6.04 Introduction to geometric progressions

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 $\left(\frac{u_{n+1}}{u_n}\right)$(un+1​un​​).

We denote the first term in the sequence by the letter $u_1$u1​ and the common ratio by the letter $r$r. For example, the sequence $4,8,16,32\dots$4,8,16,32… is geometric with $u_1=4$u1​=4 and $r=2$r=2. The sequence $100,-50,25,-12.5,\dots$100,−50,25,−12.5,… is geometric with $u_1=100$u1​=100 and $r=\frac{1}{2}$r=12​.

We can find an explicit formula in terms of $u_1$u1​ and $r$r, this is useful for finding the $n$nth term without listing the sequence.

Let's look at a table of a concrete example to see the pattern for the explicit formula. For the sequence $5,10,20,40,\ldots$5,10,20,40,…, we have starting term of $5$5 and a common ratio of $2$2, that is $u_1=5$u1​=5 and $r=2$r=2. A table of the sequence is show below:

$n$n	$u_n$un​	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
...
$n$n	$u_n$un​	$5\times2^{n-1}$5×2n−1

The pattern starts to become clear and we could guess that the tenth term becomes $u_{10}=5\times2^9$u10​=5×29  and the one-hundredth term $u_{100}=5\times2^{99}$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 $u_1$u1​ and common ratio $r$r has the terms given by: $u_1,u_1r,u_1r^2,u_1r^3,...$u1​,u1​r,u1​r2,u1​r3,... We see a similar pattern to our previous table and can write down the formula for the $n$nth term:

$u_{1n}=u_1r^{n-1}$u1n​=u1​rn−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 $u_1$u1​ and common ratio $r$r and substitute these into the general form: $u_n=u_1r^{n-1}$un​=u1​rn−1

Do: Dividing the second term by the first we get, $\frac{u_2}{u_1}=\frac{1}{3}$u2​u1​​=13​. Checking the ratio between the successive pairs we also get $\frac{1}{3}$13​.  So we have a geometric sequence with: $u_1=810$u1​=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 $u_1=5$u1​=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.

 

Practice question

QUESTION 1

 

Applications of geometric sequences

There are many real-life applications of geometric sequences such as compound interest, exponential growth of bacteria, exponential 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.

Worked example

Example 3

After receiving $\$500$$500 for her birthday, Hayley decides to spend $10%$10% of this money each week.

(a) Find a model for $B_n$Bn​, the amount of birthday money she has left at the start of the $n$nth week.

Think: At the beginning of the second week, Hayley will have spent $\$50$$50, and will only have $\$450$$450 or $90%$90% of the birthday money left. During the second week she will spend slightly less -  $10%$10% of her remaining $\$450$$450 or $\$45$$45, so at the beginning of the third week $\$405$$405 will remain.

Do: The sequence of birthday dollars that remain at the beginning of the first five weeks is given as $500,450,405,364.5,328.05$500,450,405,364.5,328.05 and we can see here that this sequence is geometric with the first term $B_1=500$B1​=500 and the common ratio $r=0.9$r=0.9.  

Hence, $B_n=500(0.9)^{n-1}$Bn​=500(0.9)n−1.

(b) If instead of spending $10%$10%, Hayley decides to double her savings by setting aside an additional $10%$10% in savings each week. How long before she reaches her savings goal?

Think: Firstly, she will add $10%$10% to the birthday money, so that at the beginning of the second week, the additional $\$50$$50 will take the total to $\$550$$550. At the beginning of week $3$3 she will add a further $10%$10% of this accrued $\$550$$550, so that the new total becomes $\$605$$605. She will keep adding $10%$10% of what ever is there at the beginning of each week until she reaches or exceeds $\$1000$$1000.

Do: The geometric progression for this plan becomes $500,550,605,665.5,732.05,...$500,550,605,665.5,732.05,...

This progression has $u_1=500$u1​=500 and $r=\frac{550}{500}=1.1$r=550500​=1.1, so that the $n$nth term of the progression is given by $u_n=500\left(1.1\right)^{n-1}$un​=500(1.1)n−1.

To solve for when the balance reaches her goal of $\$1000$$1000, solve for $n$n when $u_n=1000$un​=1000; 

$500\left(1.1\right)^{n-1}$500(1.1)n−1	$=$=	$1000$1000
$1.1^{n-1}$1.1n−1	$=$=	$2$2	Dividing both sides by 2
$\therefore(n-1)$∴(n−1)	$\approx$≈	$7.27$7.27	Solve by guess and check, technology or logarithms.
$n$n	$\approx$≈	$8.27$8.27

So $n$n must be greater than $8.27$8.27 for Hayley to have achieved her goal, so at the beginning of the $9$9th week her savings will exceed $\$1000$$1000.

Practice questions

QUESTION 2

QUESTION 3