topic badge

3.06 Geometric progressions -calculator assumed

Lesson

 

We began learning about Geometric sequences in our Geometric progressions - calculator free lesson. It is important to practice these types of questions both with and without the use of a calculator. 

We can use a CAS calculator to:

  • list the terms of a sequence from the recursive rule or the explicit form. This can help if we need to find later terms in the sequence as listing or calculating may be very time consuming.

  • graph the terms of a sequence from the recursive rule and from the explicit form. This can help to see long term patterns and trends.
 
Forms of geometric sequences

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=rtn1, where $t_1=a$t1=a or alternatively $t_{n+1}=rt_n$tn+1=rtn, 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=arn1

 

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: $t_n=ar^{n-1}$tn=arn1

Do: Remember, we have our calculators available now to help with any complex calculations. 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)n1.

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

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 $t_n$tn 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: $t_n=5\left(4\right)^{n-1}$tn=5(4)n1, substituting $t_n=327680$tn=327680, we get:

$327680$327680 $=$= $5\left(4\right)^{n-1}$5(4)n1

Use the solve feature of your calculator*

$n$n $=$= $9$9

 

*See examples with instructions specific to calculator brands below for further detail.

Hence, the $9$9th term in the sequence is $327680$327680.

Example 3

If a geometric sequence has $t_3=12$t3=12 and $t_{14}=24576$t14=24576, 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=arn1.

Do:

$t_3$t3: $ar^2=12$ar2=12 $.....\left(1\right)$.....(1)

and

$t_{14}$t14: $ar^{13}=24576$ar13=24576 $.....\left(2\right)$.....(2)

 

If we now divide equation $\left(2\right)$(2) by equation $\left(1\right)$(1), and using our calculators, we obtain the following:

$\frac{ar^{13}}{ar^2}$ar13ar2 $=$= $\frac{24576}{12}$2457612
$r^{11}$r11 $=$= $2048$2048
$\therefore r$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=2tn1,t1=3

Select the brand of calculator you use below to work through an example of using a calculator for geometric sequences and then try the practice questions.

 

Casio Classpad

How to use the CASIO Classpad to complete the following tasks regarding geometic sequences.

  1. Generate the first $10$10 terms of the sequence with the recursive relationship: $t_n=0.5t_{n-1},t_1=128$tn=0.5tn1,t1=128.

  2. Generate the first $10$10 terms of the sequence with the explicit relationship: $t_n=2(3)^{n-1}$tn=2(3)n1.

  3. Find the $20$20th term of the sequence $t_n=1.5t_{n-1}$tn=1.5tn1, $t_1=4$t1=4.

    Give the answer to one decimal place.

  4. For the geometric sequence $3,12,48,\dots$3,12,48,. Use your calculator to find $n$n if the $n$nth term is $196608$196608.

 

TI Nspire

How to use the TI Nspire to complete the following tasks regarding geometric sequences.

  1. Generate the first $10$10 terms of the sequence with the recursive relationship: $t_n=0.5t_{n-1},t_1=128$tn=0.5tn1,t1=128

  2. Generate the first $10$10 terms of the sequence with the explicit relationship: $t_n=2\left(3\right)^{n-1}$tn=2(3)n1.

  3. Find the $20$20th term of the sequence $t_n=1.5t_{n-1}$tn=1.5tn1, $t_1=4$t1=4.

    Give the answer to one decimal place.

  4. For the geometric sequence $3,12,48,\dots$3,12,48, Use your calculator to find $n$n if the $n$nth term is $196608$196608.

 

Practice questions

Question 1

Consider the first four terms in the following geometric sequence.

$-9,-10.8,-12.96,-15.552,\ldots$9,10.8,12.96,15.552,

  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 2

In a geometric progression, $T_4=-192$T4=192 and $T_7=12288$T7=12288.

  1. Find the value of $r$r, the common ratio in the sequence.

  2. Find $a$a, the first term in the progression.

  3. Find an expression for $T_n$Tn, the general $n$nth term.

Question 3

The $n$nth term of a geometric progression is given by the equation $T_n=25\times\left(\frac{1}{5}\right)^{n-1}$Tn=25×(15)n1.

  1. Complete the table of values:

    $n$n $1$1 $2$2 $3$3 $4$4 $10$10
    $T_n$Tn $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
  2. What is the common ratio between consecutive terms?

Question 4

Consider the following sequence.

$54,18,6,2,\ldots$54,18,6,2,

  1. If the sequence starts from $n=1$n=1, plot the first four terms below.

    Loading Graph...

  2. The relationship depicted by this graph is:

    linear

    A

    exponential

    B

    neither

    C
  3. Write the recursive rule for $T_n$Tn in terms of $T_{n-1}$Tn1, including the initial term $T_1$T1.

    Enter both parts on the same line, separated by a comma.

  4. What is the sum of the first $10$10 terms?

    Round your answer to the nearest whole number.

 

Outcomes

3.2.5

use recursion to generate a geometric sequence

3.2.6

display the terms of a geometric sequence in both tabular and graphical form and demonstrate that geometric sequences can be used to model exponential growth and decay in discrete situations

3.2.7

deduce a rule for the nth term of a particular geometric sequence from the pattern of the terms in the sequence, and use this rule to make predictions

What is Mathspace

About Mathspace