Geometric Sequences and Series

Intuition

An arithmetic sequence takes the same step every time: add $d$ , add $d$ , add $d$ . A geometric sequence takes the same stretch every time: multiply by $r$ , multiply by $r$ , multiply by $r$ . Swapping “add” for “multiply” changes the entire character of the sequence. Adding $3$ over and over climbs slowly and evenly; multiplying by $3$ over and over doubles, then quadruples, then races off the page. This pattern drives compound interest, a population that grows by a fixed percentage each year, and a drug that clears a fixed fraction of itself from the blood each hour. Every other quantity that changes by a factor rather than by an amount follows the same pattern. Adding up a run of such terms looks hopeless, until you see a two-line trick that collapses the whole sum at once.

Learning goals

Divide consecutive terms for a constant ratio
Apply $a_n = a_1 r^{n-1}$ , with $n - 1$ multiplications
Prove the sum by shifting and subtracting
Exclude $r = 1$ , where the sum is $na$ instead
Separate growth in size from growth in value
Reach for logarithms when $n$ sits in the exponent

What makes a sequence geometric

A sequence is geometric when each term is the same fixed multiple of the term before it:

a_{n+1} = r \cdot a_n .

The number $r$ is the common ratio. Dividing both sides by $a_n$ shows where the name comes from:

r = \frac{a_{n+1}}{a_n},

and this ratio is the same everywhere in the sequence. Arithmetic sequences keep a constant difference; geometric sequences keep a constant ratio. That is the whole distinction.

Two conditions ride along with the definition, and neither is a technicality. The first term $a_1$ must not be $0$ , and the ratio $r$ must not be $0$ . If either one were $0$ , then every term from some point on would be $0$ , and the quotient $a_{n+1}/a_n$ would be a division by zero. With $a_1 \neq 0$ and $r \neq 0$ , no term is ever $0$ and every ratio is defined.

Sequence	Common ratio $r$	Behaviour
$3,\; 6,\; 12,\; 24,\; 48$	$2$	the terms double
$81,\; 27,\; 9,\; 3,\; 1$	$\tfrac{1}{3}$	the terms shrink
$5,\; -10,\; 20,\; -40$	$-2$	the signs alternate, the size grows
$7,\; 7,\; 7,\; 7$	$1$	nothing changes (watch this one)

To test a sequence, divide each term by the one before it. Every quotient the same means geometric; one mismatch means not. In $2, 6, 18, 55$ the first two quotients are $6/2 = 3$ and $18/6 = 3$ , but $55/18$ is not $3$ , so the sequence is not geometric. Note also that $3, 6, 9, 12$ fails the test even though it looks orderly: $6/3 = 2$ but $9/6 = 1.5$ . It is arithmetic (constant difference $3$ ), not geometric.

The nth term, and the off-by-one hiding in it

The $n$ th term of a geometric sequence is $a_n = a_1 r^{\,n-1}$ #

Start at $a_1$ and walk forward. The definition says each step multiplies by $r$ , so

a_2 = a_1 r, \qquad a_3 = a_2 r = (a_1 r) r = a_1 r^2, \qquad a_4 = a_3 r = a_1 r^3 .

The exponent is not the term number. It is one less, and counting the steps instead of the terms shows why. Between $a_1$ and $a_n$ there are $n - 1$ gaps. Term $1$ to term $2$ is the first multiplication, term $2$ to term $3$ is the second, and the last one, term $n-1$ to term $n$ , is the $(n-1)$ th. So travelling from $a_1$ to $a_n$ multiplies by $r$ exactly $n - 1$ times, not $n$ times:

a_n = a_1 \underbrace{\cdot r \cdot r \cdots r}_{n-1 \text{ factors}} = a_1 r^{\,n-1} .

Check the formula at the near end, where mistakes show up fastest. At $n = 1$ it gives $a_1 r^0 = a_1$ , which is exactly right: you have not moved yet, so you have multiplied by nothing. An exponent of $n$ instead would report the first term as $a_1 r$ , which is the second term.

This is the same off-by-one you met in arithmetic sequences, where $a_n = a_1 + (n-1)d$ for precisely the same reason: $n$ terms have only $n - 1$ gaps between them. Only the operation changed, from $n-1$ additions of $d$ to $n-1$ multiplications by $r$ .

Five terms, but only four arrows. Each arrow is one multiplication by r, so reaching the fifth term takes four of them, and the fifth term is 3 times 2 to the fourth power.

Worked example 1 Find a formula and the 8th term of $5,\; 15,\; 45,\; 135,\ldots$

First confirm the sequence really is geometric by dividing consecutive terms:

\frac{15}{5} = 3, \qquad \frac{45}{15} = 3, \qquad \frac{135}{45} = 3 .

Every quotient is $3$ , so $r = 3$ and $a_1 = 5$ . Substituting into $a_n = a_1 r^{\,n-1}$ gives the explicit formula:

a_n = 5 \cdot 3^{\,n-1} .

For the eighth term, the exponent is $8 - 1 = 7$ , not $8$ :

a_8 = 5 \cdot 3^7 = 5 \cdot 2187 = 10{,}935 .

Check your understanding

In the geometric sequence $2,\; -6,\; 18,\; -54,\ldots$ , what is $a_7$ ?

-4374

1458

-1458

486

Summing a finite geometric series

A geometric series is what you get when you add up the terms of a geometric sequence. Writing $a$ for the first term, the sum of the first $n$ terms is

S_n = a + ar + ar^2 + \cdots + ar^{\,n-1} .

The same off-by-one appears again: there are $n$ terms, and the last exponent is $n - 1$ . You could add them one at a time, but for $n = 100$ that is a wasted afternoon. There is a better way, and it is worth seeing in full.

For $r \neq 1$ , the sum of the first $n$ terms is $S_n = \dfrac{a(1 - r^n)}{1 - r}$ #

Write the sum, and underneath it write $r$ times the same sum. Multiplying by $r$ turns each term into the next one, so the second line is the first line shifted one place to the right:

\begin{aligned} S_n \;&=\; a + ar + ar^2 + \cdots + ar^{\,n-1} \\[2pt] rS_n \;&=\; \phantom{a + {}} ar + ar^2 + \cdots + ar^{\,n-1} + ar^{\,n} \end{aligned}

Now subtract the second line from the first. Every term of $S_n$ except the leading $a$ also appears in $rS_n$ , and every term of $rS_n$ except the trailing $ar^{\,n}$ also appears in $S_n$ . Those shared terms, from $ar$ all the way up to $ar^{\,n-1}$ , cancel in pairs, and only the two ends survive:

S_n - rS_n = a - ar^{\,n} .

Factor each side:

S_n(1 - r) = a(1 - r^{\,n}) .

This equation is true for every ratio $r$ , with no exceptions. The condition enters only at the final step. To isolate $S_n$ you divide both sides by $1 - r$ , and division is legal only when $1 - r \neq 0$ , that is, only when $r \neq 1$ . Assuming $r \neq 1$ :

S_n = \frac{a(1 - r^{\,n})}{1 - r} .

Test it on a sum you can check by hand. With $a = 3$ , $r = 2$ , and $n = 4$ the formula gives $3(1 - 16)/(1 - 2) = 3(-15)/(-1) = 45$ , and indeed $3 + 6 + 12 + 24 = 45$ .

Multiplying the sum by r shifts every term one place to the right. Subtract the rows and each shared column cancels, leaving only the first term of the top row and the last term of the bottom row.

Multiplying the numerator and denominator by $-1$ gives the same number in a second common form:

S_n = \frac{a(1 - r^{\,n})}{1 - r} = \frac{a(r^{\,n} - 1)}{r - 1} .

They are equal, so use whichever keeps the minus signs out of your way. When $r > 1$ the right-hand form is tidier, because the denominator $r - 1$ and the factor $r^{\,n} - 1$ are both positive there. Under that same condition the left-hand form makes both of those pieces negative, and relies on the two minus signs cancelling. In summation notation the result reads

\sum_{k=1}^{n} a r^{\,k-1} = \frac{a(1 - r^{\,n})}{1 - r}, \qquad r \neq 1 .

Worked example 2 Evaluate $3 + 6 + 12 + \cdots + 384$

The terms double, so $a = 3$ and $r = 2$ . The formula needs the number of terms, and what you are given is the last term, so find $n$ first. Set $a_n = 384$ and solve:

3 \cdot 2^{\,n-1} = 384 \quad\Longrightarrow\quad 2^{\,n-1} = 128 = 2^7 \quad\Longrightarrow\quad n - 1 = 7,

so $n = 8$ . There are eight terms, not $384$ of them, and not $7$ . Now apply the sum formula in the form built for $r > 1$ :

S_8 = \frac{3(2^8 - 1)}{2 - 1} = \frac{3(256 - 1)}{1} = 3 \cdot 255 = 765 .

Adding by hand confirms it: $3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 = 765$ .

Worked example 3 Sum the first six terms of $2,\; -6,\; 18,\; -54,\ldots$

Each term is $-3$ times the one before it, so $a = 2$ , $r = -3$ , and $n = 6$ . Substitute carefully, because a negative ratio sets two sign traps at once:

S_6 = \frac{2\bigl(1 - (-3)^6\bigr)}{1 - (-3)} .

The first trap is the exponent. Since $6$ is even, $(-3)^6 = +729$ , not $-729$ . The second trap is the denominator: $1 - (-3) = 1 + 3 = 4$ , not $-2$ . With both handled,

S_6 = \frac{2(1 - 729)}{4} = \frac{2(-728)}{4} = \frac{-1456}{4} = -364 .

Checking directly, $2 - 6 + 18 - 54 + 162 - 486 = -364$ . The sum is negative because the large negative terms outweigh the positive ones.

Check your understanding

What is the sum of the first $5$ terms of the geometric sequence $4,\; 12,\; 36,\ldots$ ?

968

484

160

1456

When the ratio is 1

The sum formula has a hole in it, and the hole is worth staring at rather than stepping around.

Suppose $r = 1$ . Then every term is $a \cdot 1^{\,n-1} = a$ , so the sequence is $a, a, a, \ldots$ , which is genuinely geometric: the ratio really is constant. Adding $n$ copies of $a$ needs no clever trick at all:

S_n = \underbrace{a + a + \cdots + a}_{n \text{ terms}} = na .

Now try the formula on it. It reads

S_n = \frac{a(1 - 1^{\,n})}{1 - 1} = \frac{a \cdot 0}{0},

which is undefined. Nothing went wrong in the derivation; it simply stops one step short. The identity the derivation proved,

S_n(1 - r) = a(1 - r^{\,n}),

is still perfectly true at $r = 1$ , where it says $S_n \cdot 0 = 0$ . The trouble is that this statement is true for every value of $S_n$ , so it pins down nothing at all. Multiplying by $1 - r$ threw the information away, and no rearranging can bring it back. To get the sum you must return to the definition, which hands it over immediately: $S_n = na$ .

So the geometric sum formula carries a side condition, and it is not fine print.

S_n = \begin{cases} \dfrac{a(1 - r^{\,n})}{1 - r}, & r \neq 1, \\[10pt] na, & r = 1. \end{cases}

Only $r = 1$ is excluded. The neighbouring case $r = -1$ is perfectly fine, because the denominator becomes $1 - (-1) = 2$ , which is nowhere near zero. For instance $5 - 5 + 5 - 5$ has $a = 5$ , $r = -1$ , $n = 4$ , and the formula gives $5(1 - 1)/2 = 0$ , which is right: the four terms cancel in pairs.

Check your understanding

A geometric sequence has $a_1 = 7$ and common ratio $r = 1$ . What is the sum of its first $12$ terms?

0

7

84

D undefined, because

1 - r = 0

Growth, decay, and the curve behind the sequence

Take absolute values of the defining rule $a_{n+1} = r a_n$ and you get $|a_{n+1}| = |r| \cdot |a_n|$ , so the sizes of the terms are themselves a geometric sequence, with ratio $|r|$ . That single line settles the growth question. If $|r| > 1$ , each term is larger in size than the last, and the sizes grow without bound. If $|r| < 1$ , each term is smaller in size than the last, and the sizes decay. If $|r| = 1$ , the size never changes at all: the terms stay put when $r = 1$ and flip sign forever when $r = -1$ .

Read every word of that as a statement about size, not about which term is greater. With $a_1 = -8$ and $r = \tfrac{1}{2}$ the terms run $-8, -4, -2, -1$ , and each one is greater than the last while steadily shrinking. The sign pattern is a separate question, decided by the signs of $a_1$ and $r$ . A positive ratio $r$ keeps every term on the same side of zero as $a_1$ , while a negative ratio $r$ makes the terms alternate.

There is a familiar function hiding behind all this. Since $r \neq 0$ you may write

a_n = a_1 r^{\,n-1} = \frac{a_1}{r} \cdot r^{\,n} = C r^{\,n}, \qquad C = \frac{a_1}{r},

and when $r > 0$ and $r \neq 1$ that is precisely the exponential function $f(x) = C r^{\,x}$ , evaluated at $x = 1, 2, 3, \ldots$ instead of at every real number. A geometric sequence with $r > 0$ and $r \neq 1$ is an exponential function sampled at the integers. Both exclusions are real, and they fail for different reasons. At $r = 1$ the rule $C \cdot 1^{\,x}$ is a constant function, and an exponential base is never allowed to be $1$ . At $r < 0$ the power $r^{\,x}$ has no real value at most exponents (there is no real $(-2)^{1/2}$ ), so no real exponential curve passes through the terms at all. With a negative ratio the terms simply alternate in sign, while their sizes march along the curve belonging to $|r|$ . Everything you know about exponential growth and decay transfers straight across: a quantity growing by a fixed percentage each period has ratio $r = 1 + \frac{p}{100}$ for a $p$ percent rise. Reading the value of that quantity once per period produces a geometric sequence.

The five terms 3, 6, 12, 24, 48 sit exactly on one smooth exponential curve, read off only at the whole numbers n equals 1 through 5. A geometric sequence is an exponential function sampled at the integers.

Finding the number of terms with logarithms

In $a_n = a_1 r^{\,n-1}$ the unknown $n$ sits in the exponent, which is exactly the situation logarithms were built for. Divide by $a_1$ , take the logarithm of both sides, and bring the exponent down:

r^{\,n-1} = \frac{a_n}{a_1} \quad\Longrightarrow\quad (n-1)\log r = \log\!\left(\frac{a_n}{a_1}\right) \quad\Longrightarrow\quad n = 1 + \frac{\log(a_n / a_1)}{\log r},

valid whenever $r > 0$ and $r \neq 1$ , so that $\log r$ exists and is not zero. When the target is a nice power ( $2^{\,n-1} = 128$ , say) you can read the exponent off by inspection. When it is not, the logarithm does the work.

Worked example 4 How many terms of $5,\; 10,\; 20,\; 40,\ldots$ must be added before the sum first exceeds $10{,}000$ ?

Here $a = 5$ and $r = 2$ , so the sum of the first $n$ terms simplifies neatly:

S_n = \frac{5(2^{\,n} - 1)}{2 - 1} = 5(2^{\,n} - 1).

Set up the requirement and clear away everything around the power of $2$ :

5(2^{\,n} - 1) > 10{,}000 \quad\Longrightarrow\quad 2^{\,n} - 1 > 2000 \quad\Longrightarrow\quad 2^{\,n} > 2001 .

Now take logarithms. The logarithm is an increasing function, so it preserves the direction of the inequality:

n \log 2 > \log 2001 \quad\Longrightarrow\quad n > \frac{\log 2001}{\log 2} \approx \frac{3.3012}{0.3010} \approx 10.97 .

The number of terms must be a whole number, so the first one that works is $n = 11$ . Check both sides of the boundary: $S_{10} = 5(1024 - 1) = 5115$ , which falls short, and $S_{11} = 5(2048 - 1) = 10{,}235$ , which clears the target.

Worked example 5 A culture of $400$ bacteria triples every hour. After how many whole hours does it first pass one million?

Watch the indexing, because this is where the off-by-one bites in applications. The count at the start is $a_1 = 400$ , which is the population after zero hours. One hour later you have $a_2$ , so the population after $h$ hours is the $(h+1)$ th term:

a_{h+1} = 400 \cdot 3^{\,h} .

Require that this pass one million and isolate the power:

400 \cdot 3^{\,h} > 1{,}000{,}000 \quad\Longrightarrow\quad 3^{\,h} > 2500 .

Take logarithms of both sides and solve for $h$ :

h > \frac{\log 2500}{\log 3} \approx \frac{3.3979}{0.4771} \approx 7.12 .

Hours are whole numbers here, so the population first passes one million at $h = 8$ . Confirm it: $400 \cdot 3^7 = 874{,}800$ , still under, while $400 \cdot 3^8 = 2{,}624{,}400$ , comfortably over. Note that this eighth hour corresponds to the ninth term of the sequence, which is exactly the trap.

What happens if you never stop adding, and let the terms of a decaying geometric sequence pile up forever, is a genuinely different question. That question is the subject of the next lesson.

Common mistakes

Writing $r^{\,n}$ instead of $r^{\,n-1}$ in the nth term. The fifth term of $2, 6, 18, \ldots$ is $2 \cdot 3^4 = 162$ , not $2 \cdot 3^5 = 486$ . Five terms are separated by only four multiplications.
Using the sum formula when $r = 1$ . Dividing by $1 - r$ divides by zero. If every term is the same, the sum is $na$ . Always glance at the ratio before reaching for the formula.
Confusing the last term with the number of terms. In $3 + 6 + \cdots + 384$ the final term is $384$ , but $n = 8$ . Solve $3 \cdot 2^{\,n-1} = 384$ for $n$ before you use the formula.
Sign slips with a negative ratio. With $r = -3$ , an even exponent gives a positive power, so $(-3)^6 = 729$ , and the denominator is $1 - (-3) = 4$ , not $-2$ .
Reading a growth rate as the ratio. A quantity rising by $6$ percent a year multiplies by $r = 1.06$ each year, not by $0.06$ . A drop of $20$ percent means $r = 0.80$ .
Averaging to find a middle term. The number between $4$ and $25$ in a geometric sequence is not $14.5$ . The ratio must be constant, which forces $x^2 = 4 \cdot 25 = 100$ , so $x = 10$ or $x = -10$ , and both give a valid geometric sequence.

Key takeaways

A sequence is geometric when consecutive terms have a constant ratio, $r = a_{n+1} / a_n$ , with $a_1 \neq 0$ and $r \neq 0$ .
The nth term is $a_n = a_1 r^{\,n-1}$ . The exponent is $n - 1$ because reaching the $n$ th term from the first takes $n - 1$ multiplications.
Shift and subtract proves the sum formula: multiplying $S_n$ by $r$ shifts every term one place, so subtracting cancels the middle and leaves $S_n(1 - r) = a(1 - r^{\,n})$ , an identity valid for every $r$ .
Dividing by $1 - r$ then gives $S_n = \dfrac{a(1 - r^{\,n})}{1 - r} = \dfrac{a(r^{\,n} - 1)}{r - 1}$ , which is legal only when $r \neq 1$ .
If $r = 1$ the sum is $S_n = na$ , since every term equals $a$ . The formula cannot reach this case, and no algebra rescues it.
Since $|a_{n+1}| = |r| \cdot |a_n|$ , the terms grow in size when $|r| > 1$ and shrink in size when $|r| < 1$ . A negative $r$ alternates the signs of those terms, so “bigger” and “greater” are not the same thing here. When $r > 0$ and $r \neq 1$ , a geometric sequence is an exponential function sampled at the whole numbers.
Because $n$ sits in an exponent, logarithms are the tool for finding how many terms a condition requires.

Geometric Sequences and Series

Learning goals

What makes a sequence geometric

The nth term, and the off-by-one hiding in it

Summing a finite geometric series

When the ratio is 1

Growth, decay, and the curve behind the sequence

Finding the number of terms with logarithms

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)