# chapter-9-solutions ```Section 9.1: Sequences
#1-12: Write the first 4 terms of the sequence. (Assume n begins with 1.)
1) an = 3n
a1 = 3*1 = 3
a2 = 3*2 = 6
a3 = 3*3 = 9
a4 = 3*4 = 12
3) an = 3n – 5
a1 = 3*1-5 = -2
a2 = 3*2 – 5 = 1
a3 = 3*3-5 = 4
a4 = 3*4 – 5 = 7
5) an = 3n
a1 = 31= 3
a2 = 32 = 9
a3 = 33 = 27
a4 = 34 = 81
7) an = (-3)n
a1 = (-3)1 = -3
a2 = (-3)2 = 9
a3 = (-3)3 = -27
a4 = (-3)4 = 81 Answer: -3,9,-27,81
9) an = (-1)n
a1 = (-1)1 = -1
a2 = (-1)2 = 1
a3 = (-1)3 = -1
a4 = (-1)4 = 1
Section 9.1: Sequences
11) an =
a1 =
a2 =
a3 =
a4 =
(−1)𝑛+1
2𝑛
(−1)1+1
1
=
2∗1
2
(−1)2+1
−1
= 4
2∗2
(−1)3+1
1
=
2∗3
6
(−1)4+1
−1
= 8
2∗4
𝟏 −𝟏 𝟏 −𝟏
, , ,
𝟐 𝟒 𝟔 𝟖
#13-24: Find the indicated term
13) an = 3n; a8
Answer: a8 = 3*8 = 24
15) an = 3n – 5; a16
Answer: a16 = 3*16-5 = 43
17) an = 3n; a5
Answer: a5 = 35 = 243
19) an = (-3)n; a6
Answer: a6 = (-3)6 = 729
21) an = (-1)n; a50
Answer: a50 = (-1)50 = 1
23) an =
(−1)𝑛+1
2𝑛
; a12
(−𝟏)𝟏𝟐+𝟏
𝟐∗𝟏𝟐
=
−𝟏
𝟐𝟒
#25-40: write an expression for the nth term of the sequence
25) 2,4,6,8,10….
Section 9.1: Sequences
27) 3,5,7,9,11,13….
I know I need a 2n since each term is 2 apart.
Solve 2(1) – x = 3 to figure out the number that goes after the 2n.
29) -3, -1, 1, 3,5…
Each term is 2 apart, so again I need a 2n
Solve 2(1) – x = -3 to figure out what to write after the 2n
Answer: an = 2n - 5
31) 6,10,14,18,22…
Each term is 4 apart, so I need a 4n.
Solve 4(1) + x = 6 to figure out what number goes after the 4n.
Answer: an = 4n + 2
33) 7,13,19,25…
Each term is 6 apart, so I need a 6n.
Solve 6(1) + x = 7 to figure out what number goes after the 6n.
35) 2,4,8,16,32…
I have to multiply by 2 to get from one number to the next. Since the first number is 2 the formula,
there isn’t any algebra needed to find a number to put in front of the 2.
Section 9.1: Sequences
37) 6,12,24,48…
Each number is a multiple of 2 apart. I need a 2n in the formula. I will need a number in front of the 2n
to get the correct answer. I find the number by solving
x(21) = 6
x=3
39) 1,2,4,8,16…
Each number is a multiple of 2 apart. I need a 2n in the formula. I will need a number in front of the 2n
to get the correct answer. I find the number by solving
x(21) = 1
x = 1/2
𝟏
Answer: an = 𝟐(2n) it would also be correct to write 2n-1
#41-48: Write the first 3 terms of the recursively defined sequence.
41) a1 = 6, ak+1 = ak + 3
a2 = a1 + 3
a2 = 6 + 3
a2 = 9
a3 = a2 + 3
a3 = 9 + 3
a3 = 12
Section 9.1: Sequences
43) a1 = 6, ak+1 = 2(ak – 1)
a2 = 2(a1 – 1)
a2 = 2(6-1)
a2 = 10
a3 = 2(a2 – 1)
a3 = 2(10-1)
a3 = 18
Answer: first three terms are 6,10,18
45) a1 = -3, ak+1 = 2ak + 3
a2 = 2a1 + 3
a2 = 2*(-3) + 3
a2 = -3
a3 = 2a2 + 3
a3 = 2(-3) + 3
a3 = -3
1
47) a1 = 6, ak+1 = 2ak + 4
1
𝑎2 = 2 𝑎1 + 4
1
2
𝑎2 = ∗ 6 + 4
a2 = 7
1
𝑎3 = 2 𝑎2 + 4
1
2
𝑎3 = ∗ 7 + 4
𝑎3 =
15
2
𝟏𝟓
𝟐
Section 9.1 sequences
#49-56: find the indicated sum
49) ∑4𝑘=1 2𝑘
= 2*1 + 2*2 + 2*3 + 2*4
=2+4+6+8
6
51) ∑(2𝑘 − 3)
𝑘=2
= (2*2 -3) + (2*3 – 3) + (2*4 – 3) + (2*5 – 3)
= 1+3+5+7
4
53) ∑ 2𝑘
𝑘=0
= 20 + 21 + 22 + 23 + 24
= 1 + 2 + 4 +8 + 16
8
55) ∑(−1)𝑘
𝑘=4
= (-1)4 + (-1)5 + (-1)6 + (-1)7+(-1)8
= 1 + (-1) + 1 + (-1) + 1
2
57) ∑3𝑘=1 𝑘
2
2
2
=1+2+3
𝟏𝟏
𝟑
Section 9.1 sequences
6
59) ∑ 5
𝑘=1
=5+5+5+5+5+5+5
Section 9.2: Arithmetic sequences:
#1-8: show that each sequence is arithmetic. Find the common difference and write out the first 4
terms.
1) an = 3n + 5
a1 = 3(1) + 5 = 8
a2 = 3(2) + 5 = 11
a3 = 3(3) + 5 = 14
a4 = 3(4) + 5 = 17
common difference = 3
first 4 terms; 8,11,14,17
3) an = 5n - 2
a1 = 5(1) – 2 = 3
a2 = 5(2) – 2 = 8
a3 = 5(3) – 2 = 13
a4 = 5(4) – 2 = 18
common difference = 5
first 4 terms; 3,8,13,18
5) an = 3 – 4n
a1 = 3-4(1) = -1
a2 = 3 – 4(2) = -5
a3 = 3 – 4(3) = -9
a4 = 3- 4(4) = -16
common difference = -4
first 4 terms; -1,-5,-9,-16
Section 9.2: Arithmetic sequences:
7) an = 4 – 6n
a1 = 4 – 6(1) = -2
a2 = 4 – 6(2)= -8
a3 = 4 – 6(3) = -14
a4 = 4 – 6(4) = -20
common difference = -6
first 4 terms; -2,-8,-14,-20
#9-18: Find a formula for the nth term of the arithmetic sequence with initial term a1 and common
difference d.
9) a1 = 4; d = 3
I know I need a 3n for the common difference to be 3. I just need to solve for the number after the 3n.
an = 3n+x
a1= 3(1) +x
4=3+x
1=x
11) a1 = -2; d = 1
I need a 1n, for the common difference of 1. I need to solve for the number after the 1n.
an = 1n+x
a1 = 1(1) + x
-2 = 1 + x
-3 = x
Answer: an = 1n – 3 or just n - 3
Section 9.2: Arithmetic sequences:
13) a1 = 4; d = -3
I need a -3n, for the common difference of -3. I need to solve for the number after the -3n.
an = -3n+x
a1 = -3(1) + x
4 = -3 + x
7=x
Answer: an = -3n + 7
15) a1 = -2; d = -1
I need a -1n, for the common difference of -1. I need to solve for the number after the -1n.
an = -1n+x
a1 = -1(1) + x
-2 = -1 + x
-1 = x
Answer: an = -1n – 1 or just -n - 1
#17-24: Find the indicated term in each arithmetic sequence.
17) 50th term of, 3,6,9,12…
First I need a formula for an;
an = 3n + x
a1 = 3(1) +x
3=3+x
0=x
Formula for sequence: an = 3n
computation of 50th term: a50 = 3(50) = 150
Section 9.2: Arithmetic sequences:
19) 100th term of 5, 8, 11, 14…
First I need a formula for an:
an = 3n + x
a1 = 3(1) +x
5=3+x
2=x
Formula for sequence: an = 3n +2
computation of 100th term: a100 = 3(100)+2 = 302
21) 80th term of 5,2,-1,-4…
First I need a formula for an:
an = -3n + x
a1 = -3(1) +x
5 = -3 + x
8=x
Formula for sequence: an = -3n + 8
computation of 80th term: a80 = -3(80)+8 = -232
23) 25th term of 20, 18,16,14…
First I need a formula for an:
an = -2n + x
a1 = -2(1) +x
20 = -2 + x
22 = x
Formula for sequence: an = -2n +22
computation of 25th term: a25 = -2(25)+22
Section 9.2: Arithmetic sequences:
25) A local theater has 30 seats in the first row and 50 rows in all. Each successive row contains two
additional seats. How many seats are in the 50th row of the theater?
This is just a arithmetic sequence problem with a1 = 30, and d = 2. I need to find a50.
First I will get a formula for an
an = 2n+x
a1 = 2(1) + x
30 = 2 + x
28 = x
Formula for sequence: an = 2n+28
now to find the number of seats in the 50th row: a50 = 2(50) + 28 = 128
Answer: There are 128 seats in the 50th row.
27) An outdoor amphitheater has 40 seats in the first row, 41 seats in the second row, 42 seats in the
third row. The pattern continues. The amphitheater has 40 rows. How many seats are in the 30th row
of the theater?
This is just a arithmetic sequence problem with a1 = 40, and d = 1. I need to find a40.
First I will get a formula for an
an = 1n+x
a1 = 1(1) + x
40 = 1 + x
39 = x
Formula for sequence: an = n+39
now to find the number of seats in the 40th row: a40 = 40+39 = 79
Answer: There are 79 seats in the 40th row.
Section 9.2: Arithmetic sequences:
#29 – 40 Find the indicated sum
40
29) ∑ 2𝑘
𝑘=1
I can write out the problem without summation notation as follows;
2(1) + 2(2) + 2(3) +….+ 2(40)
2 + 4 + 6+ …. + 80
This is asking me to add the first 40 terms of an arithmetic sequence. I will use the formula
𝑛
Sn = 2 (a1 + an)
In this case: S40 =
40
(2
2
+ 80)=20(82)=1640
60
31) ∑(2𝑘 − 3)
𝑘=1
I can write out the problem without summation notation as follows;
(2*1-3) + (2*2-3) + (2*3 – 3) +….+ (2*60-3)
-1+ 1 + 3+ …. + 117
This is asking me to add the first 60 terms of an arithmetic sequence. I will use the formula
𝑛
Sn = 2 (a1 + an)
In this case: S60 =
60
(−1 +
2
117)=30(116)=3480
Section 9.2: Arithmetic sequences:
40
33) ∑(5 − 2𝑘)
𝑘=1
I can write out the problem without summation notation as follows;
(5-2*1) + (5-2*2) + (5-2*3) +….+ (5-2*40)
3+ 1 + (-1)+ …. + (-75)
This is asking me to add the first 40 terms of an arithmetic sequence. I will use the formula
𝑛
Sn = 2 (a1 + an)
In this case: S40 =
40
(3
2
+ (−75))=20(-72)= -1440
20
35) ∑(60 − 5𝑘 )
𝑘=1
I can write out the problem without summation notation as follows;
(60-5*1) + (60-5*2) + (60-5*3) +….+ (60-5*20)
55+ 50 + 45+ …. + (-40)
This is asking me to add the first 20 terms of an arithmetic sequence. I will use the formula
𝑛
Sn = 2 (a1 + an)
In this case: S20 =
20
(55 +
2
(−40))=10(15)=150
Section 9.2: Arithmetic sequences:
37) A local theater has 30 seats in the first row and 50 rows in all. Each successive row contains two
additional seats. How many seats are in the theater?
I need to know how many seats are in the 50th row.
I will find a formula for an to get started.
an = 2n + x
a1 = 2(1) + x
30 = 2 + x
28 = x
formula for sequence: an = 2n + 28
computation of number of seats in 50th row: a50 = 2(50) + 28 = 128
I need to add these numbers to get my answer: 30 + 32 + 34+…..+ 128
𝑛
2
I will use the formula Sn = (a1 + an)
S50 =
50
(30 +
2
128) = 25(158) = 3950
39) An outdoor amphitheater has 40 seats in the first row, 41 seats in the second row, 42 seats in the
third row. The pattern continues. The amphitheater has 70 rows. How many seats does the
amphitheater contain?
I need to know how many seats are in the 70th row.
I will find a formula for an to get started.
an = 1n + x
a1 = 1(1) + x
40 = 1 + x
39 = x
formula for sequence: an = n + 39
computation of number of seats in 70th row: a70 = 70 + 39 = 109
I need to add these numbers to get my answer: 40 + 41 + 42+…..+ 109
𝑛
2
I will use the formula Sn = (a1 + an)
S50 =
70
(40 +
2
109) = 35(149) = 5215
Section 9.3: Geometric Sequences and Series:
#1-10: List the first 4 terms of the geometric series and find the common ratio (r).
1) an = 2n
a1 = 21 = 2
a2 = 22 = 4
a3 = 23 = 8
a4 = 24 = 16
Each term is a multiple of 2 apart so the common ratio is 2.
Answer: First 4 terms 2,4,8,16 common ratio 2
3) an = 3(2n)
a1 = 3(21) = 6
a2 = 3(22) = 12
a3 =3( 23) = 24
a4 = 3(24) = 48
Each term is a multiple of 2 apart so the common ratio is 2.
Answer: First 4 terms 6,12,24,48 common ratio 2
1 𝑛
5) an = (2)
1 1
1
1 2
2
1 3
1
4
1 4
2
1
16
𝑎1 = (2) = 2
𝑎2 = ( ) =
1
𝑎3 = (2) = 8
𝑎4 = ( ) =
1
Each term is a multiple of &frac12; apart so the common ratio is 2
𝟏 𝟏 𝟏
𝟏
Answer: first 4 terms 𝟐 , 𝟒 , 𝟖 , 𝟏𝟔 common ratio &frac12;
Section 9.3: Geometric Sequences and Series:
7) an=23n
a1 = 23*1 = 8
a2 = 23*2 = 64
a3 = 23*3 = 512
a4 = 23*4 = 4096
Each term is a multiple of 8 apart so the common ratio is 8.
Answer: first 4 terms 8,64,512,4096 common ratio 8
9) an = 2(32n)
a1 = 2(32*1) = 18
a2 = 2(32*2) = 162
a3 = 2(32*3) = 1458
a4 = 2(32*4) = 13122
Each term is a multiple of 9 apart so the common ratio is 9.
Answer: first 4 terms 18,162,1458,13122 common ratio 9
#11-18: Find a formula for the nth term of the geometric sequence with first term a1 and common ratio
(r).
11) a1 = 2; r = 3
an = x(3n)
a1 = x(31)
2 = 3x
2
3
=𝑥
𝟐
Answer: an = 𝟑 (𝟑𝒏 )
13) a1 = 6; r = 2
an = x(2n)
a1 = x(21)
6 = 2x
3=x
Section 9.3: Geometric Sequences and Series:
1
15) a1 = 1; r = 2
1 𝑛
an = 𝑥 (2)
1 1
1 = 𝑥 (2)
1
1 = 2𝑥
1
2*1 = 2*2 𝑥
2=x
𝟏 𝒏
𝟐
Answer: an = 𝟐 ( ) or 21-n
17) a1 = 4; r =
1
2
1 𝑛
2
1 1
an = x( )
a1 = x(2)
1
4 = 2𝑥
1
2*4 = 2*2 𝑥
8=x
𝟏 𝒏
Answer: an = 8(𝟐) 𝒐𝒓 𝟐𝟑−𝒏
#19-26: Find the indicated term of the geometric sequence
19) The 10th term; 6,18,54,162,…
First develop a formula for an; first term is 6 and common ratio is 3
an = x(3n)
a1 = x(31)
6 = x(31)
2=x
formula for an = 2(3n)
Now find the 10th term: a10 = 2(310) = 118098
Section 9.3: Geometric Sequences and Series:
21) The 9th term; 5,20,80,160…
First develop a formula for an; first term is 5 and common ratio is 4
an = x(4n)
a1 = x(41)
5 = 4x
5
=
4
x
5
formula for an = 4 (4𝑛 )
5
Now find the 9th term: a9 = 4 (49 ) = 327680
23) The 14th term; 200,100,50,25…
First develop a formula for an; first term 200, common ratio &frac12;
1 𝑛
an = x(2)
1 1
2
200 = x( )
1
2
200 = 𝑥
1
2*200 = 2*2 𝑥
400 = x
1 𝑛
formula for an = 400(2)
1 14
now find the 14th term: a14 = 400(2)
𝟐𝟓
𝟏𝟎𝟐𝟒
25
= 1024
Section 9.3: Geometric Sequences and Series:
25) The 20th term; 39366, 13122, 4374,…
1
First develop a formula for an; first term 39366 common ratio 3
1 𝑛
an = x(3)
1 1
39366 = x(3)
1
3
39366 = 𝑥
1
3*39366 = 3*3 𝑥
118098 = x
1 𝑛
an = 118098(3)
1 20
2
now I can find a20 = 118098(3) = 59049
𝟐
𝟓𝟗𝟎𝟒𝟗
27) Jess currently has a salary of \$40,000 per year. She expects to get an annual raise of 2%. What will
her salary be when she begins her 10th year?
First term of sequence 40,000 common ratio 1.02
formula for an = 40000(1+02)n
a10=40000(1.02)10 = 48759.78