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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: A-SSE.B.4, F-BF.A.2, F-IF.A.3 · Level: Grade 11 Algebra 2
Problem 1 Classify the sequence
9 − 14 = −5, 4 − 9 = −5, −1 − 4 = −5. Arithmetic with d = −5; the next term is −6.
Problem 2 Explicit rule from a first term
an = 11 + 6(n − 1) = 6n + 5. Then a12 = 77. an = 6n + 5; a12 = 77.
Problem 3 Find the common ratio
135 ÷ 405 = 1/3 and 45 ÷ 135 = 1/3. The next term is 45 · 1/3. r = 1/3; next term 15.
Problem 4 List terms from a recursive rule
a2 = 2(3) + 1 = 7, a3 = 15, a4 = 31, a5 = 63. 3, 7, 15, 31, 63.
Problem 5 Arithmetic term from two terms
From 4 to 9 there are 5 steps adding 25, so d = 5. Then a1 = 17 − 3(5) = 2. And a30 = 2 + 29(5) = 147. a1 = 2, d = 5, a30 = 147.
Problem 6 Geometric term
r = 18 ÷ 12 = 1.5. a7 = 12 · 1.56 = 12 · 11.390625 = 136.6875. a7 = 136.6875.
Problem 7 Evaluate a sigma sum
The terms are 1, 4, 7, 10, 13 and 1 + 4 + 7 + 10 + 13 = 35.
Problem 8 Geometric sigma sum
It is geometric with a1 = 2, r = 3, n = 6. S6 = 2(1 − 36)/(1 − 3) = 2(−728)/(−2) = 728.
Problem 9 Sum of an arithmetic series
an = 6 + 4(n − 1) = 70 gives n = 17. S17 = 17(6 + 70)/2 = 646. 646.
Problem 10 Alternating geometric series
r = −1.5, so S5 = 8(1 − (−1.5)5)/(1 − (−1.5)) = 8(8.59375)/2.5 = 27.5. 27.5.
Problem 11 Write sigma notation
The rule is 3k + 1 (first term 4 at k = 1). Solve 3k + 1 = 40 to get k = 13. ∑k=113 (3k + 1).
Problem 12 Rule from a graph (arithmetic)
The terms drop by 3, so d = −3 and a1 = 14. an = 14 − 3(n − 1) = 17 − 3n. an = 17 − 3n.
Problem 13 Rule from a graph (geometric)
80 ÷ 160 = 1/2 and 40 ÷ 80 = 1/2. r = 1/2 and an = 160(1/2)n−1.
Problem 14 Number of terms
3 · 2n−1 = 1536, so 2n−1 = 512 = 29. Then n − 1 = 9. n = 10.
Problem 15 Theater seating
a22 = 18 + 21(3) = 81. S22 = 22(18 + 81)/2 = 11 · 99 = 1089. 81 seats in row 22; 1089 seats in total.
Problem 16 Monthly savings plan
The deposits grow to 200(1.005)23, …, 200. So S = 200 · (1.00524 − 1)/(1.005 − 1) = 200 · 25.4320. ≈ $5,086.39.
Problem 17 Salary with yearly raises
The yearly pay is geometric with a1 = 42000, r = 1.03. S10 = 42000(1.0310 − 1)/0.03 = 42000 · 11.46388. ≈ $481,483.
Problem 18 Bouncing ball heights
Heights: 4.5, 3.375, … The explicit rule is hn = 6(0.75)n, so h5 = 6 · 0.237305 = 1.4238. Recursive: h0 = 6, hn = 0.75hn−1. ≈ 1.42 feet.
Problem 19 Car loan payment
r = 0.06/12 = 0.005 and n = 48. (1.005)−48 ≈ 0.78710, so the denominator is 0.21290. M = 75/0.21290. ≈ $352.28.
Problem 20 Sum of odd numbers
The odd numbers are arithmetic with a1 = 1 and d = 2, so an = 2n − 1. Then Sn = n(1 + 2n − 1)/2 = n2. Since 49 = 2(25) − 1, n = 25. Sum = 252 = 625.