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Grade 11 · MCR3U · Functions

MCR3U: Geometric Sequences and Series

A geometric sequence multiplies by the same number at every step, so it grows or shrinks faster and faster. As with arithmetic sequences, one formula gives any term and one adds up the terms, and the sum has a neat derivation worth knowing.

How It Works

Geometric sequences. Each term is the one before times a common ratio rr: 3,6,12,24,…3, 6, 12, 24, \dots has a=3a = 3 and r=2r = 2. To check a sequence, divide each term by the one before. A negative rr makes the signs alternate, and rr between 00 and 11 makes the terms shrink.

The general term. tn=arn−1t_n = ar^{n - 1}, since you multiply by rr only n−1n - 1 times. As a recursive formula, t1=at_1 = a and tn=r⋅tn−1t_n = r \cdot t_{n - 1}.

How many terms. Set arn−1ar^{n - 1} equal to the last term, divide by aa, and write both sides as powers of rr. For 3,6,…,7683, 6, \dots, 768: 2n−1=256=282^{n - 1} = 256 = 2^8, so n−1=8n - 1 = 8 and n=9n = 9.

Geometric series. Write Sn=a+ar+⋯+arn−1S_n = a + ar + \dots + ar^{n - 1}, multiply both sides by rr, and subtract the first line from the second. Every term cancels except two, leaving rSn−Sn=arn−arS_n - S_n = ar^n - a. So Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1} for any r≠1r \ne 1. If you know the last term tnt_n rather than nn, the same algebra gives Sn=rtn−ar−1S_n = \frac{rt_n - a}{r - 1}.

Two terms given. Divide one general term by the other and aa cancels: t6t3=ar5ar2=r3\frac{t_6}{t_3} = \frac{ar^5}{ar^2} = r^3. Find rr, then aa.

5 Worked Examples

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

  1. Easy

    For the sequence 5,15,45,…5, 15, 45, \dots find t8t_8.
    Show solution
    1. a=5a = 5 and r=155=3r = \frac{15}{5} = 3.
    2. t8=5(3)7=5(2187)t_8 = 5(3)^7 = 5(2187).

    Answer: t8=10935t_8 = 10935

  2. Easy

    Find the sum of the first 1010 terms of 2+6+18+…2 + 6 + 18 + \dots
    Show solution
    1. a=2a = 2, r=3r = 3 and n=10n = 10.
    2. S10=2(310−1)3−1=310−1=59049−1S_{10} = \frac{2(3^{10} - 1)}{3 - 1} = 3^{10} - 1 = 59049 - 1.

    Answer: S10=59048S_{10} = 59048

  3. Medium

    In a geometric sequence, t3=12t_3 = 12 and t6=324t_6 = 324. Find aa, rr and t9t_9.
    Show solution
    1. t6t3=r3=32412=27\frac{t_6}{t_3} = r^3 = \frac{324}{12} = 27, so r=3r = 3.
    2. ar2=12ar^2 = 12 gives 9a=129a = 12, so a=43a = \frac{4}{3}.
    3. t9=t6⋅r3=324(27)=8748t_9 = t_6 \cdot r^3 = 324(27) = 8748.

    Answer: a=43a = \frac{4}{3}, r=3r = 3, t9=8748t_9 = 8748

  4. Medium

    Find the sum 3−6+12−24+⋯+30723 - 6 + 12 - 24 + \dots + 3072.
    Show solution
    1. a=3a = 3 and r=−2r = -2. Number of terms: 3(−2)n−1=30723(-2)^{n - 1} = 3072, so (−2)n−1=1024=(−2)10(-2)^{n - 1} = 1024 = (-2)^{10} and n=11n = 11.
    2. S11=3((−2)11−1)−2−1=3(−2049)−3=2049S_{11} = \frac{3\left((-2)^{11} - 1\right)}{-2 - 1} = \frac{3(-2049)}{-3} = 2049.
    3. Check with the last term: rtn−ar−1=−2(3072)−3−3=−6147−3=2049\frac{rt_n - a}{r - 1} = \frac{-2(3072) - 3}{-3} = \frac{-6147}{-3} = 2049.

    Answer: 20492049

  5. Test level

    A ball is dropped from 1010 m and each bounce reaches 80%80\% of the height of the one before. How far has it travelled up and down when it hits the ground for the 6th time?
    Show solution
    1. The first fall is 1010 m. After that, each bounce goes up and comes back down the same distance.
    2. Between the 1st and 6th landing there are 55 bounces, with heights 8,6.4,5.12,…8, 6.4, 5.12, \dots: a geometric sequence with a=8a = 8 and r=0.8r = 0.8.
    3. S5=8(1−0.85)1−0.8=40(1−0.32768)=26.8928S_5 = \frac{8(1 - 0.8^5)}{1 - 0.8} = 40(1 - 0.32768) = 26.8928.
    4. Total: 10+2(26.8928)=63.785610 + 2(26.8928) = 63.7856.

    Answer: About 63.7963.79 m

Learn It on Martian Lab

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

Watch the Geometric Sequences lesson free

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