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Grade 11 · MCR3U · Functions

MCR3U: Arithmetic Sequences and Series

An arithmetic sequence goes up or down by the same amount at every step. Two formulas cover the topic: one gives any term without listing the ones before it, and one adds up as many terms as you like.

How It Works

Arithmetic sequences. Each term is the one before plus a common difference dd: 5,8,11,14,…5, 8, 11, 14, \dots has first term a=5a = 5 and d=3d = 3. To check a sequence, subtract each term from the next. If every difference is the same, it is arithmetic.

The general term. tn=a+(n−1)dt_n = a + (n - 1)d. To get to term nn you add dd only n−1n - 1 times, because the first term is already there. The same sequence as a recursive formula is t1=at_1 = a and tn=tn−1+dt_n = t_{n - 1} + d.

How many terms. Set the general term equal to the last term and solve for nn. For 5,8,…,1525, 8, \dots, 152: 5+3(n−1)=1525 + 3(n - 1) = 152, so n−1=49n - 1 = 49 and n=50n = 50.

Arithmetic series. A series is the sum of the terms of a sequence. Write the sum forward and backward and add the two lines: each pair of terms adds to a+tna + t_n, and there are nn pairs, so 2Sn=n(a+tn)2S_n = n(a + t_n). That gives Sn=n2(a+tn)S_n = \frac{n}{2}(a + t_n), or with tnt_n written out, Sn=n2(2a+(n−1)d)S_n = \frac{n}{2}\left(2a + (n - 1)d\right). Use the first form when you know the last term and the second when you know dd.

Two terms given. If you know two terms but not aa or dd, write each one with the general term and solve the pair of equations. Subtracting them removes aa and leaves dd.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

  1. Easy

    For the sequence 7,11,15,19,…7, 11, 15, 19, \dots find t20t_{20}.
    Show solution
    1. a=7a = 7 and d=11−7=4d = 11 - 7 = 4.
    2. t20=7+(20−1)(4)=7+76t_{20} = 7 + (20 - 1)(4) = 7 + 76.

    Answer: t20=83t_{20} = 83

  2. Easy

    How many terms are in the sequence −4,2,8,…,152-4, 2, 8, \dots, 152?
    Show solution
    1. a=−4a = -4 and d=6d = 6.
    2. −4+6(n−1)=152-4 + 6(n - 1) = 152, so 6(n−1)=1566(n - 1) = 156 and n−1=26n - 1 = 26.

    Answer: 2727 terms

  3. Medium

    Find the sum of the first 3030 terms of 3+10+17+…3 + 10 + 17 + \dots
    Show solution
    1. a=3a = 3, d=7d = 7 and n=30n = 30.
    2. S30=302(2(3)+29(7))=15(6+203)=15(209)S_{30} = \frac{30}{2}\left(2(3) + 29(7)\right) = 15(6 + 203) = 15(209).

    Answer: S30=3135S_{30} = 3135

  4. Medium

    In an arithmetic sequence, t5=23t_5 = 23 and t12=58t_{12} = 58. Find aa, dd and t40t_{40}.
    Show solution
    1. a+4d=23a + 4d = 23 and a+11d=58a + 11d = 58.
    2. Subtract: 7d=357d = 35, so d=5d = 5 and a=23−4(5)=3a = 23 - 4(5) = 3.
    3. t40=3+39(5)=198t_{40} = 3 + 39(5) = 198.

    Answer: a=3a = 3, d=5d = 5, t40=198t_{40} = 198

  5. Test level

    A theatre has 2424 rows. The first row has 1818 seats and each row has 22 more seats than the row in front of it. How many seats are there, and how many rows would it need to seat at least 10001000 people?
    Show solution
    1. The rows form an arithmetic sequence with a=18a = 18 and d=2d = 2.
    2. S24=242(2(18)+23(2))=12(82)=984S_{24} = \frac{24}{2}\left(2(18) + 23(2)\right) = 12(82) = 984.
    3. For nn rows, Sn=n2(36+2(n−1))=n(n+17)S_n = \frac{n}{2}\left(36 + 2(n - 1)\right) = n(n + 17). Solve n2+17n=1000n^2 + 17n = 1000: n=−17+42892≈24.25n = \frac{-17 + \sqrt{4289}}{2} \approx 24.25.
    4. nn is a whole number, so round up and check: S25=25(42)=1050≥1000S_{25} = 25(42) = 1050 \ge 1000, while S24=984S_{24} = 984 falls short.

    Answer: 984984 seats; 2525 rows to seat at least 10001000.

Learn It on Martian Lab

Martian Lab teaches this in the Arithmetic Sequences lesson of its MCR3U Discrete Functions unit. Watch it, then take the Discrete Functions practice tests to see if it stuck.

Watch the Arithmetic Sequences lesson free

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