A sequence in which each term increases or decreases from the last by a constant factor is called a geometric sequence. We refer to the constant factor the terms are changing by as the common ratio, which will result from dividing any two successive terms \left(\dfrac{t_{n+1}}{t_n}\right).
We denote the first term in the sequence by the letter a and the common ratio by the letter r. For example, the sequence 4,8,16,32\ldots is geometric with a=4 and r=2. The sequence 100,-50,25,-12.5,\ldots is geometric with a=100 and r=\dfrac{-1}{2}.
To describe the rule in words we say "next term is r multiplied by previous term". Therefore we can write any geometric sequence as the recurrence relation: t_n=rt_{n-1},t_1=a or alternatively t_{n+1}=rt_n, where t_1=a.
Note that it is also possible to define the initial term as t_0, this is particularly useful in financial applications of sequences.
We can also find an explicit formula in terms of a and r, this is useful for finding the nth term without listing the sequence.
n | t_{n} | \text{Pattern} |
---|---|---|
1 | 5 | 5\times 2^0 |
2 | 10 | 5\times 2^1 |
3 | 20 | 5\times 2^2 |
4 | 40 | 5\times 2^3 |
... | ||
n | t_n | 5\times 2^{n-1} |
For any geometric progression with starting value a and common ratio r has the terms given by: a,ar,ar^2,ar^3, \ldots. We see a similar pattern to our previous table and can write down the formula for the nth term: t_n=ar^{n-1}.
For any geometric sequence with starting value a and common ratio r, we can express it in either of the following two forms:
Recursive form: t_n=rt_{n-1}, where t_1=a or alternatively t_{n+1}=rt_n, where t_1=a
Explicit form: t_n=ar^{n-1}
Study the pattern for the following geometric sequence, and write down the next two terms.4,\, 12, \,36, \,\ldots
In a geometric progression, T_4=32 and T_6=128.
Solve for r, the common ratio in the sequence.
For the case where r=2, solve for a, the first term in the progression.
Consider the sequence in which the first term is positive. Find an expression for T_n, the general nth term of this sequence.
For any geometric sequence with starting value a and common ratio r, we can express it in either of the following two forms:
Recursive form: t_n=rt_{n-1}, where t_1=a or alternatively t_{n+1}=rt_n, where t_1=a
Explicit form: t_n=ar^{n-1}
When given a formula for the nth term we can generate a table of values for the sequence. For example in the sequence given by the formula t_n=12\times \left(1.5\right)^{n-1}, by substituting for n appropriately and using a calculator, we can generate the following table of the first 6 terms of the sequence:
n | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|
T_n | 12 | 18 | 27 | 40.5 | 60.75 | 91.125 |
Perhaps more interesting though is the different types of graphs that geometric sequences correspond to. The graphs are not linear like arithmetic progressions, except for the trivial case of r=1. The path of points plotted from a geometric sequence follow an exponential curve for positive values of r.
Sequences of the form t_n=ar^{n-1} are exponential graphs, and where a>0 they will follow:
The path of an exponential growth function for r>1.
The path of an exponential decay function for 0<r<1.
If a is negative the path will be reflected about the x-axis.
What happens when r is negative? The values of successive terms flip their sign so that the graph is depicted as either a growing (r<-1) or diminishing (-1<r<0) zig-zag path - alternating between points on the graph f(n)=ar^{n-1} and f(n)=-ar^{n-1}, depending on the power being odd or even.
Consider the geometric progression with starting value 12 and ratio r=-1.5. This is the same as the example in the previous table but the ratio is now negative. The nth term is given by t_n=12 \times (-1.5)^{n-1}, the table will be the same but the sign of the terms will alternate.
The new table becomes:
n | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|
T_n | 12 | -18 | 27 | -40.5 | 60.75 | -91.125 |
Below is the comparison of the graphs of T_n=12 \times 1.5^{n-1} and T_n=12\times (-1.5)^{n-1}. The graph in red illustrates T_n=12 \times 1.5^{n-1} and the zig-zag graph illustrates T_n=12\times (-1.5)^{n-1}.
Based on the graph above, it can be observed that the odd terms of the zig-zag graph coincide with the terms of the first geometric progression.
Try adjusting the values of a and r in the applet below to observe the effect on the plotted points.
If r \gt 0, the points trend upwards along an exponential function. If r \lt 0, the points zig-zag about the x-axis.
If a \gt 0, the plotted points trend upwards. If a \lt 0, the plotted points trend downwards.
The nth term of a geometric progression is given by the equation T_n=2\times 3^{n-1}.
Complete the table of values:
n | 1 | 2 | 3 | 4 | 10 |
---|---|---|---|---|---|
T_n |
What is the common ratio between consecutive terms?
Plot the points in the table that correspond to n=1,\,n=2,\,n=3, and n=4.
If the plots on the graph were joined they would form:
Consider the following graph of the first 4 terms of a sequence.
Write a recursive rule for T_n in terms of T_{n-1}, including the initial term T_1.
The given table of values represents terms in a geometric sequence.
n | 1 | 2 | 3 | 4 |
---|---|---|---|---|
T_n | 7 | -21 | 63 | -189 |
Identify r, the common ratio between consecutive terms.
Write a simplified expression for the general nth term of the sequence, T_n.
Find the 12th term of the sequence.
We can represent geometric sequences in both tables and graphs.
From a graph of points, the sequence can be read from the y-coordinates.