Geometric Sequences

Intuition

A single bacterium in a warm dish splits into two, then each of those splits again, so the count runs $1, 2, 4, 8, 16$ , doubling on every round. A car worth 20,000 dollars loses a fifth of its value each year, so its price slides to $16{,}000$ , then $12{,}800$ , shrinking by the same factor annually. A ball dropped on a hard floor rebounds to a fixed fraction of its last height on every bounce. None of these lists is arithmetic, because the terms do not change by a fixed amount; they change by a fixed multiple. A pattern in which each term is the one before it times a constant is a geometric sequence. Such a sequence is the natural partner to the arithmetic sequence you just met. This lesson builds the formula that jumps straight to any term, and shows exactly why it works.

Learning goals

Divide consecutive terms to find the common ratio
Apply $a_n = a_1 r^{n-1}$ for any term
Move between terms with $a_k = a_m r^{k-m}$
Test membership by solving for $n$ , or with a logarithm
Read the ratio for growth, decay, alternation or constancy
Exclude $r = 0$ , since no term is ever zero

What makes a sequence geometric

In an arithmetic sequence you reach the next term by adding a fixed number, the common difference $d$ . In a geometric sequence you reach it by multiplying by a fixed number. That fixed multiplier is the common ratio, written $r$ . To find it, divide any term by the term before it, and in a geometric sequence that quotient is always the same:

r = \frac{a_2}{a_1} = \frac{a_3}{a_2} = \frac{a_4}{a_3} = \cdots

Take $3, 6, 12, 24, 48$ . Dividing each term by the one before it gives $\tfrac{6}{3} = 2$ , $\tfrac{12}{6} = 2$ , $\tfrac{24}{12} = 2$ , and $\tfrac{48}{24} = 2$ . Every ratio is $2$ , so the sequence is geometric with $r = 2$ . This is the mirror image of the arithmetic test, where you subtract to look for a constant difference. Here the differences are $3, 6, 12, 24$ , which are nowhere near constant, so the list is not arithmetic. Multiplying by a constant is a completely different engine from adding one, and it is what makes the terms speed up.

The common ratio can be almost any number. When $r$ is greater than $1$ the terms grow, and when $r$ is between $0$ and $1$ they shrink toward zero. When $r$ is negative the terms flip sign back and forth, and when $r$ is exactly $1$ every term is the same. The single value ruled out is $r = 0$ , because then every term after the first would collapse to zero, leaving nothing to multiply. For the same reason no term of a geometric sequence is ever zero. If you start away from zero and only ever multiply by a nonzero $r$ , you can never land on zero.

To test whether a sequence is geometric, check that all the consecutive ratios agree. The sequence $2, 4, 6, 8$ is not geometric: the first ratio is $\tfrac{4}{2} = 2$ but the second is $\tfrac{6}{4} = 1.5$ , and a single mismatch disqualifies it. That list is arithmetic instead, since it is the differences that stay constant there.

A geometric sequence advances by multiplying by the common ratio r at every step, here r = 2. Because each jump multiplies rather than adds, the terms 3, 6, 12, 24, 48 grow faster and faster, and reaching the 5th term takes 4 multiplications from the first.

A formula for the nth term

Multiplying by $r$ over and over reaches any term, but it is slow: to find the $20$ th term you would multiply nineteen times. The picture above already shows the shortcut. To land on a later term you apply the ratio a whole number of times from the first term, so you only need to count the multiplications.

The $n$ th term is $a_n = a_1 r^{n-1}$ #

Start at the first term and multiply by the common ratio one step at a time. The second term is the first term times one factor of $r$ :

a_2 = a_1 r.

The third term is the second times another $r$ , which makes two factors of $r$ on the first term:

a_3 = a_2 r = (a_1 r)\,r = a_1 r^2.

The fourth term multiplies once more, for three factors in all:

a_4 = a_3 r = a_1 r^3.

A pattern is now plain. The number of factors of $r$ is always one less than the position of the term. The second term carries one, the third carries two, and the fourth carries three. The reason is easy to say in words. To travel from the first term to the $n$ th term you multiply exactly $n - 1$ times, and every multiplication is by $r$ . That count is right because those multiplications land you on the second term, then the third, and so on up to the $n$ th. Applying $n - 1$ factors of $r$ to the first term gives

a_n = a_1 r^{n-1}.

This closed formula reaches any term straight from its position $n$ , with no need to build the terms before it.

The formula has a clean meaning. You begin at $a_1$ and apply the ratio $n - 1$ times, so the term is the starting value scaled by $n - 1$ factors of $r$ . An arithmetic sequence is pinned down by its first term and common difference. A geometric sequence is pinned down in the same way by just two numbers, the first term $a_1$ and the common ratio $r$ . Notice the exponent is $n - 1$ , not $n$ , for the same reason the arithmetic formula used $n - 1$ steps. The first term already sits in position $1$ , with no multiplications applied yet.

Worked example 1 Find the 8th term

Find the $8$ th term of the sequence $3, 6, 12, 24, \ldots$

First read off the two numbers that define the sequence. The first term is $a_1 = 3$ . The common ratio is any term divided by the one before it, $\tfrac{6}{3} = 2$ , so $r = 2$ .

Now use the formula with $n = 8$ . The number of multiplications is $n - 1 = 7$ :

a_8 = a_1 r^{8-1} = 3 \cdot 2^{7}.

Work out the power before the product, $2^7 = 128$ , then multiply:

a_8 = 3 \cdot 128 = 384.

So the $8$ th term is $384$ . Notice you never had to list the first seven terms.

Check your understanding

A geometric sequence has first term $a_1 = 7$ and common ratio $r = 3$ . What is the fourth term $a_4$ ?

567

189

63

21

The recursive rule and moving between terms

The formula $a_n = a_1 r^{n-1}$ is the explicit (or closed) form, because it gives a term directly from its position. The same sequence also has a recursive form, which describes one step at a time:

a_n = r\,a_{n-1} \qquad \text{for } n \ge 2,

together with a stated first term $a_1$ . Read aloud, it is the definition itself: each term is $r$ times the term before it. The recursive rule often matches a real process most honestly (this year’s balance is last year’s balance times the growth factor). But to reach a far-off term with the recursive rule, you must climb through every term in between. The explicit formula is the tool for jumping ahead; the recursive rule is the tool for describing the step.

The multiplying works between any two terms, not only from the first. To get from the $m$ th term to the $k$ th term you multiply by $r$ exactly $k - m$ times, so

a_k = a_m\,r^{k-m}.

This is the version to reach for when you know two terms and want to recover the ratio or a distant term.

Worked example 2 Recover the ratio and first term

In a geometric sequence the second term is $a_2 = 6$ and the fifth term is $a_5 = 48$ . Find the common ratio and the first term.

Between the second term and the fifth there are $5 - 2 = 3$ multiplications by $r$ . Using $a_k = a_m r^{k-m}$ ,

a_5 = a_2\,r^{5-2} \quad\Longrightarrow\quad 48 = 6\,r^{3}.

Divide both sides by $6$ to isolate the power, then take the cube root:

r^{3} = 8 \quad\Longrightarrow\quad r = 2.

A cube root has just one real value, so $r = 2$ is settled. Now back up one step from the second term to the first, dividing by $r$ because a geometric sequence multiplies going forward:

a_1 = \frac{a_2}{r} = \frac{6}{2} = 3.

The sequence is $3, 6, 12, 24, 48, \ldots$ , and a quick check confirms $a_2 = 6$ and $a_5 = 48$ .

One warning about recovering $r$ . In that example the two known terms sat an odd number of positions apart, so $r^3 = 8$ had a single real solution. When the two terms are an even number of positions apart, the equation for $r$ is an even power, and an even power hides a sign. From $a_1 = 3$ and $a_3 = 12$ , for instance, you get $r^2 = \tfrac{12}{3} = 4$ , which is solved by both $r = 2$ and $r = -2$ . The first gives $3, 6, 12, 24$ and the second gives $3, -6, 12, -24$ , and both genuinely have $a_3 = 12$ . With only those two terms you cannot tell which sequence is meant. To settle it you need one more fact, such as a term that fixes the sign or a statement that the ratio is positive.

Testing membership and solving for n

The explicit formula answers the reverse question too. Instead of asking “what is term number $n$ ,” you can ask “which term equals a given value,” and you find out by solving for $n$ . Because $n$ must be a counting number, this also tells you whether a value appears in the sequence at all.

Worked example 3 Is a value a term?

For the sequence $3, 6, 12, 24, \ldots$ , decide whether $384$ is a term, and whether $300$ ever appears.

Here $a_1 = 3$ and $r = 2$ , so the explicit formula is $a_n = 3 \cdot 2^{n-1}$ . Set it equal to $384$ and isolate the power:

3 \cdot 2^{n-1} = 384 \quad\Longrightarrow\quad 2^{n-1} = 128.

Now match powers of $2$ . Since $128 = 2^{7}$ , the exponents must agree, so $n - 1 = 7$ and $n = 8$ . That is a whole number, so $384$ is the $8$ th term (the same one found in the first worked example).

Try $300$ the same way:

3 \cdot 2^{n-1} = 300 \quad\Longrightarrow\quad 2^{n-1} = 100.

But $100$ is not a power of $2$ : it falls between $2^6 = 64$ and $2^7 = 128$ , so no whole number exponent works. Therefore $300$ is not a term of this sequence. A value belongs only when solving for $n$ gives a positive whole number.

Matching powers is quickest when the target is a clean multiple of a power of $r$ . When it is not, you can still solve $a_1 r^{n-1} = V$ by isolating the power and taking a logarithm, the tool from the logarithms lesson. Dividing gives $r^{n-1} = \tfrac{V}{a_1}$ , and a base- $r$ logarithm brings the exponent down to the ground:

n - 1 = \log_{r}\!\left(\frac{V}{a_1}\right) \quad\Longrightarrow\quad n = 1 + \log_{r}\!\left(\frac{V}{a_1}\right).

This logarithm route applies only when the ratio is positive, with $r > 0$ , $r \ne 1$ , and $\tfrac{V}{a_1} > 0$ , since a base- $r$ logarithm makes sense only then. When $r < 0$ the terms alternate in sign, so there is no base- $r$ logarithm to take. For a negative ratio, test membership by matching powers of $\lvert r \rvert$ and checking that the sign comes out right.

If $\tfrac{V}{a_1}$ is a recognizable power of $r$ , the logarithm is a whole number (for $2^{n-1} = 128$ it is $\log_{2}(128) = 7$ ). If it is not, change of base evaluates it, $\log_{r}(x) = \tfrac{\log(x)}{\log(r)}$ , and a non-whole result confirms the value is not a term. The same setup answers “which is the first term past a threshold”: solve for $n$ , then round up to the next whole position.

How the common ratio shapes the sequence

Everything about the long-run behavior of a geometric sequence is decided by the ratio $r$ . Four cases cover it, and the starting value $16$ makes them easy to compare.

Common ratio $r$	What the terms do	Example starting at $16$
$r > 1$	grow without bound	$r = 2$ gives $16, 32, 64, 128, \ldots$
$0 < r < 1$	shrink toward $0$ , never reaching it	$r = \tfrac{1}{2}$ gives $16, 8, 4, 2, \ldots$
$r < 0$	alternate in sign	$r = -2$ gives $16, -32, 64, -128, \ldots$
$r = 1$	stay constant	$16, 16, 16, 16, \ldots$

Two of these deserve a closer look. When $0 < r < 1$ , each term is a fraction of the one before it, so the terms fall toward zero and get arbitrarily small. Even so, no term is ever exactly zero, since you only ever multiply a nonzero start by a nonzero ratio. This is exactly the exponential decay you saw with a base between $0$ and $1$ . When $r < 0$ , multiplying by a negative number flips the sign every step, so the terms swing positive, negative, positive, negative. The sizes of those terms still follow the powers of $\lvert r \rvert$ . A negative ratio does not mean the sequence decreases; it means it alternates.

Geometric sequences around us

Geometric sequences describe any quantity that is repeatedly scaled by the same factor. A population that doubles each period has $r = 2$ . A radioactive sample that loses half its atoms in each half-life has $r = \tfrac{1}{2}$ , a decay toward zero it never quite reaches. A savings balance left at a fixed annual rate compounded once a year is multiplied by the same growth factor each year. So the yearly balances form a geometric sequence: at a rate of $5$ percent the factor is $r = 1.05$ , and the balance after $n - 1$ years is $a_1 r^{n-1}$ . That is exactly the compound-interest pattern from the earlier lesson. A ball that rebounds to a fixed fraction of its height on each bounce is the same idea with $0 < r < 1$ .

Worked example 4 A bouncing ball

A ball is dropped and each bounce rebounds to $0.6$ of the previous rebound height. The first rebound reaches $300$ cm. How high is the $4$ th rebound, and after how many bounces does the rebound first fall below $20$ cm?

The rebound heights form a geometric sequence with first term $a_1 = 300$ and common ratio $r = 0.6$ , a decay because $0 < r < 1$ . The $4$ th rebound uses $n - 1 = 3$ multiplications:

a_4 = 300 \cdot 0.6^{3} = 300 \cdot 0.216 = 64.8 \text{ cm}.

For the second question, ask when the rebound height drops below $20$ cm. That means $300 \cdot 0.6^{\,n-1} < 20$ , so first isolate the power:

0.6^{\,n-1} < \frac{20}{300} = \frac{1}{15}.

Take a logarithm of both sides and use the power law. Because $\log(0.6)$ is negative, dividing by it reverses the inequality:

(n-1)\log(0.6) < \log\!\left(\tfrac{1}{15}\right) \quad\Longrightarrow\quad n - 1 > \frac{\log(1/15)}{\log(0.6)} = \frac{-1.176}{-0.222} \approx 5.30.

So $n - 1$ must be at least $6$ , giving $n = 7$ . Checking confirms it: the $6$ th rebound is $300 \cdot 0.6^{5} \approx 23.3$ cm, still above $20$ , while the $7$ th is $300 \cdot 0.6^{6} \approx 14.0$ cm, the first below $20$ . The ball keeps bouncing forever in theory, each rebound a fixed fraction of the last, but the heights shrink toward zero.

Common mistakes

Adding when you should multiply. A geometric sequence multiplies by a constant ratio; an arithmetic sequence adds a constant difference. Find $r$ by dividing consecutive terms, not by subtracting them. In $3, 6, 12, 24$ the ratio is $2$ , even though the differences $3, 6, 12$ keep changing.
Using the exponent $n$ instead of $n - 1$ . The $n$ th term applies the ratio $n - 1$ times, so $a_n = a_1 r^{n-1}$ . Writing $a_1 r^{n}$ multiplies by one extra factor of $r$ and overshoots the term.
Losing a sign when recovering $r$ . Two terms an even number of positions apart give an even power of $r$ , and $r^2 = 4$ has both $r = 2$ and $r = -2$ as solutions. Pin the sign with extra information (an adjacent term, a term an odd distance away, or a stated sign) before committing to one ratio.
Reading the ratio upside down. The ratio is a term divided by the one before it, $r = \tfrac{a_2}{a_1}$ , not $\tfrac{a_1}{a_2}$ . Dividing the wrong way gives the reciprocal ratio.
Thinking a negative ratio means the sequence shrinks. A negative $r$ makes the terms alternate in sign, not decrease. The sizes still grow when $\lvert r \rvert > 1$ ; only the signs flip.
Expecting a decaying sequence to reach zero. With $0 < r < 1$ the terms fall toward zero and get as small as you like. No term is ever exactly zero, though, because you only ever multiply a nonzero start by a nonzero ratio.

Key takeaways

A sequence is geometric when consecutive terms have a constant common ratio $r = \tfrac{a_2}{a_1} = \tfrac{a_3}{a_2} = \cdots$ , found by dividing a term by the one before it. Each term is the previous term times $r$ .
The explicit formula is $a_n = a_1 r^{n-1}$ : start at $a_1$ and multiply by $r$ a total of $n - 1$ times. Two numbers, $a_1$ and $r$ , fix the whole sequence.
The recursive rule is $a_n = r\,a_{n-1}$ with $a_1$ given, and between any two terms $a_k = a_m r^{k-m}$ , which recovers $r$ or a distant term.
To test whether a value $V$ is a term, solve $a_1 r^{n-1} = V$ ; it belongs only if $n$ is a positive whole number. Match powers when you can, or, for a positive ratio ( $r > 0$ , $r \ne 1$ ), take a logarithm, $n = 1 + \log_{r}\!\left(\tfrac{V}{a_1}\right)$ .
The ratio sets the behavior: $r > 1$ grows, $0 < r < 1$ decays toward zero, $r < 0$ alternates in sign, and $r = 1$ is constant. The ratio $r = 0$ is not allowed, and no term is ever zero.
Geometric sequences model doubling populations, radioactive halving, rebound heights, and compound interest, wherever a quantity is repeatedly scaled by the same factor.

Geometric Sequences

Learning goals

What makes a sequence geometric

A formula for the nth term

The recursive rule and moving between terms

Testing membership and solving for n

How the common ratio shapes the sequence

Geometric sequences around us

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)