Question 1
The first four terms of an arithmetic sequence are -4, -2, 0, 2. What is the 20th term of the sequence?
- A.
34
- B.
36
- C.
38
- D.
40
Answer: A. 34
The common difference is 2. The nth term is a₁ + (n - 1)d = -4 + (20 - 1)(2) = 34.
ACT · Math · Preparing for Higher Math: Functions
Sequences
Sequences is part of Preparing for Higher Math: Functions in the ACT Math section. The questions below are written in the style of the real test, each with the answer and a worked explanation.
Question types you will see here include arithmetic sequence term, geometric sequence term, common difference or ratio from nonconsecutive terms and sum of an arithmetic series.
Common traps: using n instead of n - 1; using the wrong common difference; adding the difference once too many times.
Sample questions
Question 2
Priya saves AED 150 in her first week of saving, and each week after that she saves AED 25 more than the week before. How much does she save in total over the first 12 weeks?
- A.
AED 3,450
- B.
AED 3,600
- C.
AED 5,100
- D.
AED 6,900
Answer: A. AED 3,450
Week 12 is 150 + 11(25) = AED 425, so the total is 12 × (150 + 425) ÷ 2 = AED 3,450. B adds 25 twelve times, making week 12 AED 450; C multiplies the last week's amount by 12 instead of the average week; D forgets to divide by 2.
- A.
Question 3
The sum of the first n terms of a sequence is given by Sₙ = 3n² + 2n for every positive integer n. What is the 10th term of the sequence?
- A.
5
- B.
32
- C.
53
- D.
59
Answer: D. 59
The 10th term is the difference between the sum of the first 10 terms and the sum of the first 9: S₁₀ − S₉ = (300 + 20) − (243 + 18) = 320 − 261 = 59. Choice A is S₁, the first term; choice B is 3(10) + 2, misreading the formula as the nth term; choice C is S₉ − S₈ = 261 − 208, the 9th term.
- A.
Question 4
The nth term of a sequence is aₙ = 2n + 5. What is the sum of the 11th term through the 30th term, inclusive?
- A.
893
- B.
920
- C.
1,080
- D.
1,840
Answer: B. 920
a₁₁ = 27 and a₃₀ = 65, and there are 30 − 11 + 1 = 20 terms, so the sum is 20(27 + 65) ÷ 2 = 920. A computes S₃₀ − S₁₁, which also removes the 11th term; C is the sum of all 30 terms; D forgets to divide by 2.
- A.
Question 5
A new climbing gym in Dubai has 80 members in month 1, and the number of members grows by 15% each month. In which month does the membership first exceed 300?
- A.
Month 9
- B.
Month 10
- C.
Month 11
- D.
Month 20
Answer: C. Month 11
Membership in month n is 80(1.15)ⁿ⁻¹. Month 10 gives 80(1.15)⁹ ≈ 281 and month 11 gives 80(1.15)¹⁰ ≈ 324, so month 11. A rounds the solution of 1.15ᵏ = 3.75 (k ≈ 9.46) down and uses it as the month; B is the last month below 300; D adds a fixed 12 members a month.
- A.
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