Module 6: Practice Problems
Sequences and Series
About These Practice Problems
This set contains 20 comprehensive problems covering all topics in Module 6. Work through each problem carefully, then check your solution. Each problem includes detailed step-by-step explanations.
Problems 1-5: Arithmetic Sequences
Problem 1: Find Terms of an Arithmetic Sequence
Find the first five terms of the arithmetic sequence with first term a1 = 7 and common difference d = 4.
Solution:
Step 1: Recall the formula for the nth term
For an arithmetic sequence: an = a1 + (n - 1)d
For an arithmetic sequence: an = a1 + (n - 1)d
Step 2: Find each term using the formula
a1 = 7 (given)
a2 = 7 + (2 - 1)(4) = 7 + 4 = 11
a3 = 7 + (3 - 1)(4) = 7 + 8 = 15
a4 = 7 + (4 - 1)(4) = 7 + 12 = 19
a5 = 7 + (5 - 1)(4) = 7 + 16 = 23
a1 = 7 (given)
a2 = 7 + (2 - 1)(4) = 7 + 4 = 11
a3 = 7 + (3 - 1)(4) = 7 + 8 = 15
a4 = 7 + (4 - 1)(4) = 7 + 12 = 19
a5 = 7 + (5 - 1)(4) = 7 + 16 = 23
Step 3: Verify the common difference
11 - 7 = 4, 15 - 11 = 4, 19 - 15 = 4, 23 - 19 = 4 (Check)
11 - 7 = 4, 15 - 11 = 4, 19 - 15 = 4, 23 - 19 = 4 (Check)
Final Answer: 7, 11, 15, 19, 23
Problem 2: Find the nth Term Formula
Find the explicit formula for the nth term of the arithmetic sequence: 3, 8, 13, 18, 23, ...
Solution:
Step 1: Identify the first term and common difference
a1 = 3
d = 8 - 3 = 5
a1 = 3
d = 8 - 3 = 5
Step 2: Use the formula an = a1 + (n - 1)d
an = 3 + (n - 1)(5)
an = 3 + (n - 1)(5)
Step 3: Simplify the formula
an = 3 + 5n - 5
an = 5n - 2
an = 3 + 5n - 5
an = 5n - 2
Step 4: Verify with known terms
n = 1: a1 = 5(1) - 2 = 3 (Check)
n = 2: a2 = 5(2) - 2 = 8 (Check)
n = 3: a3 = 5(3) - 2 = 13 (Check)
n = 1: a1 = 5(1) - 2 = 3 (Check)
n = 2: a2 = 5(2) - 2 = 8 (Check)
n = 3: a3 = 5(3) - 2 = 13 (Check)
Final Answer: an = 5n - 2
Problem 3: Find a Specific Term
Find the 20th term of the arithmetic sequence: -5, -2, 1, 4, ...
Solution:
Step 1: Identify a1 and d
a1 = -5
d = -2 - (-5) = 3
a1 = -5
d = -2 - (-5) = 3
Step 2: Use the formula an = a1 + (n - 1)d
a20 = -5 + (20 - 1)(3)
a20 = -5 + (20 - 1)(3)
Step 3: Calculate
a20 = -5 + 19(3)
a20 = -5 + 57
a20 = 52
a20 = -5 + 19(3)
a20 = -5 + 57
a20 = 52
Final Answer: a20 = 52
Problem 4: Find Number of Terms
How many terms are in the arithmetic sequence 4, 10, 16, ..., 100?
Solution:
Step 1: Identify known values
a1 = 4
d = 10 - 4 = 6
an = 100 (last term)
a1 = 4
d = 10 - 4 = 6
an = 100 (last term)
Step 2: Use the formula an = a1 + (n - 1)d
100 = 4 + (n - 1)(6)
100 = 4 + (n - 1)(6)
Step 3: Solve for n
100 = 4 + 6n - 6
100 = 6n - 2
102 = 6n
n = 17
100 = 4 + 6n - 6
100 = 6n - 2
102 = 6n
n = 17
Step 4: Verify
a17 = 4 + (17 - 1)(6) = 4 + 96 = 100 (Check)
a17 = 4 + (17 - 1)(6) = 4 + 96 = 100 (Check)
Final Answer: 17 terms
Problem 5: Sum of an Arithmetic Series
Find the sum of the first 15 terms of the arithmetic sequence: 6, 11, 16, 21, ...
Solution:
Step 1: Identify a1 and d
a1 = 6
d = 11 - 6 = 5
a1 = 6
d = 11 - 6 = 5
Step 2: Find a15
a15 = 6 + (15 - 1)(5)
a15 = 6 + 70 = 76
a15 = 6 + (15 - 1)(5)
a15 = 6 + 70 = 76
Step 3: Use the sum formula Sn = n(a1 + an)/2
S15 = 15(6 + 76)/2
S15 = 15(6 + 76)/2
Step 4: Calculate
S15 = 15(82)/2
S15 = 1230/2
S15 = 615
S15 = 15(82)/2
S15 = 1230/2
S15 = 615
Final Answer: S15 = 615
Problems 6-10: Geometric Sequences and Series
Problem 6: Find Terms of a Geometric Sequence
Find the first five terms of the geometric sequence with first term a1 = 3 and common ratio r = 2.
Solution:
Step 1: Recall the formula for the nth term
For a geometric sequence: an = a1 · rn-1
For a geometric sequence: an = a1 · rn-1
Step 2: Find each term
a1 = 3 (given)
a2 = 3 · 21 = 6
a3 = 3 · 22 = 12
a4 = 3 · 23 = 24
a5 = 3 · 24 = 48
a1 = 3 (given)
a2 = 3 · 21 = 6
a3 = 3 · 22 = 12
a4 = 3 · 23 = 24
a5 = 3 · 24 = 48
Step 3: Verify the common ratio
6/3 = 2, 12/6 = 2, 24/12 = 2, 48/24 = 2 (Check)
6/3 = 2, 12/6 = 2, 24/12 = 2, 48/24 = 2 (Check)
Final Answer: 3, 6, 12, 24, 48
Problem 7: Find the nth Term Formula
Find the explicit formula for the nth term of the geometric sequence: 5, 15, 45, 135, ...
Solution:
Step 1: Identify the first term and common ratio
a1 = 5
r = 15/5 = 3
a1 = 5
r = 15/5 = 3
Step 2: Use the formula an = a1 · rn-1
an = 5 · 3n-1
an = 5 · 3n-1
Step 3: Verify with known terms
n = 1: a1 = 5 · 30 = 5 (Check)
n = 2: a2 = 5 · 31 = 15 (Check)
n = 3: a3 = 5 · 32 = 45 (Check)
n = 1: a1 = 5 · 30 = 5 (Check)
n = 2: a2 = 5 · 31 = 15 (Check)
n = 3: a3 = 5 · 32 = 45 (Check)
Final Answer: an = 5 · 3n-1
Problem 8: Find a Specific Term
Find the 8th term of the geometric sequence: 64, 32, 16, 8, ...
Solution:
Step 1: Identify a1 and r
a1 = 64
r = 32/64 = 1/2
a1 = 64
r = 32/64 = 1/2
Step 2: Use the formula an = a1 · rn-1
a8 = 64 · (1/2)7
a8 = 64 · (1/2)7
Step 3: Calculate
a8 = 64 · (1/128)
a8 = 64/128
a8 = 1/2
a8 = 64 · (1/128)
a8 = 64/128
a8 = 1/2
Final Answer: a8 = 1/2
Problem 9: Sum of a Geometric Series
Find the sum of the first 6 terms of the geometric sequence: 2, 6, 18, 54, ...
Solution:
Step 1: Identify a1 and r
a1 = 2
r = 6/2 = 3
a1 = 2
r = 6/2 = 3
Step 2: Use the sum formula Sn = a1(1 - rn)/(1 - r)
S6 = 2(1 - 36)/(1 - 3)
S6 = 2(1 - 36)/(1 - 3)
Step 3: Calculate 36
36 = 729
36 = 729
Step 4: Complete the calculation
S6 = 2(1 - 729)/(1 - 3)
S6 = 2(-728)/(-2)
S6 = -1456/(-2)
S6 = 728
S6 = 2(1 - 729)/(1 - 3)
S6 = 2(-728)/(-2)
S6 = -1456/(-2)
S6 = 728
Final Answer: S6 = 728
Problem 10: Infinite Geometric Series
Find the sum of the infinite geometric series: 12 + 6 + 3 + 1.5 + ...
Solution:
Step 1: Identify a1 and r
a1 = 12
r = 6/12 = 1/2
a1 = 12
r = 6/12 = 1/2
Step 2: Check if the series converges
Since |r| = |1/2| = 1/2 < 1, the series converges.
Since |r| = |1/2| = 1/2 < 1, the series converges.
Step 3: Use the infinite sum formula S = a1/(1 - r)
S = 12/(1 - 1/2)
S = 12/(1 - 1/2)
Step 4: Calculate
S = 12/(1/2)
S = 12 · 2
S = 24
S = 12/(1/2)
S = 12 · 2
S = 24
The sum approaches 24 as we add more and more terms. The series converges to 24.
Final Answer: S = 24
Problems 11-14: Summation Notation
Problem 11: Evaluate a Summation
Evaluate: Σ(3k + 2) from k=1 to 5
Solution:
Step 1: Expand the summation by substituting each value of k
Σ(3k + 2) from k=1 to 5 = (3·1 + 2) + (3·2 + 2) + (3·3 + 2) + (3·4 + 2) + (3·5 + 2)
Σ(3k + 2) from k=1 to 5 = (3·1 + 2) + (3·2 + 2) + (3·3 + 2) + (3·4 + 2) + (3·5 + 2)
Step 2: Evaluate each term
= 5 + 8 + 11 + 14 + 17
= 5 + 8 + 11 + 14 + 17
Step 3: Add the terms
= 55
= 55
Final Answer: 55
Problem 12: Write in Summation Notation
Write the series in summation notation: 4 + 8 + 12 + 16 + 20 + 24
Solution:
Step 1: Identify the pattern
Each term is a multiple of 4: 4(1), 4(2), 4(3), 4(4), 4(5), 4(6)
Each term is a multiple of 4: 4(1), 4(2), 4(3), 4(4), 4(5), 4(6)
Step 2: Determine the general term
The kth term is 4k
The kth term is 4k
Step 3: Determine the range
k goes from 1 to 6 (6 terms total)
k goes from 1 to 6 (6 terms total)
Step 4: Write in summation notation
Σ(4k) from k=1 to 6
Σ(4k) from k=1 to 6
Alternative notation: Σ(4k) where k = 1, 2, 3, 4, 5, 6
Final Answer: Σ(4k) from k=1 to 6
Problem 13: Evaluate Using Summation Formulas
Evaluate: Σ(k2) from k=1 to 10
Solution:
Step 1: Recall the formula for sum of squares
Σ(k2) from k=1 to n = n(n + 1)(2n + 1)/6
Σ(k2) from k=1 to n = n(n + 1)(2n + 1)/6
Step 2: Substitute n = 10
Σ(k2) from k=1 to 10 = 10(10 + 1)(2·10 + 1)/6
Σ(k2) from k=1 to 10 = 10(10 + 1)(2·10 + 1)/6
Step 3: Calculate
= 10(11)(21)/6
= 2310/6
= 385
= 10(11)(21)/6
= 2310/6
= 385
Step 4: Verify by expansion (optional)
1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 385 (Check)
1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 385 (Check)
Final Answer: 385
Problem 14: Evaluate Complex Summation
Evaluate: Σ(2k2 - k + 3) from k=1 to 4
Solution:
Step 1: Expand the summation
Σ(2k2 - k + 3) = Σ(2k2) - Σ(k) + Σ(3)
Σ(2k2 - k + 3) = Σ(2k2) - Σ(k) + Σ(3)
Step 2: Evaluate each summation separately
Σ(2k2) from k=1 to 4 = 2·Σ(k2) = 2(1 + 4 + 9 + 16) = 2(30) = 60
Σ(k) from k=1 to 4 = 1 + 2 + 3 + 4 = 10
Σ(3) from k=1 to 4 = 3 + 3 + 3 + 3 = 12
Σ(2k2) from k=1 to 4 = 2·Σ(k2) = 2(1 + 4 + 9 + 16) = 2(30) = 60
Σ(k) from k=1 to 4 = 1 + 2 + 3 + 4 = 10
Σ(3) from k=1 to 4 = 3 + 3 + 3 + 3 = 12
Step 3: Combine the results
60 - 10 + 12 = 62
60 - 10 + 12 = 62
Alternative: Direct expansion
k=1: 2(1) - 1 + 3 = 4
k=2: 2(4) - 2 + 3 = 9
k=3: 2(9) - 3 + 3 = 18
k=4: 2(16) - 4 + 3 = 31
Sum = 4 + 9 + 18 + 31 = 62 (Check)
k=1: 2(1) - 1 + 3 = 4
k=2: 2(4) - 2 + 3 = 9
k=3: 2(9) - 3 + 3 = 18
k=4: 2(16) - 4 + 3 = 31
Sum = 4 + 9 + 18 + 31 = 62 (Check)
Final Answer: 62
Problems 15-20: Applications
Problem 15: Financial Application - Simple Interest
Maria saves $50 in her first month, $75 in her second month, $100 in her third month, and so on, increasing by $25 each month. How much will she have saved after 12 months?
Solution:
Step 1: Identify the sequence type
This is an arithmetic sequence with a1 = 50 and d = 25
This is an arithmetic sequence with a1 = 50 and d = 25
Step 2: Find the 12th term (amount saved in month 12)
a12 = 50 + (12 - 1)(25)
a12 = 50 + 275 = 325
a12 = 50 + (12 - 1)(25)
a12 = 50 + 275 = 325
Step 3: Find the total saved using the sum formula
S12 = 12(a1 + a12)/2
S12 = 12(50 + 325)/2
S12 = 12(375)/2
S12 = 4500/2
S12 = 2250
S12 = 12(a1 + a12)/2
S12 = 12(50 + 325)/2
S12 = 12(375)/2
S12 = 4500/2
S12 = 2250
Final Answer: Maria will have saved $2,250 after 12 months
Problem 16: Population Growth
A bacteria culture starts with 500 bacteria and doubles every hour. How many bacteria will there be after 8 hours?
Solution:
Step 1: Identify the sequence type
This is a geometric sequence with a1 = 500 and r = 2
This is a geometric sequence with a1 = 500 and r = 2
Step 2: Determine which term to find
After 8 hours means at the beginning of the 9th hour, so we need a9
After 8 hours means at the beginning of the 9th hour, so we need a9
Step 3: Use the formula an = a1 · rn-1
a9 = 500 · 28
a9 = 500 · 28
Step 4: Calculate
28 = 256
a9 = 500 · 256 = 128,000
28 = 256
a9 = 500 · 256 = 128,000
Final Answer: 128,000 bacteria after 8 hours
Problem 17: Depreciation
A car worth $30,000 depreciates by 15% each year. What will the car be worth after 5 years? Round to the nearest dollar.
Solution:
Step 1: Identify the sequence type
This is a geometric sequence. The car retains 85% of its value each year.
a1 = 30,000
r = 0.85 (100% - 15% = 85% = 0.85)
This is a geometric sequence. The car retains 85% of its value each year.
a1 = 30,000
r = 0.85 (100% - 15% = 85% = 0.85)
Step 2: Determine which term to find
After 5 years means the value at the end of year 5, which is a6
After 5 years means the value at the end of year 5, which is a6
Step 3: Use the formula an = a1 · rn-1
a6 = 30,000 · (0.85)5
a6 = 30,000 · (0.85)5
Step 4: Calculate
(0.85)5 = 0.4437...
a6 = 30,000 · 0.4437
a6 = 13,311.07
(0.85)5 = 0.4437...
a6 = 30,000 · 0.4437
a6 = 13,311.07
Final Answer: $13,311 (rounded to nearest dollar)
Problem 18: Bouncing Ball
A ball is dropped from a height of 20 feet. Each time it bounces, it reaches 60% of its previous height. What is the total vertical distance traveled by the ball when it comes to rest?
Solution:
Step 1: Analyze the motion
Initial drop: 20 feet down
First bounce: up 20(0.6) = 12 feet, then down 12 feet
Second bounce: up 12(0.6) = 7.2 feet, then down 7.2 feet
And so on...
Initial drop: 20 feet down
First bounce: up 20(0.6) = 12 feet, then down 12 feet
Second bounce: up 12(0.6) = 7.2 feet, then down 7.2 feet
And so on...
Step 2: Set up the total distance
Total = 20 + 2(12) + 2(7.2) + 2(4.32) + ...
Total = 20 + 2[12 + 7.2 + 4.32 + ...]
Total = 20 + 2[sum of infinite geometric series]
Total = 20 + 2(12) + 2(7.2) + 2(4.32) + ...
Total = 20 + 2[12 + 7.2 + 4.32 + ...]
Total = 20 + 2[sum of infinite geometric series]
Step 3: Find the sum of the infinite geometric series
First term: a1 = 12
Common ratio: r = 0.6
S = a1/(1 - r) = 12/(1 - 0.6) = 12/0.4 = 30
First term: a1 = 12
Common ratio: r = 0.6
S = a1/(1 - r) = 12/(1 - 0.6) = 12/0.4 = 30
Step 4: Calculate total distance
Total = 20 + 2(30) = 20 + 60 = 80 feet
Total = 20 + 2(30) = 20 + 60 = 80 feet
Final Answer: 80 feet total vertical distance
Problem 19: Seating Arrangement
An auditorium has 20 rows of seats. The first row has 15 seats, the second row has 18 seats, the third row has 21 seats, and so on. How many seats are in the auditorium?
Solution:
Step 1: Identify the sequence type
This is an arithmetic sequence with a1 = 15 and d = 3
This is an arithmetic sequence with a1 = 15 and d = 3
Step 2: Find the number of seats in the 20th row
a20 = 15 + (20 - 1)(3)
a20 = 15 + 57 = 72
a20 = 15 + (20 - 1)(3)
a20 = 15 + 57 = 72
Step 3: Find the total number of seats
S20 = 20(a1 + a20)/2
S20 = 20(15 + 72)/2
S20 = 20(87)/2
S20 = 1740/2
S20 = 870
S20 = 20(a1 + a20)/2
S20 = 20(15 + 72)/2
S20 = 20(87)/2
S20 = 1740/2
S20 = 870
Final Answer: 870 seats in the auditorium
Problem 20: Investment Application
You invest $5,000 at 6% annual interest compounded annually. How much will the investment be worth after 10 years? Round to the nearest cent.
Solution:
Step 1: Identify the sequence type
This is a geometric sequence where each year the amount is multiplied by 1.06
a1 = 5000
r = 1.06
This is a geometric sequence where each year the amount is multiplied by 1.06
a1 = 5000
r = 1.06
Step 2: Determine which term to find
After 10 years means a11 (initial amount is a1)
After 10 years means a11 (initial amount is a1)
Step 3: Use the formula an = a1 · rn-1
a11 = 5000 · (1.06)10
a11 = 5000 · (1.06)10
Step 4: Calculate
(1.06)10 = 1.790847...
a11 = 5000 · 1.790847
a11 = 8954.24
(1.06)10 = 1.790847...
a11 = 5000 · 1.790847
a11 = 8954.24
This can also be solved using the compound interest formula A = P(1 + r)t
Final Answer: $8,954.24 after 10 years
Ready to Test Your Knowledge?
Now that you have practiced these problems, take the Module 6 Quiz to demonstrate your mastery!