# 7.04 Geometric sequences

Lesson

## Introduction to geometric sequences

Geometric sequences all start with a first term and then either increase or decrease by a constant factor called the common ratio. We denote the first term by the letter $a$a 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 $a=4$a=4 and $r=2$r=2. The sequence $100,-50,25,-12.5,\dots$100,50,25,12.5, is geometric with $a=100$a=100 and $r=-\frac{1}{2}$r=12.

The size and sign of the geometric ratio plays an important role in how the sequence grows. Geometric ratios greater than $r=1$r=1 will cause terms in the sequence to get larger. Ratios between $0$0 and $1$1 will cause the terms to get smaller. Negative ratios will cause sign changes across consecutive terms, just like the last example mentioned in the previous paragraph.

A great example of a geometric sequence concerns animal cell division, where a single cell divides into two ‘daughter’ cells through biological processes known as mitosis and cytokinesis. The daughter cells divide again and again, each time creating two new daughter cells so that the number of daughters in each new generation form the geometric sequence $2,4,8,16,32,\dots$2,4,8,16,32,

Clearly the first term is $a=2$a=2 and the common ratio $r=2$r=2.

Another example concerns radioactive decay – a process whereby half of a certain amount of radioactive material disappears in a specified time known as a “half-life”. If for example the half-life of a certain radioactive material is $10$10 years, then a quantity of $120$120 grams of the material reduces to $60$60 grams in the first $10$10 years, then to $30$30 grams in the next $10$10 years, and then to $15$15 grams in the next $10$10 years, and so on. Over $50$50 years, we see the quantity reduce according to the geometric sequence $120,60,30,15,7.5$120,60,30,15,7.5.

In general terms, every geometric sequence begins as $a,ar,ar^2,ar^3,...$a,ar,ar2,ar3,... so that the $n$nth term is given by:

$t_n=ar^{n-1}$tn=arn1

Having a formula for the $n$nth term allows us to quickly calculate the value of any term. For example, in the geometric sequence beginning $12,18,27,\dots$12,18,27, we might want to know what the value of the $7$7th term is. Using the formula, and noticing that $a=12$a=12 and $r=\frac{3}{2}$r=32 we can show that $t_7=12\times\left(\frac{3}{2}\right)^6$t7=12×(32)6 or $136.6875$136.6875

This applet will allow you to visualize the geometric side of geometric sequences. Play with the values of $a$a and $r$r. What happens when $r$r is less than $1$1? What about when $a$a is negative? What else do you notice?

#### Practice questions

##### QUESTION 1

Study the pattern for the following sequence, and write down the next two terms.

1. $3$3, $15$15, $75$75, $\editable{}$, $\editable{}$

##### QUESTION 2

Consider the first four terms in this geometric sequence: $-8$8, $-16$16, $-32$32, $-64$64

1. If $T_n$Tn is the $n$nth term, evaluate $\frac{T_2}{T_1}$T2T1.

2. Evaluate $\frac{T_3}{T_2}$T3T2

3. Evaluate $\frac{T_4}{T_3}$T4T3

4. Hence find the value of $T_5$T5.

##### QUESTION 3

Some of the terms in the following geometric progression are missing. Use the common ratio to find these terms.

1. $\editable{}$, $\editable{}$, $\frac{3}{25}$325, $-\frac{3}{125}$3125, $\editable{}$

## Writing the recursive formula

It is generally known as a recurrence relationship and for geometric sequences, the recurrence formula is given by:

$t_{n+1}=r\times t_n,t_1=a$tn+1=r×tn,t1=a

The equation states that the $\left(n+1\right)$(n+1)th term is $r$r times the $n$nth term with the first term equal to $a$a.

Thus the second term, $t_2$t2 is $r$r times the first term $t_1$t1, or $ar$ar

The third term $t_3$t3 is $r$r times $t_2$t2 or  $ar^2$ar2

The fourth term $t_4$t4 is $r$r times $t_3$t3, or $ar^3$ar3, and so on.

Hence, step by step, the sequence is revealed as $a$a, $ar$ar$ar^2$ar2$ar^3...$ar3... , $ar^{n-1}$arn1

Take for example the recursive relationship given as $t_{n+1}=\frac{t_n}{2}$tn+1=tn2 with $t_1=64$t1=64. From this formula, we see that

$t_2=\frac{t_1}{2}=32$t2=t12=32 and

$t_3=\frac{t_2}{2}=16$t3=t22=16, and so on.

This means that the sequence becomes $64,32,16,8,...$64,32,16,8,... which is clearly geometric with $a=64$a=64 and $r=\frac{1}{2}$r=12

Consider the recurrence relationship given as $t_{n+1}=3t_n+2$tn+1=3tn+2 with $t_1=5$t1=5.

To test whether or not the relationship is geometric, we can evaluate the first three terms.

$t_1=5$t1=5,

$t_2=3\times5+2=17$t2=3×5+2=17

$t_3=3\times17+2=53$t3=3×17+2=53.

Thus, the sequence begins $5,17,53,...$5,17,53,... and we immediately see that $\frac{53}{17}$5317 is not the same fraction as $\frac{17}{5}$175, and thus the recursive relationship is not geometric. In fact the only way the relationship given by $t_{n+1}=rt_n+k$tn+1=rtn+k is geometric is when the constant term $k$k is zero.

#### Practice questions

##### Question 4

Consider the first-order recurrence relationship defined by $T_n=2T_{n-1},T_1=2$Tn=2Tn1,T1=2.

1. Determine the next three terms of the sequence from $T_2$T2 to $T_4$T4.

Write all three terms on the same line, separated by commas.

2. Plot the first four terms on the graph below.

3. Is the sequence generated from this definition arithmetic or geometric?

Arithmetic

A

Geometric

B

Neither

C

Arithmetic

A

Geometric

B

Neither

C

##### Question 5

The first term of a geometric sequence is $5$5. The third term is $80$80.

1. Solve for the possible values of the common ratio, $r$r, of this sequence.

2. State the recursive rule, $T_n$Tn, that defines the sequence with a positive common ratio.

Write both parts of the relationship on the same line, separated by a comma.

3. State the recursive rule, $T_n$Tn, that defines the sequence with a negative common ratio.

Write both parts of the relationship on the same line, separated by a comma.

The average rate of depreciation of the value of a Ferrari is $14%$14% per year. A new Ferrari is bought for $\$90000$$90000. 1. What is the car worth after 11 year? 2. What is the car worth after 33 years? 3. Write a recursive rule for V_nVn, defining the value of the car after nn years. Write both parts of the rule on the same line, separated by a comma. ## Solving for terms Sometimes we are asked to find the common ratio of a certain geometric sequence, and then use this to find a group of terms or simply a specific term. We need to remember that the nnth term of a GP is given by: t_n=ar^{n-1}tn=arn1 Note that there are two variables in the formula – the first term aa and the common ratio rr and any set of three consecutive terms will allow us to evaluate rr Take for example the geometric sequence beginning 3,12,48,...3,12,48,... . The common ratio is simply the ratio \frac{t_2}{t_1}=\frac{t_3}{t_2}=4t2t1=t3t2=4. This means that we can write down a formula for the nnth term as  t_ntn​ == ar^{n-1}arn−1 == 3\left(4\right)^{n-1}3(4)n−1 and this in turn allows us to determine any term or sequence of terms we like. For example, we see that the 55th term is given by t_5=3\times4^4=768t5=3×44=768 As another example, we might wonder whether the sequence \sqrt{3},6,12\sqrt{3}3,6,123 is geometric. It may not be immediately obvious, but we can show that the numbers are in geometric sequence in two ways. In the first method, we note that both  \frac{6}{\sqrt{3}}6√3​ == \frac{6}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}6√3​×√3√3​ == 2\sqrt{3}2√3 and that \frac{12\sqrt{3}}{6}=2\sqrt{3}1236=23, and so it is immediately geometric with r=2\sqrt{3}r=23. In the second method we note that in any geometric sequence, \frac{t_2}{t_1}=\frac{t_3}{t_2}t2t1=t3t2 and by "cross" multiplication, we see that \left(t_2\right)^2=t_1\times t_3(t2)2=t1×t3 or that the middle term is always the square root of the product of terms on each side of it. The middle term is defined to be the geometric mean of the outer two terms so that t_2=\sqrt{t_1\times t_3}t2=t1×t3 . In a final example consider the three terms 2,x,322,x,32 where the middle term is unknown. If this sequence is geometric then we can find xx, for we know that \frac{x}{2}=\frac{32}{x}x2=32x and so x^2=64x2=64 and this means that there are two possible solutions for xx as x=8x=8 or x=-8x=8. The two sequences are thus 2,8,322,8,32 or 2,-8,322,8,32. #### Practice questions ##### QUESTION 7 Study the pattern for the following sequence. -99,, 3.63.6,, -1.441.44,, 0.5760.576 ... 1. State the common ratio between the terms. ##### QUESTION 8 Study the pattern for the following sequence, and write down the next two terms. 1. 33, 1515, 7575, \editable{}, \editable{} ##### QUESTION 9 Study the pattern for the following sequence, and write down the next two terms. 1. 1212, -4848, 192192, \editable{}, \editable{} ## Graphs and tables of geometric sequences Having a formula for the nnth term allows us to quickly generate a table of values for the sequence. For example in the sequence 12,18,27,\dots12,18,27, the first term is 1212 and the common ratio is 1.51.5 and so the general term is given by the formula t_n=12\times\left(1.5\right)^{n-1}tn=12×(1.5)n1 . By substituting for nn appropriately and using a scientific calculator, we can quickly generate the following table of the first 77 terms of the sequence: n 1 2 3 4 5 6 tn 12 18 27 40.5 60.75 91.125 Perhaps more interestingly though is the different types of graphs that geometric sequences correspond to. Usually the graphs are not linear like arithmetic sequences. Graphs of geometric sequences are best known as rising or reducing graphs where the rate of rising continually changes, resulting in a curved growth or decay path. This happens whenever the common ratio is positive like the geometric sequence depicted in the above table. However, when the common ratio is negative, the values of successive terms flip their sign so that the graph is depicted as either a growing or diminishing zig-zag path. Think, for example, about the geometric sequence that that is identical to the one in the table, but has a negative ratio r=-1.5r=1.5 so that its nnth term is given by t_n=12\times\left(-1.5\right)^{n-1}tn=12×(1.5)n1 . The new table becomes: n 1 2 3 4 5 6 tn 12 -18 27 -40.5 60.75 -91.125 Checking, for n=1n=1, we have t_1=12\times\left(-1.5\right)^{1-1}=12t1=12×(1.5)11=12 and for n=2n=2 we have t_2=12\times\left(-1.5\right)^{2-1}=-18t2=12×(1.5)21=18 so 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 sequence. Note also that had the absolute value of the ratios of both geometric sequences been less than 1, then the the absolute value of the terms in both sequences would be reducing in size. #### Practice questions ##### Question 10 The nnth term of a geometric progression is given by the equation T_n=2\times3^{n-1}Tn=2×3n1. 1. Complete the table of values:  nn T_nTn​ 11 22 33 44 1010 \editable{} \editable{} \editable{} \editable{} \editable{} 2. What is the common ratio between consecutive terms? 3. Plot the points in the table that correspond to n=1n=1, n=2n=2, n=3n=3 and n=4n=4. Loading Graph... 4. If the plots on the graph were joined they would form: a straight line A a curved line B a straight line A a curved line B ##### QUESTION 11 On Mercury the equation d=1.5t^2d=1.5t2 can be used to approximate the distance in meters, dd, that an object falls in tt seconds, if air resistance is ignored. 1. Complete the table. Do not round any values.  time distance 00 22 44 66 \editable{} \editable{} \editable{} \editable{} 2. Graph the function d=1.5t^2d=1.5t2. Loading Graph... 3. Use the equation or otherwise to determine the number of seconds, tt, that it would take an object to fall 5.65.6m. Give the value of tt to the nearest second. ##### QUESTION 12 A new car purchased for \38200$$38200 depreciates at a rate $r$r each year.

1. Use the table of values to determine the value of $r$r.

 years passed ($n$n) $0$0 $1$1 $2$2 value of car ($A$A) $38200$38200 $37818$37818 $37439.82$37439.82
2. Determine the rule for $A$A, the value of the car, $n$n years after it is purchased.

3. Assuming the rate of depreciation remains constant, how much can the car be sold for after $6$6 years? Give your answer to the nearest cent.

4. A new motorbike purchased for the same amount depreciates according to the model $V=38200\left(0.97^n\right)$V=38200(0.97n). Which vehicle depreciates more rapidly?

Car

A

Motorbike

B

Car

A

Motorbike

B