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Geometric sequence

Calculus
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Series and Sequences
schedule Mar 10, 2022
Last updated
local_offer Arthur
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In a geometric sequence, each term is the previous term multiplied by a constant. The constant we multiply each term by to get the next term is referred to as the common ratio.

General form

The general form of a geometric sequence can be expressed as follows:

$$\{a, ar, ar^2,ar^3, \ldots \}$$

where:

  • $a$ is the first term in the geometric sequence

  • $r$ is the common ratio

WARNING

Note that $r$ cannot be 0 or else the sequence is not geometric and will result in $\{a, 0, 0, 0,\ldots\}$.

Example

Consider the following geometric sequence:

$$\{2,4,8,16,32,\ldots\}$$

Here we start with 2 and multiply by 2 to get the next term in the geometric sequence.

Therefore in terms of the general form of the geometric sequence we can say:

$a$ = 2 (first term)

$r$ = 2 (common ratio)

Note that elements do not necessarily have to get larger in the sequence:

$$\{4, 2, 1, \frac{1}{2}, \frac{1}{4}, \ldots\}$$

The above is a perfectly valid geometric sequence with:

$a$ = 4

$r$ = 0.5

Sum of a geometric sequence

Finite

The general formula for calculating the sum of a finite geometric sequence:

$$\sum_{i=0}^{n-1} ar^i = \frac{a(1-r^n)}{1-r},\, r \ne 1$$

where:

  • $n$: number of terms in the geometric sequence

  • $a$: first term in the geometric sequence

  • $r$: common ratio

Example

Consider the following finite geometric sequence with 5 terms:

$$\{2,4,8,16,32\}$$

Here we can see that:

$n$ = 5

$a$ = 2

$r$ = 2

Therefore the sum can be calculated as:

$$\begin{align*} \sum_{i=0}^{4} ar^i &= \frac{2(1-2^5)}{1-2} \\ &= \frac{2(1-32)}{-1} \\ &= 62 \end{align*}$$

Derivation

Let us represent the sum of the finite geometric sequence using $S$:

$$S = a + ar +\, \ldots\, + ar^{n-2} + ar^{n-1}$$

Multiplying both sides by $r$:

$$S * r=ar+ar^2+\,\ldots\,+ar^{n-1}+ar^{n}$$

If we subtract the bottom equation from the top equation:

$$\begin{align*} S - S*r &= a - ar^n \\ S(1-r)&= a(1-r^n) \\ S &= \frac{a(1-r^n)}{1-r} \end{align*}$$
NOTE

Note that when we subtract the bottom equation from the top equation the terms such as $ar, ar^2$ etc all cancel each other out.

Infinite

The general formula for calculating the sum of a infinite geometric sequence:

$$\begin{equation}\label{eq:tUfdg31PYKYRi6pE17v} \sum_{i=0}^{\infty}ar^i=\frac{a}{1-r}, \enspace -1 < r < 1 \end{equation}$$
WARNING

We can only calculate a sum when $-1< r < 1$ (i.e. when elements in the geometric sequence progressively get closer and closer to 0). Otherwise, we are not able to calculate a sum.

Example

Consider the following infinite geometric sequence:

$$\left\{1,\frac{1}{2},\frac{1}{4},\frac{1}{8}, \ldots\right\}$$

Here we can see that:

$a$ = 1

$r$ = $\frac{1}{2}$

Therefore the sum can be calculated as:

$$\begin{align*} \sum_{i=0}^{\infty}\frac{1}{2}^i&=\frac{1}{1-0.5} \\ &= 2 \end{align*}$$
robocat
Published by Arthur Yanagisawa
Edited by 0 others
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