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Difference between revisions of "Geometric sequence"
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In [[algebra]], a '''geometric sequence''', sometimes called a '''geometric progression''', is a [[sequence]] of numbers such that the ratio between any two consecutive terms is constant. This constant is called the '''common difference''' of the sequence.
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In [[algebra]], a '''geometric sequence''', sometimes called a '''geometric progression''', is a [[sequence]] of numbers such that the ratio between any two consecutive terms is constant. This constant is called the '''common ratio''' of the sequence.
  
 
For example, <math>7, 14, 28, 56</math> is a geometric sequence with common ratio <math>2</math> and <math>100, -50, 25, -25/2, \ldots</math> is a geometric sequence with common ratio <math>-1/2</math>; however, <math>1, 3, 9, 19</math> and <math>-3, 1, 5, 9</math> are not geometric sequences, as the ratio between consecutive terms varies.
 
For example, <math>7, 14, 28, 56</math> is a geometric sequence with common ratio <math>2</math> and <math>100, -50, 25, -25/2, \ldots</math> is a geometric sequence with common ratio <math>-1/2</math>; however, <math>1, 3, 9, 19</math> and <math>-3, 1, 5, 9</math> are not geometric sequences, as the ratio between consecutive terms varies.

Revision as of 14:28, 3 November 2021

In algebra, a geometric sequence, sometimes called a geometric progression, is a sequence of numbers such that the ratio between any two consecutive terms is constant. This constant is called the common ratio of the sequence.

For example, $7, 14, 28, 56$ is a geometric sequence with common ratio $2$ and $100, -50, 25, -25/2, \ldots$ is a geometric sequence with common ratio $-1/2$; however, $1, 3, 9, 19$ and $-3, 1, 5, 9$ are not geometric sequences, as the ratio between consecutive terms varies.

More formally, the sequence $a_1, a_2, \ldots , a_n$ is a geometric progression if and only if $a_2 / a_1 = a_3 / a_2 = \cdots = a_n / a_{n-1}$. This definition appears most frequently in its three-term form: namely, that constants $a$, $b$, and $c$ are in geometric progression if and only if $b / a = c / b$.

Properties

The $n$th term has the closed-form:

$a_n = a_1\cdot r^{n-1}$

Summing a Geometric Sequence

The sum of the first $n$ terms of a geometric sequence is given by

$S_n = a_1 + a_2 + \cdots + a_n = a_1\cdot\frac{r^n-1}{r-1}$

where $a_1$ is the first term in the sequence, and $r$ is the common ratio.

Proof

The geometric sequence can be rewritten as $a_1+r \cdot a_1+r^2 \cdot a_1+ \cdots + r^{n-1} \cdot a_1=a_1(1+r+r^2+ \cdots +r^{n-1})$ where $n$ is the number of terms, $r$ is the common ratio, and $a_1$ is the first term. Multiplying by $(r-1)$ yields $r^n-1$ so $a_1 + a_2 + \cdots + a_n = a_1\cdot\frac{r^n-1}{r-1}$.

Infinite Geometric Sequences

An infinite geometric sequence is a geometric sequence with an infinite number of terms. If the common ratio is small, the terms will approach 0 and the sum of the terms will approach a fixed limit. In this case, "small" means $|r|<1$. We say that the sum of the terms of this sequence is a convergent sum.

For instance, the series $1 + \frac12 + \frac14 + \frac18 + \cdots$, sums to 2. The general formula for the sum of such a sequence is:

$S = \frac{a_1}{1-r}$.



Where $a_1$ is the first term in the sequence, and $r$ is the common ratio.

Proof

Let the sequence be

$S=a_1+a_1r+a_1r^2+a_1r^3+\cdots.$

Multiplying by $r$ yields,

$S \cdot r=a_1r+a_1r^2+a_1r^3+\cdots.$

We subtract these two equations to obtain:

$S-Sr=a_1.$

There is only one term on the right side of the equation because the rest of the terms cancel out after subtraction. Finally, we can factor and divide to get

$S(1-r)=a_1$

thus,

$S=\frac{a_1}{1-r}.$

This method of multiplying the sequence and subtracting equations, called telescoping, is a frequently used method to evaluate infinite sequences. In fact, the same method can be used to calculate the sum of a finite geometric sequence (given above).

Common uses

One common instance of summing infinite geometric sequences is the decimal expansion of most rational numbers. For instance, $0.33333\ldots = \frac 3{10} + \frac3{100} + \frac3{1000} + \frac3{10000} + \ldots$ has first term $a_0 = \frac 3{10}$ and common ratio $\frac1{10}$, so the infinite sum has value $S = \frac{\frac3{10}}{1-\frac1{10}} = \frac13$, just as we would have expected.

Problems

Intermediate

See also

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