Geometric Progression

The numbers 3, 6, 12, 24, 48 form a sequence in which each number is twice the one before it. A sequence in which every term is obtained from the previous one by multiplying by the same number is called a geometric progression, or geometric sequence.

It is called the common ratio because the ratio of any two neighboring terms is the same: ${\frac{a_2}{a_1}}={\frac{a_3}{a_2}}={\frac{a_4}{a_3}}=\ldots={r}$. An arithmetic progression is built by adding the same number each time; a geometric progression, by multiplying.

Contents: Examples · The n-th term · The role of r · Properties · Arithmetic vs geometric · Sum of n terms · Infinite series · Calculator · Common mistakes · Solved problems · Practice

Examples

• $1,\ 2,\ 4,\ 8,\ 16,\ 32,\ \ldots$ – first term $1$, common ratio ${r=2}$.
• $81,\ 27,\ 9,\ 3,\ 1,\ \ldots$ – first term $81$, common ratio ${r=\frac{1}{3}}$, because each term is three times smaller than the one before.
• $5,\ -5,\ 5,\ -5,\ \ldots$ – first term $5$, common ratio ${r=-1}$.
• $1,\ -2,\ 4,\ -8,\ 16,\ \ldots$ – common ratio ${r=-2}$: the signs alternate.
• $7,\ 7,\ 7,\ 7,\ \ldots$ – common ratio ${r=1}$: all the terms are equal.

The sequence $2,\ 4,\ 6,\ 8,\ \ldots$ is not a geometric progression: ${\frac{4}{2}}={2}$, but ${\frac{6}{4}}={1.5}$. It is an arithmetic progression, because the difference between neighboring terms is constant (2).

Why must the terms and the common ratio be nonzero? If ${r=0}$, the sequence looks like $4,\ 0,\ 0,\ 0,\ \ldots$ and the ratio ${\frac{a_3}{a_2}}={\frac{0}{0}}$ makes no sense. That is why a sequence such as $4,\ 0,\ 0,\ \ldots$ is not counted as a geometric progression.

The formula for the n-th term

Start from $a_1$ and keep multiplying by $r$:

${a_2}={a_1r},\quad {a_3}={a_2r}={a_1r^2},\quad {a_4}={a_1r^3},\quad \ldots$

To reach the $n$-th term we multiply by $r$ a total of $n-1$ times. This gives the formula for the general term:

Example. The progression $3,\ 6,\ 12,\ \ldots$ has ${a_1}={3}$ and ${r}={2}$. Its eighth term is ${a_8}={3\times 2^7}={3\times 128}={384}$.

If we know some term $a_k$ rather than the first one, we can start from it: ${a_n}={a_k\cdot r^{\,n-k}}$. For example, ${a_{10}}={a_7\cdot r^3}$.

How to find the common ratio

The common ratio is any term divided by the term before it:

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

If two terms that are not neighbors are given, use ${a_n}={a_k\cdot r^{\,n-k}}$.

• ${a_3}={12}$ and ${a_6}={96}$: ${a_6}={a_3\cdot r^3}$, so ${12r^3}={96}$, ${r^3}={8}$ and ${r}={2}$.
• ${a_2}={6}$ and ${a_4}={54}$: ${6r^2}={54}$, ${r^2}={9}$. Here there are two possibilities: ${r}={3}$ (the progression $2,\ 6,\ 18,\ 54,\ \ldots$) or ${r}={-3}$ (the progression $-2,\ 6,\ -18,\ 54,\ \ldots$).

How the common ratio shapes the sequence

How the terms behave depends on the common ratio $r$ and on the sign of the first term.

Common ratio	What happens to the terms	Example
${r\gt 1}$	If ${a_1\gt 0}$, the progression is increasing and its terms grow without bound. If ${a_1\lt 0}$, it is decreasing.	$1,\ 5,\ 25,\ 125,\ \ldots$ $-1,\ -2,\ -4,\ -8,\ \ldots$
${r}={1}$	All the terms are equal (a constant sequence).	$7,\ 7,\ 7,\ \ldots$
${0\lt r\lt 1}$	If ${a_1\gt 0}$, the progression is decreasing; if ${a_1\lt 0}$, increasing. Either way the terms approach $0$.	$16,\ 8,\ 4,\ 2,\ 1,\ \ldots$
${-1\lt r\lt 0}$	The signs alternate and the terms approach $0$.	$4,\ -2,\ 1,\ -0.5,\ 0.25,\ \ldots$
${r}={-1}$	The terms alternate: $a_1,\ -a_1,\ a_1,\ \ldots$	$5,\ -5,\ 5,\ -5,\ \ldots$
${r\lt -1}$	The signs alternate and the absolute values of the terms grow without bound. The terms in even places head toward one infinity, those in odd places toward the other.	$1,\ -5,\ 25,\ -125,\ \ldots$

In short: when ${|r|\lt 1}$ the terms approach zero, and when ${|r|\gt 1}$ their absolute values grow without bound. With a negative common ratio the signs always alternate.

Properties

The geometric mean property

Every term except the first is linked to its neighbors like this:

Why it is true: ${a_{k-1}\cdot a_{k+1}}={\frac{a_k}{r}\cdot a_kr}={a_k^2}$.

If the terms are positive, ${a_k}={\sqrt{a_{k-1}\cdot a_{k+1}}}$: each term is the geometric mean of its neighbors. This is where the name “geometric” progression comes from. For example, in $4,\ 6,\ 9$ we have ${6}={\sqrt{4\cdot 9}}$.

The converse is also true: if the numbers $a$, $b$, $c$ are nonzero and ${b^2}={ac}$, then $a$, $b$, $c$ are consecutive terms of a geometric progression. For example, the numbers $2$, $x$, $18$ are consecutive terms of a geometric progression when ${x^2}={36}$, that is, for ${x}={6}$ or ${x}={-6}$.

Terms equally far from the ends

Two terms that are the same distance from the ends have the same product:

${a_1\cdot a_n}={a_2\cdot a_{n-1}}={a_3\cdot a_{n-2}}=\ldots={a_k\cdot a_{n-k+1}}$

More generally: if ${m+n}={p+s}$, then ${a_m\cdot a_n}={a_p\cdot a_s}$, because both products equal ${a_1^2\,r^{\,m+n-2}}$.

Example. In a geometric progression ${a_2\cdot a_8}={36}$. Then ${a_4\cdot a_6}={36}$ (because ${4+6}={2+8}$) and ${a_5^2}={36}$, so ${a_5}={6}$ or ${a_5}={-6}$.

Arithmetic and geometric progressions compared

	Arithmetic progression	Geometric progression
Rule	${a_{n+1}}={a_n+d}$	${a_{n+1}}={a_n\cdot r}$
General term	${a_n}={a_1+(n-1)d}$	${a_n}={a_1\cdot r^{n-1}}$
Middle term	${a_k}={\frac{a_{k-1}+a_{k+1}}{2}}$ – the arithmetic mean	${a_k^2}={a_{k-1}\cdot a_{k+1}}$ – the geometric mean
Sum of $n$ terms	${S_n}={\frac{(a_1+a_n)\,n}{2}}$	${S_n}={a_1\cdot\frac{r^n-1}{r-1}}$

An arithmetic progression grows “linearly”, by the same step each time, while a geometric one grows “exponentially”, by the same factor each time. So when ${r\gt 1}$ a geometric progression sooner or later overtakes any arithmetic progression.

The sum of the first n terms

Let ${S_n}={a_1+a_2+\ldots+a_n}={a_1+a_1r+a_1r^2+\ldots+a_1r^{n-1}}$. A sum like this is called a geometric series.

Deriving the formula. Multiply the sum by $r$ and subtract:

${rS_n}={a_1r+a_1r^2+\ldots+a_1r^{n-1}+a_1r^n}$

${rS_n-S_n}={a_1r^n-a_1}$

All the terms in the middle cancel. What is left is ${S_n(r-1)}={a_1(r^n-1)}$. If ${r\ne 1}$, divide by ${r-1}$:

The two fractions are equal (the numerator and the denominator are both multiplied by $-1$). The first is handier when ${r\gt 1}$, the second when ${r\lt 1}$. Since ${a_1r^n}={a_n r}$, the formula can also be written with the last term: ${S_n}={\frac{a_nr-a_1}{r-1}}$.

Example 1. ${1+2+4+\ldots+2^9}={1\times\frac{2^{10}-1}{2-1}}={1023}$.

Example 2. ${3-6+12-24+48-96}=\,?$ Here ${a_1}={3}$, ${r}={-2}$, ${n}={6}$: ${S_6}={3\times\frac{(-2)^6-1}{-2-1}}={3\times\frac{63}{-3}}={-63}$.

The chessboard legend. The inventor of chess asked for 1 grain of wheat for the first square, 2 for the second, 4 for the third, and so on, doubling each time, up to the 64th square. The number of grains is

${1+2+2^2+\ldots+2^{63}}={2^{64}-1}={18{,}446{,}744{,}073{,}709{,}551{,}615}.$

That is far more wheat than the whole world grows in a year.

Money. If you deposit $\$1{,}000$ at the start of every year into an account that pays 4% interest compounded yearly, at the end of the fifth year you will have

${\$1{,}000\times 1.04}+{\$1{,}000\times 1.04^2}+{\ldots+\$1{,}000\times 1.04^5}={\$1{,}040\times\frac{1.04^5-1}{0.04}}\approx{\$5{,}632.98}.$

The terms form a geometric progression with ${a_1}={1{,}040}$ and ${r}={1.04}$. You can work out compound interest with the compound interest calculator. The same sum gives the monthly payment of a loan: see where the loan payment formula comes from.

Infinite geometric series

When ${|r|\lt 1}$, the absolute values of the terms shrink toward zero. Even though there are infinitely many terms, their sum is a finite number. A progression like this is sometimes called an infinite decreasing geometric progression.

${\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots}={1}$: the pieces fill the whole square.

Where the formula comes from. We know that ${S_n}={a_1\cdot\frac{1-r^n}{1-r}}$. When ${|r|\lt 1}$ and $n$ grows, $r^n$ approaches $0$. So $S_n$ approaches ${\frac{a_1}{1-r}}$.

Examples.

• ${1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots}={\frac{1}{1-\frac{1}{2}}}={2}$.
• ${1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots}={\frac{1}{1-\left(-\frac{1}{2}\right)}}={\frac{2}{3}}$.
• Repeating decimals are sums of such series: ${0.\overline{3}}={\frac{3}{10}+\frac{3}{100}+\frac{3}{1000}+\ldots}={\frac{\frac{3}{10}}{1-\frac{1}{10}}}={\frac{1}{3}}$.
• In the same way ${0.\overline{27}}={\frac{\frac{27}{100}}{1-\frac{1}{100}}}={\frac{27}{99}}={\frac{3}{11}}$, and ${0.\overline{9}}={\frac{\frac{9}{10}}{1-\frac{1}{10}}}={1}$.

The formula ${S}={\frac{a_1}{1-r}}$ holds only when ${|r|\lt 1}$. The sum ${1+2+4+8+\ldots}$ has no finite value, yet the formula would give ${\frac{1}{1-2}}={-1}$, which is absurd.

Geometric Progression Calculator

Enter the first term, the common ratio and the number of terms. The ratio can be a fraction such as $1/2$ or a decimal such as $0.5$. The calculator shows the terms, the $n$-th term, the sum of the first $n$ terms and, when ${|r|\lt 1}$, the sum of the infinite series.

Common mistakes

Solved problems

Problem 1. Is the sequence $2,\ 4,\ 6,\ 8,\ \ldots$ a geometric progression?

Solution. No. ${\frac{4}{2}}={2}$, but ${\frac{6}{4}}={1.5}$: the ratios of neighboring terms are not equal. (It is an arithmetic progression with common difference 2.)

Problem 2. The numbers $2,\ 4,\ 8,\ \ldots$ form a geometric progression. Find its 10th term.

Solution. ${a_1}={2}$ and ${r}={2}$, so ${a_{10}}={a_1r^9}={2\times 2^9}={2\times 512}={1024}$.

Problem 3. Find the first term and the common ratio of a geometric progression if ${a_5-a_1}={15}$ and ${a_4-a_2}={6}$.

Solution. Write the terms using $a_1$ and $r$:

${a_1r^4-a_1}={15}\ \Rightarrow\ {a_1(r^2-1)(r^2+1)}={15}$

${a_1r^3-a_1r}={6}\ \Rightarrow\ {a_1r(r^2-1)}={6}$

Divide the first equation by the second (${r^2-1}\ne{0}$, otherwise the first would say ${0}={15}$): ${\frac{r^2+1}{r}}={\frac{5}{2}}$, so ${2r^2-5r+2}={0}$, which gives ${r}={2}$ or ${r}={\frac{1}{2}}$.

• If ${r}={2}$: ${a_1\times 2\times 3}={6}$, ${a_1}={1}$. The progression is $1,\ 2,\ 4,\ 8,\ 16,\ \ldots$
• If ${r}={\frac{1}{2}}$: ${a_1\times\frac{1}{2}\times\left(-\frac{3}{4}\right)}={6}$, ${a_1}={-16}$. The progression is $-16,\ -8,\ -4,\ -2,\ -1,\ \ldots$

Answer: ${a_1}={1}$, ${r}={2}$, or ${a_1}={-16}$, ${r}={\frac{1}{2}}$. Check the second one: ${-1-(-16)}={15}$ and ${-2-(-8)}={6}$.

Problem 4. For which values of $x$ are the numbers ${x-1}$, ${x+1}$ and ${2x+5}$ consecutive terms of a geometric progression?

Solution. By the geometric mean property, ${(x+1)^2}={(x-1)(2x+5)}$, that is, ${x^2+2x+1}={2x^2+3x-5}$ and ${x^2+x-6}={0}$. The roots are ${x}={2}$ and ${x}={-3}$.

• ${x}={2}$: the numbers $1,\ 3,\ 9$, with common ratio $3$.
• ${x}={-3}$: the numbers $-4,\ -2,\ -1$, with common ratio $\frac{1}{2}$.

In both cases none of the numbers is zero, so the answer is ${x}={2}$ or ${x}={-3}$.

Problem 5. Three numbers form a geometric progression. Their sum is 21 and their product is 216. Find the numbers.

Solution. Call the numbers $\frac{b}{r}$, $b$, $br$. Their product is ${b^3}={216}$, so ${b}={6}$. Then ${\frac{6}{r}+6r}={15}$, that is, ${2r^2-5r+2}={0}$, which gives ${r}={2}$ or ${r}={\frac{1}{2}}$. Answer: $3,\ 6,\ 12$ or $12,\ 6,\ 3$.

Problem 6. The number of bacteria in a colony doubles every 20 minutes. At the start there is 1 bacterium. How many bacteria will there be after 3 hours?

Solution. In 3 hours there are ${\frac{180}{20}}={9}$ doublings. The number at the start is ${a_1}={1}$, after the first doubling it is $a_2$, and after the ninth it is $a_{10}$: ${a_{10}}={1\times 2^9}={512}$ bacteria.

Problem 7. A ball is dropped from a height of 8 m. After each bounce it rises to $\frac{3}{4}$ of its previous height. How far does the ball travel before it comes to rest?

Solution. Going down, the ball travels ${8+6+4.5+\ldots}$, and going up, ${6+4.5+\ldots}$ The sum ${6+4.5+\ldots}$ is an infinite geometric series with ${a_1}={6}$ and ${r}={\frac{3}{4}}$: ${\frac{6}{1-\frac{3}{4}}}={24}$. The total distance is ${8+2\times 24}={56}$ m.

Problem 8. Write $0.\overline{36}$ as a fraction.

Solution. ${0.\overline{36}}={\frac{36}{100}+\frac{36}{10{,}000}+\ldots}={\frac{\frac{36}{100}}{1-\frac{1}{100}}}={\frac{36}{99}}={\frac{4}{11}}$.

Practice problems

Try them yourself, then open the solution.

1. ${a_1}={5}$, ${r}={3}$. Find $a_6$.

${a_6}={5\times 3^5}={5\times 243}={1215}$.

2. ${a_1}={64}$, ${r}={-\frac{1}{2}}$. Find $a_7$.

${a_7}={64\times\left(-\frac{1}{2}\right)^6}={64\times\frac{1}{64}}={1}$.

3. ${a_2}={6}$ and ${a_5}={162}$. Find $a_1$ and $r$.

${a_5}={a_2r^3}$, so ${r^3}={\frac{162}{6}}={27}$ and ${r}={3}$. Then ${a_1}={\frac{6}{3}}={2}$.

4. ${a_3}={4}$ and ${a_5}={36}$. Find $a_4$.

${r^2}={\frac{36}{4}}={9}$, so ${r}={3}$ or ${r}={-3}$. Then ${a_4}={12}$ or ${a_4}={-12}$. You can also see this from ${a_4^2}={a_3\cdot a_5}={144}$.

5. Find ${3+6+12+\ldots+768}$.

${a_1}={3}$, ${r}={2}$. From ${3\times 2^{n-1}}={768}$ we get ${2^{n-1}}={256}$, so ${n}={9}$. ${S_9}={3\times\frac{2^9-1}{2-1}}={3\times 511}={1533}$.

6. ${a_1}={2}$, ${r}={3}$ and ${S_n}={242}$. Find $n$.

${2\times\frac{3^n-1}{3-1}}={242}$, so ${3^n-1}={242}$, ${3^n}={243}$ and ${n}={5}$.

7. Find ${9-3+1-\frac{1}{3}+\ldots}$

${a_1}={9}$, ${r}={-\frac{1}{3}}$ and ${|r|\lt 1}$. ${S}={\frac{9}{1+\frac{1}{3}}}={\frac{9}{\frac{4}{3}}}={\frac{27}{4}}$.

8. An infinite geometric series has sum 12 and first term 8. Find the common ratio and the third term.

${\frac{8}{1-r}}={12}$, so ${1-r}={\frac{2}{3}}$ and ${r}={\frac{1}{3}}$. ${a_3}={8\times\frac{1}{9}}={\frac{8}{9}}$.

9. Write $0.1\overline{6}$ as a fraction.

${0.1\overline{6}}={\frac{1}{10}+\frac{6}{100}+\frac{6}{1000}+\ldots}={\frac{1}{10}+\frac{\frac{6}{100}}{1-\frac{1}{10}}}={\frac{1}{10}+\frac{1}{15}}={\frac{1}{6}}$.

10. A town has 50,000 people and its population grows by 10% a year. What will it be in 3 years?

Each year the population is multiplied by $1.1$: ${50{,}000\times 1.1^3}={50{,}000\times 1.331}={66{,}550}$ people.

11. A geometric progression has positive terms and ${a_3\cdot a_9}={144}$. Find $a_6$.

${3+9}={6+6}$, so ${a_6^2}={a_3\cdot a_9}={144}$. The terms are positive, so ${a_6}={12}$.

More: Arithmetic Progression · Geometric progression problems · Problems involving progressions