are 14.7, 29.4, 44.1, 58.8. The general term of this sequence is t (^) n = 14.7 n , where n is the number of 30-ft descents.
b) First determine n for 1000 ft. 1000 1 33 30 3
Water pressure
= + ⎛^ − ⎞
The pressure at a depth of 1000 ft is 490 psi.
Determine n for 2000 ft. 2000 2 66 30 3
Water pressure
The pressure at a depth of 1000 ft is 980 psi.
c)
d) The y -intercept is 14.7.
e) The slope is 14.7.
f) The y -intercept is t 1 and the slope is the common difference.
Section 1.1 Page 19 Question 20
Let a represent the length of the shortest side and d the common difference. Then, the four sides are: t 1, t 1 + d , t 1 + 2 d , and t 1 + 3 d. Use the perimeter: t 1 + t 1 + d + t 1 + 2 d + t 1 + 3 d = 60 4 t 1 + 6 d = 60 2 t 1 + 3 d = 30 c Given that the longest side is 24 cm, t 1 + 3 d = 24 d Subtract d from c. t 1 = 6 Substitute t 1 = 6 in (2) to find d. 6 + 3 d = 24 3 d = 18 d = 6
The other three sides lengths are 6 cm, 12 cm and 18 cm. Check: 6 + 12 + 18 + 24 = 60.
Section 1.1 Page 19 Question 21
a) Term Number 1 2 3 4 5 Number of minutes 4 8 12 16 20 Number of degrees^1 2 3 4
The sequence of number of minutes is 4, 8, 12, 16, 20, …
b) The number of minutes, t (^) n , is given by t (^) n = 4 n where n is the number of degrees of turn.
c) The time for a rotation of 80º is 4(80) or 320 min. This is 5 h 20 min.
Section 1.1 Page 19 Question 22
Year 1986 1987 … 2006 2007 Term Number 1 22 Number of Beekeepers 1657 1048
t 1 = 1657, tn = 1048, n = 22 t (^) n = t 1 + ( n − 1) d 1048 = 1657 + (21) d −609 = 21 d d = − 29 From 1986 to 2007, the number of beekeepers decreased by about 29 each year.
b) An arithmetic sequence is an decreasing sequence if and only the difference between one term and the next is a negative number. An arithmetic sequence is a decreasing sequence if and only if d < 0.
c) An arithmetic sequence is constant if and only all the terms are the same, or in other words there is no difference between successive terms. An arithmetic sequence is constant if and only if d = 0.
d) The first term of a sequence is t 1.
e) The symbol for the general term of a sequence is t (^) n.
Section 1.1 Page 21 Question 27
Answers may vary. Example:
Section 1.1 Page 21 Question 28
Answers may vary. Example: Step 1 : The graph that models an arithmetic sequence is always a straight line. Successive terms are obtained by adding a constant, so successive points form steps of a line. No other shape is possible.
Step 2: a) As the value of the first term increases, the points are raised vertically. As the value of the first term decreases, each point is lowered by that amount. b) The graph is still a line with the same slope.
Arithmetic Sequence
Consecutive terms differ by the same amount, the common difference.
The general, n th, term is tn = t 1 + ( n − 1) d , where d is the common difference.
2, 5, 8, 11, 14, … is an arithmetic sequece with first term 2 and common difference 3. For this sequence, tn = 2 + ( n − 1)3.
Each successive term in an ordered list is formed by adding a constant to the previous terms.
Step 3: a) As the value of the common difference increases, successive points are raised vertically more and more. As the value of the common difference decreases, successive points are lowered vertically more and more. b) As the value of the common difference increases, the slope of the line is steeper. As the value of the common difference decreases, the line is less steep.
Step 4: The slope would be the same as the common difference.
Step 5: The slope is the same as the value of d in the general term for the sequence. This is the coefficient of the variable n in t (^) n.
Section 1.2 Arithmetic Series
Section 1.2 Page 27 Question 1
a) t 1 = 5, d = 3, t (^) n = 53 Step 1 : Determine n. Substitute into t (^) n = t 1 + ( n − 1) d. 53 = 5 + ( n − 1) 53 = 5 + 3 n − 3 51 = 3 n n = 17
Step 2 : Determine the sum of the series.
Substitute into ( 1 ) n (^) 2 n
n S = t + t.
17
17
S
S
b) t 1 = 7, d = 7, t (^) n = 98 Step 1 : Determine n. Substitute into t (^) n = t 1 + ( n − 1) d. 98 = 7 + ( n − 1) 98 = 7 + 7 n − 7 98 = 7 n n = 14
Step 2 : Determine the sum of the series.
Substitute into ( 1 ) n (^) 2 n
n S = t + t.
14
14
S
S
b) t 1 = 40, d = −5, n = 11
Substitute into [2 1 ( 1) ] n 2
n S = t + n − d.
1
11
1
11
[2( ) ( 1)( )]
[
0 50]
S 0 11 5
S
S
c) t 1 =
, d = 1, n = 7
Substitute into [2 1 ( 1) ] n 2
n S = t + n − d.
8
7
8
[1 6]
or 24. 2
S
S
S
⎣ ⎝^ ⎠ ⎦
d) t 1 = −3.5, d = 2.25, n = 6
Substitute into [2 1 ( 1) ] 2 n
n S = t + n − d.
6
6
6
[2( ) ( 1) ]
3[ 7
11.25]
S 2.
S
S
Section 1.2 Page 27 Question 3
a) t 1 = 7, t (^) n = 79, n = 8
Substitute into ( 1 ) 2 n n
n S = t + t.
8 8
8
S
S
S
b) t 1 = 58, t (^) n = −7, n = 26
Substitute into ( 1 ) n (^) 2 n
n S = t + t.
2
26
6 26
S
S
S
c) t 1 = −12, tn = 51, n = 10
Substitute into ( 1 ) 2 n n
n S = t + t.
1
10
0 10
S
S
S
d) t 1 = 12, d = 8, n = 9
Substitute into [2 1 ( 1) ] n 2
n S = t + n − d.
9
9
9
[2( ) ( 1)( )]
[24 64]
S
S
S
e) t 1 = 42, d = −5, n = 14
Substitute into [2 1 ( 1) ] n 2
n S = t + n − d.
14 1
14
4
[2( 42 ) ( 14 1)( 5 )]
7[84 65]
S
S
S
d) d = −3, Sn = 279, n = 18
Substitute into [2 1 ( 1) ] n 2
n S = t + n − d.
1
1 1 1 1
[2 ( 1)( )]
79 9[2 51]
t 3
t t t t
Section 1.2 Page 27 Question 5
a) t 1 = 8, t (^) n = 68, Sn = 608
Substitute into ( 1 ) n (^) 2 n
n S = t + t.
n
n n
b) t 1 = −6, t (^) n = 21, Sn = 75
Substitute into ( 1 ) n (^) 2 n
n S = t + t.
n
n n
Section 1.2 Page 27 Question 6
a) t 1 = 5, d = 5, n = 10
For t 10 , substitute into tn = t 1 + ( n − 1) d. t 10 = 5 + (10 − 1) t 10 = 45
For S 10 , substitute into [2 1 ( 1) ] n 2
n S = t + n − d.
1
10
0 10
[2( ) ( 1) ]
5[10 45]
S 0
S
S
b) t 1 = 10, d = −3, n = 10
For t 10 , substitute into tn = t 1 + ( n − 1) d. t 10 = 10 + (10 − 1)(−3) t 10 = − 17
For S 10 , substitute into [2 1 ( 1) ] n 2
n S = t + n − d.
10 1
10
0
[2( 10 ) ( 10 1)( 3 )]
5[20 27]
S
S
S
c) t 1 = −10, d = −4, n = 10
For t 10 , substitute into tn = t 1 + ( n − 1) d. t 10 = −10 + (10 − 1)(−4) t 10 = − 10 − 36 t 10 = − 46
For S 10 , substitute into [2 1 ( 1) ] 2 n
n S = t + n − d.
10 1
10
0
[2( 10 ) ( 10 1)( 4 )]
5[ 20 36]
S
S
S
d) t 1 = 2.5, d = 0.5, n = 10
For t 10 , substitute into tn = t 1 + ( n − 1) d. t 10 = 2.5 + (10 − 1)(0.5) t 10 = 7
For S 10 , substitute into [2 1 ( 1) ] n 2
n S = t + n − d.
10
10
10
[2( 2.5 ) ( 10 1)(0. )]
5[5 4.5]
S
S
S
