Geometric sequences

Sequences and series: Geometric sequences and series

Geometric sequences

We will now check out geometric sequences.

Geometric sequence

A geometric sequence is a sequence in which each term is obtained by multiplying the previous term with a fixed number.

This fixed number is called the common ratio. It is often denoted with $r$ .

Hence, for every $n$ greater than $1$ we have: $t_n=r \cdot t_{n-1}$ .

Hence, the terms in the sequence increase exponentially if $r>1$ .

If $0<r<1$ then the terms in the sequence decrease exponentially.

Examples

Give the initial term $t_1$ , the common ratio $r$ and the number of terms $n$ of the following sequence $t$ .

$2, 8, 32, 128, \ldots, 131072, 524288$

$t_1=2$
$r=4$
$n=10$

The initial term is the first term and is equal to $2$ . Hence, $t_1=2$ .
To calculate the common ratio, we divide to subsequent terms with each other: $\frac{8}{2}=4$ (note: we divide the second term $t_2$ by the first one $t_1$ and not the other way around). We see that this common ratio also holds for other subsequent terms. Hence, it's a geometric series. In this case, we have $r=4$ .
Since the sequence starts at $2$ and ends at $524288$ , in which each time is multiplied by $4$ , this sequence has $10$ terms. Hence, $n=10$ .

New example

Each sequence which is both arithmetic and geometric is constant.

This can be seen by using the first three terms of the sequence $t_1$ , $t_2$ , $t_3$ :

Since the sequence is arithmetic, there is a difference $d$ , such that $t_2=t_1+d$ and $t_3=t_1+2v$ .
Since the sequence is geometric, there is a common ratio $r$ such that $t_2= t_1\cdot r$ and $t_3=t_1\cdot r^2$ .

If we compare the two expressions for $t_2$ and next the two expressions for $t_3$ , and after moving all terms with $t_1$ to the left-hand side and all terms without $t_1$ to the right-hand side, we find the system of equations:

$\eqs{t_1\cdot(r-1) &=& d\\ t_1\cdot(r^2-1) &=& 2 d}$

After multiplying both terms of the first equation by $(r+1)$ , both equations are equal on the left-hand side. Hence, after equalizing the right-hand sides we get

$(r+1)\cdot d = 2d$

Hence, $d=0$ or $r=1$ . In both cases, the sequence is constant.

The formula in the definition for finding the $n$ -th term is called a recursive formula since the previous term is needed. We can also compose a direct formula. That is a formula for finding the $n$ -th term by means of the rank number and the common ratio. There we do not need the previous term.

Direct formula geometric sequence

The direct formula for calculating the $n$ -th term of an geometric sequence $t_1, t_2, \ldots$ , is:

$t_n=t_1\cdot r^{n-1}$ where $r$ is the common ratio.

In general, the terms of a geometric sequence can be described as follows:

$\begin{array}{lcl} t_1&=&t_1 \\ t_2=t_1 \cdot r&=&t_1 \cdot r\\ t_3=t_2 \cdot r = t_1\cdot r \cdot r &=&t_1 \cdot r^2\\ t_4=t_3 \cdot r=t_1 \cdot r^2 \cdot r &=&t_1 \cdot r^3\\ \vdots \phantom{=t_2 \cdot r = t_1xx} &&\vdots\\ t_{20}=t_{19} \cdot r&=&t_1 \cdot r^{19}\\ \vdots \phantom{=t_2 \cdot r = t_1xx} &&\vdots \end{array}$

This leads us to the direct formula:

$t_n=t_1\cdot r^{n-1}$

In this case, the sequence starts at $t_1$ . We can also let the sequence start at $t_0$ . In that case, the direct formula is:

$t_n=t_0\cdot r^n$

This follows from the given formula since there are now $n+1$ terms, and the initial term is $t_0$ .

Notice that if we replace the addition by multiplication and multiplication by exponentiation in the direct formula for the term with rank number $n$ of an arithmetic sequence, we end up with the direct formula for the term with rank number $n$ of a geometric sequence.

In other words: the logarithm of the $n$ -th term of a geometric sequence is the $n$ -th term of an arithmetic sequence. In there, the difference of the arithmetic sequence is equal to the logarithm of the common ratio.

Here are a couple of examples which show how we can compose the direct formula and how it can be used to calculate a term.

Which direct formula corresponds with the following geometric sequence?

$a_1=2,\quad a_2=16{,}\quad a_3=128{,}\quad a_4=1024{,}\quad\ldots$

$a_k=$ $2\cdot 8^{k-1}$

Given is the fact that $a_k$ is a geometric sequence. This means that the direct formula is equal to $a_k=a_1 \cdot r^{k-1}$ , in which $r$ is the common ratio. hence, we have to determine $a_1$ and the common ratio.

The initial term $a_1$ is given and equal to $2$ .
The common ratio is equal to $r=\frac{16}{2}=8$ .

Consequently, the direct formula gives

$a_k=2\cdot 8^{k-1}$

New example