MCR3U Geometric Sequences: Notes and Practice

Where an arithmetic sequence adds the same amount every step, a geometric sequence MULTIPLIES by the same amount every step. A population doubling every year, a bouncing ball losing 20% of its height on every bounce, money compounding in a bank account — all geometric. That constant multiplier is the common ratio, $r$ . The difference matters enormously over time. Start at 100 and add 100 each step, and after 10 steps you have 1100. Start at 100 and double each step, and after 10 steps you have 102 400. Geometric sequences are the discrete version of the exponential functions from the previous unit, and they are the foundation of all the financial mathematics later in this unit.

The common ratio

Find $r$ by dividing any term by the one right before it: $r = \frac{t _{n}}{t _{n - 1}}$ . This must come out the same for every consecutive pair for the sequence to be geometric. Divide in the right order, later term over earlier term: for $80, 40, 20$ the ratio is $\frac{40}{80} = \frac{1}{2}$ , not 2. The ratio can be a whole number, a fraction, a decimal or a negative number, and it is often hidden in a percentage. A $k %$ increase keeps the whole amount and adds $k %$ , so $r = 1 + \frac{k}{100}$ ; a $k %$ decrease keeps $100% - k %$ , so $r = 1 - \frac{k}{100}$ .

Each step the quantity...	Common ratio r	Example terms
grows by 6%	1.06	1000, 1060, 1123.60
shrinks by 20%	0.8	500, 400, 320
doubles	2	3, 6, 12
halves	0.5	64, 32, 16
is multiplied by −3	−3	1, −3, 9

The explicit formula

Starting from $a$ , to reach the $n$ th term you multiply by $r$ a total of $(n - 1)$ times. This is the same fence-post idea as for arithmetic sequences, but with multiplication: $t_{1} = a$ , $t_{2} = a r$ , $t_{3} = a r^{2}$ , $t_{4} = a r^{3}$ , so the exponent is always one less than the term number. Rewriting it as $t_{n} = \frac{a}{r} \cdot r^{n}$ shows that a geometric sequence is an exponential function of $n$ , sampled only at whole numbers: its dots lie on an exponential curve. The recursive version is simply $t_{1} = a$ , $t_{n} = r t_{n - 1}$ .

t_{n} = a r^{n - 1}

Fig 1Geometric sequences curve away from a straight line — that's the visual hallmark of repeated multiplication instead of repeated addition.

VideoMario uses the explicit formula to find a far-off term of a geometric sequence and also writes the sequence recursively. His aₙ = a₁rⁿ⁻¹ is this lesson's tₙ = arⁿ⁻¹ with different letters.Watch on YouTube

Growth vs. decay

AGrowth (|r| > 1)

Terms get larger in size
Example: $r = 2$ — doubling
Models population growth, compound interest
Each term is further from 0 than the last, and the gaps keep widening
Graph of the dots bends upward like an exponential growth curve

BDecay (0 < |r| < 1)

Terms shrink toward 0
Example: $r = 0.9$ — losing 10% each step
Models depreciation, radioactive decay, bounce heights
The terms get closer and closer to 0 but never reach it
Graph of the dots flattens toward the horizontal axis

Watch out

A negative

r

makes the sequence ALTERNATE sign every term (e.g.

r = - 2

3, - 6, 12, - 24, \dots

) — that's still geometric, just oscillating. The sign of any term follows from the parity of its exponent:

t_{n} = 3 (- 2)^{n - 1}

is positive when

n - 1

is even (odd-numbered terms) and negative when

n - 1

is odd. So

t_{10} = 3 (- 2)^{9} = - 1536

without listing anything. Size and sign are separate questions:

∣ r ∣

decides whether the terms grow or shrink, and the sign of

r

decides whether they alternate. With

r = - 0.5

, the terms

8, - 4, 2, - 1, 0.5

alternate and shrink toward 0 at the same time.

Key idea

Never divide by zero and never let

a

r

equal 0 — a true geometric sequence needs every term nonzero, since you must be able to divide consecutive terms to find

r

. The value

r = 1

is allowed but dull (every term equals

a

), and it is the only ratio for which the sequence is both arithmetic and geometric. Any nonzero constant sequence such as

7, 7, 7

has

d = 0

and

r = 1

at once.

A strobe photograph of a basketball on a black background, captured at many positions as it travels through two arcs, the second lower than the first. — PhotoA bouncing ball photographed with a strobe light. Each bounce peaks lower than the one before; in a simple model each peak is a fixed fraction of the previous one, so the peak heights form a geometric sequence with a ratio between 0 and 1.MichaelMaggs Edit by Richard Bartz, CC BY-SA 3.0, via Wikimedia Commons

Solving for unknowns

Given two terms that are not consecutive, divide them: the later term over the earlier term equals $r$ raised to the gap in position. If $t_{2} = 6$ and $t_{5} = 162$ , then $\frac{162}{6} = 27 = r^{3}$ , so $r = 3$ . Watch the parity of the gap. An odd gap has one real root ( $r^{3} = - 8$ gives $r = - 2$ ), but an even gap has two: $r^{2} = 9$ gives $r = 3$ or $r = - 3$ , and both produce valid sequences unless the question rules one out. To find which term has a given value, isolate the power and match bases, just like the exponential equations you already solved: $3 (2)^{n - 1} = 1536$ gives $2^{n - 1} = 512 = 2^{9}$ , so $n - 1 = 9$ and $n = 10$ .

Confirm the sequence is geometric: check that at least two consecutive ratios agree.
Identify $a$ and $r$ , converting any percentage to a ratio first.
In context, decide which term is the starting value. If $t_{1}$ is the value at time 0, then the value after $k$ steps is $t_{k + 1}$ .
Substitute into $t_{n} = a r^{n - 1}$ , or set it equal to a known value and solve for $n$ or $r$ .
Check your answer on one of the given terms.

That third step causes the most lost marks. A car worth 24 000 dollars that loses 15% each year is worth $24000 (0.85)^{3}$ after 3 years. If you list the values as a sequence starting with the purchase price, $t_{1} = 24000$ , then "after 3 years" is $t_{4}$ , because three multiplications have happened. The formula exponent $n - 1$ always counts the number of multiplications, so whenever a context counts steps from zero, the step number is the exponent. A bouncing ball dropped from 2 m that rebounds to 75% of its height reaches $2 (0.75)^{1} = 1.5$ m on the first bounce and $2 (0.75)^{4} \approx 0.63$ m on the fourth.

Geometric Sequences

Key things to remember

The common ratio

The explicit formula

Growth vs. decay

Solving for unknowns

Worked examples

Where marks get lost

Flashcards 8

Practice