Geometric Sequences

In a geometric sequence, you multiply by the same number to get each term, like $3, 6, 12, 24, \dots$ Bouncing balls, doubling bacteria, and money earning interest all follow this pattern. Geometric sequences are the step-by-step version of exponential functions.

Key ideas

The common ratio

A sequence is geometric if the ratio of consecutive terms is always the same. That ratio is the common ratio, $r$:

$$r = \frac{t_2}{t_1} = \frac{t_3}{t_2} = \dots$$

The first term is $a$. For $3, 6, 12, 24, \dots$: $a = 3$ and $r = 2$.

• $r \gt 1$: the terms grow.
• $0 \lt r \lt 1$: the terms shrink toward $0$.
• $r \lt 0$: the signs alternate, as in $1, -3, 9, -27, \dots$

The general term

To reach the $n$th term, start at $a$ and multiply by $r$ a total of $n - 1$ times:

$$t_n = ar^{n - 1}$$

The recursion formula is $t_1 = a$, $t_n = r \cdot t_{n - 1}$.

Arithmetic vs. geometric

Arithmetic sequences grow in a straight line; geometric sequences grow exponentially.

Worked examples

Example 1: The general term

For $3, 6, 12, 24, \dots$, find the general term and $t_{10}$.

Solution. $a = 3$ and $r = \dfrac{6}{3} = 2$:

$$t_n = 3(2)^{n - 1}, \qquad t_{10} = 3(2)^9 = 3(512) = 1536$$

Example 2: A shrinking sequence

Find $t_8$ for $80, 40, 20, \dots$

Solution. $a = 80$ and $r = \dfrac{40}{80} = \dfrac{1}{2}$:

$$t_8 = 80\left(\frac{1}{2}\right)^7 = \frac{80}{128} = \frac{5}{8}$$

Example 3: Which term?

Which term of $2, 6, 18, \dots$ is $4374$?

Solution. $a = 2$, $r = 3$:

$$2(3)^{n - 1} = 4374 \quad\Rightarrow\quad 3^{n - 1} = 2187$$

Since $2187 = 3^7$, $n - 1 = 7$ and $n = 8$. It’s the $8$th term.

Example 4: From two terms

In a geometric sequence, $t_2 = 12$ and $t_5 = 324$. Find the general term.

Solution. Going from $t_2$ to $t_5$ multiplies by $r$ three times:

$$r^3 = \frac{324}{12} = 27 \quad\Rightarrow\quad r = 3$$

Then $t_2 = ar$ gives $12 = 3a$, so $a = 4$:

$$t_n = 4(3)^{n - 1}$$

Common mistakes

Using $r^n$ instead of $r^{n - 1}$. The first term is $a$, with no factor of $r$ yet.

Multiplying $a$ and $r$ before applying the exponent. $3(2)^9$ means $2^9 = 512$ first, then times $3$. It isn’t $6^9$.

Dividing in the wrong order. $r = \dfrac{t_2}{t_1}$, later term over earlier term. For $80, 40, 20$, $r = \tfrac{1}{2}$, not $2$.

Losing the sign of a negative ratio. For $1, -3, 9, \dots$, $r = -3$, and $t_7 = (-3)^6 = 729$ is positive.

Checking differences instead of ratios. Geometric sequences have a constant ratio. Their differences keep changing.

Practice

1. (Warm-up) Is each sequence geometric? If so, give $r$.

• (a) $5, 15, 45, \dots$
• (b) $2, 4, 6, \dots$
• (c) $64, -32, 16, -8, \dots$

Solution

(a) Yes, $r = 3$.

(b) No. It’s arithmetic: the ratios $2$ and $1.5$ aren’t equal.

(c) Yes, $r = -\tfrac{1}{2}$.

2. (Warm-up) Write the general term of the geometric sequence with $a = 7$ and $r = 2$, and find $t_6$.

Solution

$t_n = 7(2)^{n - 1}$, so $t_6 = 7(32) = 224$.

3. (Warm-up) Find $t_7$ for $1, -3, 9, \dots$

Solution

$r = -3$, so $t_7 = 1(-3)^6 = 729$.

4. (Core) How many terms are in the sequence $3, 6, 12, \dots, 3072$?

Solution

$$3(2)^{n - 1} = 3072 \quad\Rightarrow\quad 2^{n - 1} = 1024 = 2^{10} \quad\Rightarrow\quad n = 11$$

There are $11$ terms.

5. (Core) In a geometric sequence, $t_3 = 18$ and $t_6 = 486$. Find the general term.

Solution

$r^3 = \dfrac{486}{18} = 27$, so $r = 3$. Then $18 = a(3)^2$ gives $a = 2$.

$$t_n = 2(3)^{n - 1}$$

6. (Core) A ball dropped from $2$ m rebounds to $60\%$ of its previous height each time. Find its height after the $5$th bounce, to the nearest tenth of a centimetre.

Solution

The heights after each bounce are $2(0.6),\ 2(0.6)^2, \dots$, so after the $n$th bounce the height is $2(0.6)^n$.

$$2(0.6)^5 = 0.15552 \text{ m} \approx 15.6 \text{ cm}$$

7. (Core) For which values of $x$ is $4, x, 25$ a geometric sequence?

Solution

The ratios must match: $\dfrac{x}{4} = \dfrac{25}{x}$, so $x^2 = 100$ and $x = 10$ or $x = -10$.

($4, 10, 25$ has $r = 2.5$; $4, -10, 25$ has $r = -2.5$.)

8. (Challenge) A sheet of paper is $0.1$ mm thick. Each fold doubles the thickness. How thick is it after $10$ folds? How many folds would it take to be thicker than $1$ m?

Solution

After $n$ folds the thickness is $0.1(2)^n$ mm.

After $10$ folds: $0.1(1024) = 102.4$ mm, about $10$ cm.

$1$ m $= 1000$ mm. After $13$ folds: $0.1(8192) = 819.2$ mm. After $14$ folds: $0.1(16\,384) = 1638.4$ mm. So it takes $14$ folds. (In real life, paper is very hard to fold more than about $7$ times!)

9. (Challenge) Show that $t_n = ar^{n - 1}$ can be written as $t_n = \dfrac{a}{r} \cdot r^n$, and explain what kind of function this is.

Solution

$$ar^{n - 1} = a \cdot r^n \cdot r^{-1} = \frac{a}{r} \cdot r^n$$

That’s a constant times $r^n$: an exponential function of $n$, evaluated only at whole numbers. So a geometric sequence is a discrete exponential function, just as an arithmetic sequence is a discrete linear function.