Consider the sequence $4,8,16,32,\ldots$4,8,16,32,…. Rather than adding a constant value to each term like the arithmetic sequences we have encountered, the pattern here is generated by multiplying each term by a constant.
Common ratio and the first term
A sequence in which each term increases or decreases from the last by a constant factor is called a geometric sequence. We refer to this constant factor as the common ratio. We still denote the first term in the sequence, or progression, by the letter $a$a and we will refer to the common ratio by the letter $r$r. Thus, our sequence $4,8,16,32\dots$4,8,16,32… is geometric with $a=4$a=4 and $r=2$r=2. Given a geometric sequence, we can find $r$r by dividing any two successive terms $\frac{T_n}{T_{n-1}}$TnTn−1.
It is possible to have a sequence with a negative $r$r, such as $100,-50,25,-12.5,\ldots$100,−50,25,−12.5,…. This is particularly interesting, as you can see the sign of each term alternates. This is a geometric sequence with $a=100$a=100 and $r=-\frac{1}{2}$r=−12.
Recurrence relation
Each term is a multiple of the previous term, and so, we can write a recurrence relation for any geometric sequence as:
$T_n=rT_{n-1},T_1=a$Tn=rTn−1,T1=a
Explicit formula
We recall that the benefit of an explicit formula for a sequence means we can find specific terms without listing out all of the preceding terms, so let's look at a table of values for a concrete example to see the pattern that develops for geometric sequences. For the sequence $5,10,20,40,\ldots$5,10,20,40,…, we have a starting term of $5$5 and a common ratio of $2$2, that is $a=5$a=5 and $r=2$r=2. This gives us:
| $n$n | $T_n$Tn | Pattern |
|---|---|---|
| $1$1 | $5$5 | $5\times2^0$5×20 |
| $2$2 | $10$10 | $5\times2^1$5×21 |
| $3$3 | $20$20 | $5\times2^2$5×22 |
| $4$4 | $40$40 | $5\times2^3$5×23 |
| $\ldots$… | ||
| $n$n | $t_n$tn | $5\times2^{n-1}$5×2n−1 |
From these examples, we could guess that the tenth term becomes:
$T_{10}=10\times2^9$T10=10×29
And the one-hundredth term:
$T_{100}=5\times2^{99}$T100=5×299
The explicit formula for the $n$nth term here is $T_n=5\times2^{n-1}$Tn=5×2n−1.
Any geometric sequence with starting value $a$a and common ratio $r$r has the terms given by:
$a,ar,ar^2,ar^3,\ldots$a,ar,ar2,ar3,…
We see a similar pattern to that of our previous table, and can write down the formula for the $n$nth term as:
$T_n=ar^{n-1}$Tn=arn−1
For any geometric sequence with starting value $a$a and common ratio $r$r, we can express it in either of the following two forms:
- Recursive form is a way to express any term in relation to the previous term:
$T_n=rT_{n-1}$Tn=rTn−1, where $T_1=a$T1=a
- Explicit form is a way to express any term in relation to the term number:
$T_n=ar^{n-1}$Tn=arn−1
Remember, a geometric sequence is defined by the presence of a common ratio, that is, $r=\frac{T_n}{T_{n-1}}$r=TnTn−1. In practice, we confirm this by using any $3$3 consecutive terms, for example, $\frac{T_2}{T_1}=\frac{T_3}{T_2}$T2T1=T3T2.
Worked examples
Example 1
For the sequence $810,270,90,30,\ldots$810,270,90,30,…, find an explicit rule for the $n$nth term and hence, find the $8$8th term.
Think: Check that the sequence is geometric, does each term differ from the last by a constant factor? Then write down the starting value $a$a and common ratio $r$r and substitute these into the general form: $T_n=ar^{n-1}$Tn=arn−1
Do: Dividing the second term by the first we get, $\frac{T_2}{T_1}=\frac{1}{3}$T2T1=13. Checking the ratio between the successive pairs we also get $\frac{1}{3}$13. So we have a geometric sequence with $a=810$a=810 and $r=\frac{1}{3}$r=13. The general formula for this sequence is: $T_n=810\left(\frac{1}{3}\right)^{n-1}$Tn=810(13)n−1.
Hence, the $8$8th term is: $T_8=810\left(\frac{1}{3}\right)^7=\frac{10}{27}$T8=810(13)7=1027.
Example 2
If a geometric sequence has $T_3=12$T3=12 and $T_6=96$T6=96, find the recurrence relation for the sequence.
Think: To find the recurrence relation we need the starting value and common ratio. As we have two terms we can set up two equations in terms of $a$a and $r$r using $T_n=ar^{n-1}$Tn=arn−1.
Do:
| $T_3$T3: | $ar^2=12$ar2=12 |
equation $\left(1\right)$(1)
|
| $T_6$T6: | $ar^5=96$ar5=96 |
equation $\left(2\right)$(2)
|
If we now divide equation $\left(2\right)$(2) by equation $\left(1\right)$(1), we obtain the following:
| $\frac{ar^5}{ar^2}$ar5ar2 | $=$= | $\frac{96}{12}$9612 |
| $r^3$r3 | $=$= | $8$8 |
| $\therefore r$∴r | $=$= | $2$2 |
With the common ratio found to be $2$2, then we know that, using equation $\left(1\right)$(1) $a\times2^2=12$a×22=12 and so $a$a is $3$3. The recurrence relation for this sequence is given by:
$T_n=2T_{n-1},T_1=3$Tn=2Tn−1,T1=3
Geometric sequences in tables and graphs
An arithmetic sequence will always define a straight line graph. For a geometric series the graphical representation is not linear like an arithmetic sequence, except for the trivial case of $r=1$r=1.
The terms of a geometric sequence take the form $T_n=ar^{n-1}$Tn=arn−1, which looks very similar to the form $y=ka^x$y=kax that we have previously encountered when sketching functions. Hence, the path of points plotted according to a geometric sequence follow an exponential curve for positive values of $r$r. More specifically, the graph of $T_n=ar^{n-1}$Tn=arn−1 will trace:
- the path of an exponential growth function for $r>1$r>1
- the path of an exponential decay function for $0
and if $a$a is negative the path will be reflected across the $x$x-axis.
What we haven't previously considered when graphing exponentials is the impact of a negative value of $a$a in $y=ka^x$y=kax. This corresponds to the value of $r$r in $T_n=ar^{n-1}$Tn=arn−1, and we know from our previous examples that it is possible to have a geometric sequence with a negative common ratio.
The table below shows the value of several terms of the geometric sequence $T_n=12\times\left(1.5\right)^{n-1}$Tn=12×(1.5)n−1, which has a starting value of $12$12 and ratio $r=1.5$r=1.5.
| $n$n | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 | $6$6 |
|---|---|---|---|---|---|---|
| $T_n$Tn | $12$12 | $18$18 | $27$27 | $40.5$40.5 | $60.75$60.75 | $91.125$91.125 |
Compare that to this second table shows the value of several terms of the geometric sequence $T_n=12\times\left(-1.5\right)^{n-1}$Tn=12×(−1.5)n−1, which has a starting value of $12$12 and ratio $r=-1.5$r=−1.5.
| $n$n | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 | $6$6 |
|---|---|---|---|---|---|---|
| $T_n$Tn | $12$12 | $-18$−18 | $27$27 | $-40.5$−40.5 | $60.75$60.75 | $-91.125$−91.125 |
As you can see, the magnitude of the values of $T_n$Tn are all increasing, it is only the sign that changes for every second value of $n$n.
Here is a graph of the two geometric sequences depicted in both tables. Note that the odd terms of the zig-zag graph coincide with the terms of the first geometric progression.
And so we can see that the graph of a geometric sequence with a negative $r$r forms a zig-zag pattern, rather than the exponential function we see for positive values. In our example, the zig-zag is growing, since $|r|>1$|r|>1. If $|r|<1$|r|<1 we would see a zig-zag path that was diminishing in size.
Applications of geometric sequences
Since geometric sequences grow or decay exponentially for positive common ratios, we can use them to model situations we have already encountered when studying exponential functions such as the growth of bacteria or the decay of radioactive elements. If we have an amount increasing or decreasing by a constant factor at set time periods, then you can consider that process as being geometric.
A particular application of interest is the growth of an investment due to compound interest. The compound interest formula is $A=P(1+r)^n$A=P(1+r)n. The form of this equation is again very similar to the formula $T_n=ar^{n-1}$Tn=arn−1. The principal $P$P, which is the initial value of the investment, is equivalent to the first term of a geometric sequence. The compound interest rate $r$r is expressed as a percentage and added to $1$1 to give the common ratio of a geometric sequence.
- If a quantity is increasing by $r%$r%, you must multiply by $(1+r%)$(1+r%) to find successive terms. The $1$1 represents $100%$100% of the original quantity and ensures that your calculation gives you the next term rather than just the amount by which it has increased.
- If a given interest rate in a question is compounding at a different frequency, you will need to change both the rate and the time periods. For example, $8%$8% p.a. compounding quarterly for three years becomes a rate of $\frac{8}{4}=2%$84=2% per quarter and the number of time periods becomes $4\times3=12$4×3=12 quarters.
Practice questions
QUESTION 1
Some of the terms in the following geometric sequence are missing. Use the common ratio to find these terms.
$-9$−9, $\editable{}$, $-144$−144, $-576$−576, $\editable{}$
QUESTION 2
The $n$nth term of a geometric sequence is given by the equation $T_n=2\times3^{n-1}$Tn=2×3n−1.
Complete the table of values:
$n$n $1$1 $2$2 $3$3 $4$4 $11$11 $T_n$Tn $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ What is the common ratio between consecutive terms?
Plot the points in the table that correspond to $n=1$n=1, $n=2$n=2, $n=3$n=3 and $n=4$n=4.
Loading Graph...If the plots on the graph were joined they would form:
a straight line
Aa curve
B
QUESTION 3
Consider the finite sequence: $4$4, $4\sqrt{2}$4√2, $8$8, . . . , $256$256
Find the common ratio.
Find $T_6$T6, the $6$6th term.
Solve for $n$n, the number of terms in the sequence.
QUESTION 4
Caitlin invests $\$21000$$21000 in an account that earns $7%$7% p.a. with interest calculated at the end of each year.
Complete the table, calculating the value of the investment, $A$A, at the end of each year for the first four years. Round all values to the nearest dollar.
$n$n $0$0 $1$1 $2$2 $3$3 $4$4 $A$A (in dollars) $21000$21000 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ What is the common ratio between consecutive values of the investment from year to year?
Which of the following equations models the value of the investment, $A$A, after $n$n years?
$A=21000\times\left(\frac{7}{100}\right)^{n-1}$A=21000×(7100)n−1
A$A=21000\left(1+\frac{7}{100}\right)^n$A=21000(1+7100)n
B$A=21000\times\left(\frac{7}{100}\right)^n$A=21000×(7100)n
C$A=21000\left(1+\frac{7}{100}\right)^{n-1}$A=21000(1+7100)n−1
D
QUESTION 5
The average annual rate of inflation in Kazakhstan is $2.6%$2.6%. Bread cost $\$3.65$$3.65 in 2015.
What would the bread cost in 2016?
Give your answer to the nearest cent.
At this rate, what would bread cost in 2018?
Give your answer to the nearest cent.
Write a recursive rule for $V_n$Vn defining the cost of bread $n$n years after 2015, and an initial condition $V_0$V0.
Write both parts of the rule on the same line, separated by a comma.