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MCR3U Sequences & Series
Grade 11 Unit 4 — practice questions with full solutions

Arithmetic and geometric sequences, general terms, series formulas, and reading a pattern out of a diagram.

Free MCR3U sequences & series practice for Grade 11 students in Ontario. These 5 questions cover the same material as a typical unit test or exam review on this topic, and every one comes with a complete worked solution — not just an answer key. Useful whether you are preparing for a MCR3U unit test, catching up on a lesson, or reviewing before the final exam.
MCR3U · Grade 11 Discrete Functions 4 core 1 challenge
1Core
Figure 1Figure 2Figure 3Each figure is built from unit squares
Study the growing pattern of squares. (a) Count the squares in figures 1, 2 and 3. (b) Find a formula for the number of squares in figure n. (c) How many squares are in figure 20?
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  1. (a) Each figure has two full rows of n squares plus one extra square below: figure 1 has 3, figure 2 has 5, figure 3 has 7.
  2. (b) The first differences are 2 each time, so the pattern is arithmetic with common difference 2.
  3. Using tn = a + (n−1)d with a = 3, d = 2: tn = 3 + 2(n−1) = 2n + 1.
  4. (c) t₂₀ = 2(20) + 1 = 41.
  5. Check the formula on figure 3: 2(3) + 1 = 7. ✔
Final answer(a) 3, 5, 7   (b) tn = 2n + 1   (c) 41 squares
2Core
An arithmetic sequence has t₃ = 14 and t₇ = 30. Find a, d, and t₂₅.
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  1. Between the 3rd and 7th terms there are 4 steps: 30 − 14 = 16, so 4d = 16 and d = 4.
  2. Work back from t₃ = 14: a = 14 − 2(4) = 6.
  3. General term: tn = 6 + 4(n − 1) = 4n + 2.
  4. So t₂₅ = 4(25) + 2 = 102.
Final answera = 6, d = 4, t₂₅ = 102
3Core
For the geometric sequence 3, 12, 48, …, find (a) the common ratio, (b) the 8th term, and (c) the sum of the first 8 terms.
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  1. (a) Divide consecutive terms: 12 ÷ 3 = 4, so r = 4.
  2. (b) tn = arn−1, so t₈ = 3(4)⁷ = 3 × 16384 = 49152.
  3. (c) Sn = a(rn − 1) ÷ (r − 1).
  4. S₈ = 3(4₈ − 1) ÷ 3 = 4₈ − 1 = 65536 − 1 = 65535.
Final answer(a) r = 4   (b) 49152   (c) 65535
4Core
Find the sum of the arithmetic series 7 + 11 + 15 + … + 103.
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  1. Identify a = 7 and d = 4.
  2. Find how many terms: 103 = 7 + 4(n − 1), so 96 = 4(n−1) and n − 1 = 24, giving n = 25.
  3. Use Sn = n(a + tn) ÷ 2.
  4. S₂₅ = 25(7 + 103) ÷ 2 = 25 × 110 ÷ 2 = 1375.
Final answer1375
5Challenge
A ball is dropped from 20 m and rebounds to 60% of its previous height each bounce. Find the total vertical distance travelled before it comes to rest.
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  1. The initial drop is 20 m and happens once.
  2. After that, each bounce goes up and comes back down, so those heights count twice.
  3. The rebound heights form a geometric series: 12, 7.2, 4.32, … with a = 12 and r = 0.6.
  4. Since |r| < 1, the infinite sum is S = a ÷ (1 − r) = 12 ÷ 0.4 = 30.
  5. Total distance = 20 + 2(30) = 80 m.
Final answer80 m

MCR3U Sequences & Series — common questions

Short answers to the things students ask most about this unit.

How do you tell an arithmetic sequence from a geometric one?

Subtract consecutive terms: if the difference is constant it is arithmetic. Divide consecutive terms: if the ratio is constant it is geometric.

What is the formula for the general term?

Arithmetic uses tₙ = a + (n − 1)d. Geometric uses tₙ = a · r^(n−1), where a is the first term.

More MCR3U practice — grade 11 functions

MCR3U Functions & TransformationsMCR3U Quadratic FunctionsMCR3U Exponential FunctionsMCR3U Financial ApplicationsAll MCR3U units
← Unit 3: Exponential FunctionsUnit 5: Financial Applications →
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