ACT Math study guide · Chapter 17
Sequences and Series
A sequence lists terms one at a time; a series adds them together. Finding the sum of the first n terms of a sequence, instead of just an individual term, requires a dedicated sum formula, and using the term formula in its place is one of the most common errors on this topic.
An arithmetic series sums a sequence with a constant common difference. Its sum is Sₙ = (n/2)(2a₁ + (n−1)d), or equivalently (n/2)(a₁ + aₙ) once the last term is known. A geometric series sums a sequence with a constant common ratio, and its sum is Sₙ = a₁(1−rⁿ)/(1−r) for r ≠ 1.
Sigma notation, Σ, is shorthand for a sum written out term by term. Σ(k=1 to n) f(k) means: substitute k=1, then k=2, and so on through k=n, into the expression f(k), and add the results. Recognizing sigma notation as nothing more than a compact way to write a sum removes most of its difficulty. Many sums that appear this way are arithmetic or geometric series once expanded, but not all are; a sum like Σk² follows neither pattern, so when no familiar series structure appears, expand the indicated terms directly instead of forcing a formula that doesn't apply.
Where this appears as a model: a savings plan that increases by a fixed amount each period. Saving $50 in the first month and increasing by $10 each month after that totals $1,260 after 12 months, found with the arithmetic series formula instead of adding twelve terms individually.
▸ Summing an arithmetic series
Worked Example Find the sum of the first 9 terms of the arithmetic sequence with a₁ = 6 and d = 3.
Apply the arithmetic series formula directly.
S₉ = (9/2)(2(6) + 8(3))
= (9/2)(12 + 24)
= (9/2)(36) = 162
▸ Summing a geometric series
Worked Example Find the sum of the geometric series 5 + 10 + 20 + 40 + 80.
This is a 5-term geometric series with a₁ = 5 and r = 2.
S₅ = 5(1 − 2⁵)/(1 − 2)
= 5(−31)/(−1) = 155
▸ Reading sigma notation
Worked Example Evaluate Σ(k=1 to 4) (2k + 5).
Substitute each value of k from 1 to 4, then add the results.
k=1: 7; k=2: 9; k=3: 11; k=4: 13
7 + 9 + 11 + 13 = 40
Worked Example A sequence begins 2, 6, 18, 54. Find the sum of the first 6 terms.
Recognize. Each term is 3 times the previous term, a constant ratio instead of a constant difference. This is a geometric series, not an arithmetic one, so the arithmetic sum formula does not apply.
Set up.
a₁ = 2, r = 3, n = 6
S₆ = a₁(1 − r⁶)/(1 − r)
Solve.
S₆ = 2(1 − 729)/(1 − 3)
S₆ = 2(−728)/(−2) = 728
Check. Listing all six terms directly: 2, 6, 18, 54, 162, 486. Their sum is 728, matching the formula.
Best route. The formula by hand. A calculator helps evaluate 3⁶ quickly but does not decide which series formula applies.
Common error. Applying the arithmetic series formula using the first difference (6−2=4) as if it were constant, instead of recognizing the constant ratio.
▸ Practice
Question 1
- AA. 220
- BB. 195
- CC. 210
- DD. 230
Question 2
- FF. 240
- GG. 230
- HH. 207
- JJ. 225
Question 3
- AA. 50
- BB. 45
- CC. 40
- DD. 55
Question 4
- FF. 1458
- GG. 726
- HH. 728
- JJ. 364
Question 5
- AA. 62
- BB. 30
- CC. 16
- DD. 15
Question 6
- FF. 111
- GG. 136
- HH. 121
- JJ. 116
Question 7
- AA. 93
- BB. 45
- CC. 96
- DD. 90
Question 8
- FF. 60
- GG. 21
- HH. 24
- JJ. 27
Question 9
- AA. −44
- BB. −60
- CC. −52
- DD. −48
Question 10
- FF. 15/16
- GG. 7/8
- HH. 1
- JJ. 31/16
Question 11
- AA. −4
- BB. −3.5
- CC. 4
- DD. −1
Question 12
- FF. 130
- GG. 135
- HH. 140
- JJ. 55
Question 13
- AA. 196.875
- BB. 187.5
- CC. 193.75
- DD. 200
Question 14
- FF. 200
- GG. 210
- HH. 220
- JJ. 190
Question 15
- AA. 96
- BB. 189
- CC. 90
- DD. 93
Question 16
- FF. 645
- GG. 660
- HH. 630
- JJ. 690
Question 17
- AA. 144
- BB. 130.56
- CC. 126
- DD. 117.6
Question 18
- FF. −10
- GG. 4
- HH. −4
- JJ. 0
Question 19
- AA. 40
- BB. 44
- CC. 124
- DD. 48
Question 20
- FF. 980
- GG. 931
- HH. 960
- JJ. 1000
▸ Answers and Explanations
Show the answers and explanations
1: C. The sum of an arithmetic series is Sₙ = (n/2)(2a₁ + (n−1)d): S₁₀ = (10/2)(2(3) + 9(4)) = 5(6+36) = 5(42) = 210. Choice D uses n instead of (n−1) in the formula, computing (10/2)(6+40)=230.
2: G. First find n: 41 = 5 + (n−1)(4), so n=10. Sum = (n/2)(a₁+aₙ) = (10/2)(5+41) = 5(46) = 230. Choice H uses n=9 instead of the correct 10, an off-by-one error in counting the terms.
3: A. Σ(3k+1) for k=1 to 5: (4)+(7)+(10)+(13)+(16) = 50. Choice B omits the +1 in each term, summing only 3k for k=1 to 5.
4: H. The sum of a finite geometric series is Sₙ = a₁(1−rⁿ)/(1−r): S₆ = 2(1−3⁶)/(1−3) = 2(−728)/(−2) = 728. Choice F reports a₁·rⁿ directly (2×3⁶=1,458) instead of applying the full sum formula.
5: B. Σ(2ᵏ) for k=1 to 4: 2+4+8+16 = 30. Choice F starts the sum at k=0 instead of k=1, adding an extra term (1) and stopping one term early.
6: J. S₈ = (8/2)(2(−3) + 7(5)) = 4(−6+35) = 4(29) = 116. Choice G uses n instead of (n−1) in the formula, computing (8/2)(−6+40)=136.
7: A. This 5-term geometric series has a₁=3, r=2: S₅ = 3(1−2⁵)/(1−2) = 3(−31)/(−1) = 93. Choice C comes from an arithmetic slip while applying the geometric series formula.
8: H. Σ(2k−3) for k=1 to 6: (−1)+(1)+(3)+(5)+(7)+(9) = 24. Choice F reverses the sign in each term, summing (2k+3) instead of (2k−3).
9: D. S₁₂ = (12/2)(2(7) + 11(−2)) = 6(14−22) = 6(−8) = −48. Choice G uses n instead of (n−1) in the formula, computing (12/2)(14−24)=−60.
10: F. This 4-term geometric series has a₁=1/2, r=1/2: S₄ = (1/2)(1−(1/2)⁴)/(1−1/2) = (1/2)(15/16)/(1/2) = 15/16. Choice G stops after only 3 terms, summing 1/2+1/4+1/8=7/8 instead of all four terms.
11: A. First find n: −11 = 10 + (n−1)(−3), so n=8. Sum = (n/2)(a₁+aₙ) = (8/2)(10+(−11)) = 4(−1) = −4. Choice G uses n=7 instead of the correct 8, an off-by-one error in counting the terms.
12: G. Σ(k²) for k=3 to 7: 9+16+25+36+49 = 135. Choice F sums k² for k=1 to 5 instead of k=3 to 7, using the same number of terms but the wrong starting index.
13: C. S₅ = 100(1−0.5⁵)/(1−0.5) = 100(0.96875)/(0.5) = 193.75. Choice B uses n=4 instead of the correct n=5, an off-by-one error in the number of terms.
14: G. This is an arithmetic series with a₁=20, aₙ=1, and n=20 rows: S = (n/2)(a₁+aₙ) = (20/2)(20+1) = 10(21) = 210. Choice J sums only 19 rows (1 through 19) instead of all 20, giving 19(20)/2=190, an off-by-one error in counting the rows.
15: D. This is geometric with a₁=3 (at k=1), r=2: Σ for k=1 to 5 = 3+6+12+24+48 = 93. Choice F includes an extra term beyond k=5, extending the sum one step too far.
16: F. S₁₅ = (15/2)(2(1) + 14(6)) = 7.5(2+84) = 7.5(86) = 645. Choice J uses n instead of (n−1) in the formula, computing (15/2)(2+90)=690.
17: B. This is a 4-term geometric series with a₁=60, r=0.6: S₄ = 60(1−0.6⁴)/(1−0.6) = 60(0.8704)/(0.4) = 130.56. Choice A comes from an arithmetic slip while applying the geometric series formula, treating the four bounces as a simple multiplicative pattern (60×4×0.6=144) instead of a proper sum.
18: J. Σ(5−2k) for k=1 to 4: 3+1+(−1)+(−3) = 0. Choice F reverses the sign of the constant, summing (−5−2k) instead of (5−2k).
19: B. This 5-term geometric series has a₁=4, r=−2: S₅ = 4(1−(−2)⁵)/(1−(−2)) = 4(33)/(3) = 44. Choice H treats the ratio as positive 2 instead of −2, computing 4(1−2⁵)/(1−2)=124.
20: F. S₂₀ = (20/2)(2(30) + 19(2)) = 10(60+38) = 10(98) = 980. Choice G uses n=19 instead of the correct 20 in the outer factor only, giving (19/2)(98)=931, an off-by-one error in counting the rows.