What is a geometric sequence?
A sequence in which each term increases or decreases from the last by multiplying by a constant factor is called a geometric sequence. The constant factor multiplying each term in the sequence to create the next term is referred to as the common ratio, which will result from dividing any two successive terms $\frac{u_{n+1}}{u_n}$un+1un.
The first term in the sequence is denoted by the letter $a$a and the common ratio is denoted by $R$R. For example, the sequence $4,8,16,32$4,8,16,32 is geometric with $a=4$a=4 and $R=2$R=2. The sequence $100,-50,25,-12.5$100,−50,25,−12.5 is geometric with $a=100$a=100 and $R=-\frac{1}{2}$R=−12.
Since, $u_2=R\times u_1$u2=R×u1, $u_3=R\times u_2$u3=R×u2 and so on, any geometric sequence can be written as a recurrence relation, which was found previously in this chapter:
$u_{n+1}=Ru_n,u_0=a$un+1=Run,u0=a
An explicit generating rule can be found in terms of $a$a and $R$R. This is useful for finding the $n$nth term without listing the sequence or having to use the previous term in the sequence each time to find the next term.
Consider the following table to see the pattern for the explicit formula. For the sequence $5,10,20,40,\ldots$5,10,20,40,…, the starting term is $5$5 and there is a common ratio of $2$2, that is $a=5$a=5 and $R=2$R=2. A table of the sequence is show below:
| $n$n | $u_{n+1}$un+1 | Pattern |
|---|---|---|
| $0$0 | $5$5 | $5\times2^0$5×20 |
| $1$1 | $10$10 | $5\times2^1$5×21 |
| $2$2 | $20$20 | $5\times2^2$5×22 |
| $3$3 | $40$40 | $5\times2^3$5×23 |
| ... | ||
| $n$n | $u_{n+1}$un+1 | $5\times2^n$5×2n |
By correctly identifying the pattern, the tenth term becomes $u_{10}=10\times2^9$u10=10×29 and the one-hundredth term would be $u_{100}=5\times2^{99}$u100=5×299. And following the pattern, the explicit formula for the $n$nth term is $u_n=5\times2^{n-1}$un=5×2n−1.
For any geometric progression with starting value $a$a and common ratio $R$R and with terms given by: $a,aR,aR^2,aR^3,...$a,aR,aR2,aR3,... , a similar pattern can be observed. Hence, the explicit generating rule for the $n$nth term in any geometric sequence is given by:
$u_n=aR^{n-1}$un=aRn−1
For any geometric sequence with starting value $a$a and common ratio $R$R, the sequence can be expressed in either of the following two forms:
- Recursive form is a way to express any term in relation to the previous term:
$u_{n+1}=Ru_n$un+1=Run, where $u_0=a$u0=a
- Explicit form is a way to express any term in relation to the term number:
$u_n=aR^{n-1}$un=aRn−1
Worked examples
Example 1
For the sequence $810,270,90,30...$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 the starting value $a$a and common ratio $R$R and substitute these into the general form: $u_n=aR^{n-1}$un=aRn−1
Do: Dividing the second term by the first we get, $\frac{u_2}{u_1}=\frac{1}{3}$u2u1=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: $u_n=810\left(\frac{1}{3}\right)^{n-1}$un=810(13)n−1.
Hence, the $8$8th term is: $u_8=810\left(\frac{1}{3}\right)^7=\frac{10}{27}$u8=810(13)7=1027.
Example 2
For the sequence $5,20,80,320,...$5,20,80,320,..., find $n$n if the $n$nth term is $327680$327680.
Think: Find a general rule for the sequence, substitute in $327680$327680 for $u_n$un and rearrange for $n$n.
Do: This is a geometric sequence with $a=5$a=5 and common ratio $R=4$R=4. Hence, the general rule is: $u_n=5\left(4\right)^{n-1}$un=5(4)n−1, substituting $u_n=327680$un=327680, we get:
| $327680$327680 | $=$= | $5\left(4\right)^{n-1}$5(4)n−1 | |
| $\therefore4^{n-1}$∴4n−1 | $=$= | $65536$65536 | |
| $4^{n-1}$4n−1 | $=$= | $4^8$48 | Solve by guess and check, technology or logarithms. |
| Hence, $n-1$n−1 | $=$= | $8$8 | |
| $n$n | $=$= | $9$9 |
Hence, the $9$9th term in the sequence is $327680$327680.
Example 3
If a geometric sequence has $u_3=12$u3=12 and $u_6=96$u6=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 $u_n=aR^{n-1}$un=aRn−1.
Do:
| $u_3$u3: | $aR^2=12$aR2=12 | $.....\left(1\right)$.....(1) |
and
| $u_6$u6: | $aR^5=96$aR5=96 | $.....\left(2\right)$.....(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:
$u_{n+1}=2u_n,u_1=3$un+1=2un,u1=3
Practice questions
QUESTION 1
Study the pattern for the following sequence.
$-9$−9$,$, $3.6$3.6$,$, $-1.44$−1.44$,$, $0.576$0.576 ...
State the common ratio between the terms.
QUESTION 2
In a geometric progression, $T_4=54$T4=54 and $T_6=486$T6=486.
Solve for $r$r, the common ratio in the sequence. Write both solutions on the same line separated by a comma.
For the case where $r=3$r=3, solve for $a$a, the first term in the progression.
Consider the sequence in which the first term is positive. Find an expression for $T_n$Tn, the general $n$nth term of this sequence.
Geometric sequences in tables and graphs
When given a formula for the $n$nth term, a table of values can be generated for the sequence. For example, in the sequence given by the formula $u_n=12\times\left(1.5\right)^{n-1}$un=12×(1.5)n−1, by substituting for $n$n appropriately and using a calculator, the following table can be created, listing the first $6$6 terms of the sequence:
| $n$n | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 | $6$6 |
|---|---|---|---|---|---|---|
| $u_n$un | $12$12 | $18$18 | $27$27 | $40.5$40.5 | $60.75$60.75 | $91.125$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 $R=1$R=1. The path of points plotted from a geometric sequence follow an exponential curve for positive values of $R$R (graph can be seen below in the next worked example).
Geometric sequences of the form $u_n=aR^{n-1}$un=aRn−1, where $a>0$a>0 will follow:
- The path of an exponential growth function for $R>1$R>1
- The path of an exponential decay function for $0
- If $a$a is negative the path will be reflected about the $x$x-axis
What if $R$R is negative? The values of successive terms flip their sign so that the graph is depicted as either a growing ($|R|>1$|R|>1) or diminishing ($|R|<1$|R|<1) zig-zag path - alternating between points on the graph$f(n)=a|R|^{n-1}$f(n)=a|R|n−1 and $f(n)=-a|R|^{n-1}$f(n)=−a|R|n−1, depending on the power being odd or even.
Worked examples
Example 4
For the geometric progression with starting value $12$12 and ratio $R=-1.5$R=−1.5, create a table and plot a graph of the sequence.
Think: This is the same as the example in the previous table but the ratio is now negative. The $n$nth term is given by $u_n=12\times\left(-1.5\right)^{n-1}$un=12×(−1.5)n−1, the table will be the same but the sign of the terms will alternate.
Do: The new table becomes:
| $n$n | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 | $6$6 |
|---|---|---|---|---|---|---|
| $u_n$un | $12$12 | $-18$−18 | $27$27 | $-40.5$−40.5 | $60.75$60.75 | $-91.125$−91.125 |
Checking, for $n=1$n=1, we have $u_1=12\times\left(-1.5\right)^{1-1}=12$u1=12×(−1.5)1−1=12 and for $n=2$n=2 we have $u_2=12\times\left(-1.5\right)^{2-1}=-18$u2=12×(−1.5)2−1=−18, continuing on even numbered terms become negative and odd numbered terms become positive.
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.
Try adjusting the values of $a$a and $R$R in the applet below to observe the effect on the plotted points.
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For $|R|>1$|R|>1 the values will diverge as $n$n increases, the terms will keep getting larger in size without bound.
For $|R|<1$|R|<1 the values will converge as $n$n increases, each term getting smaller and smaller and approaching a limit of $0$0.
Practice questions
QUESTION 3
The $n$nth term of a geometric progression 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 $10$10 $T_n$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 curved line
B