Applications of Geometric Sequences

Now that we have learnt how to how geometric sequences and series, we can apply them to real life scenarios.

Growth

Geometric growth is found in many real life scenarios such as population growth and the growth of an investment.

Geometric growth occurs when the common ratio is greater than 1, that is $r>1$ .

The common ratio can be found by adding the percentage increase and 1, that is $r=1+%$ increase.

Say that the percentage increase was 3%. The common ratio can be found by adding the percentage increase (of 3%) and 1.

This would be the original amount plus an extra 3%.

The reasoning for this is as follows:

Original amount + 3% of the original amount
= original amount (1+3%)
=original amount (1+0.03)
=1.03 x original amount.

Example 1

The rabbit population in a Victorian town was estimated to be 320,000 in 2012. Scientists believe that this will increase by two percent each year.

a) What will the rabbit population be in 2015? Round your answer to the nearest decimal place.

Start by finding the first term.

$a=320,000$

Find the common ratio.

$1+2%=1+0.02$
$r=1.02$

Find the value of $n$ .

2012 is the first term, so $n=1$
2013 is the second term, so $n=2$
2015 is the fourth term, so $n=4$

Use the rule.

$t_n=ar^{n-1}$
$t_4=320,000\times 1.02^{4-1}$
$t_4=320,000\times 1.0612$
$t_4=339,586.56$

Make sure that you write your answer as a full sentence.

There will be 339,587 rabbits in 2015.

b) In which year will the rabbit population reach 400,000?

Use the equation where $t_n=400,000$

$t_n=ar^{n-1}$
$400,000=320,000\times (1.02)^{n-1}$
$1.25=(1.02)^{n-1}$
$log_{10}(1.02)^{n-1}=log_{10}(1.25)$
$n-1=\dfrac {log_{10}(1.25)}{log_{10}(1.02)}$
$n-1=11.26838$
$n=12.27$

Make sure that you write your answer as a full sentence.

The rabbit population will reach 400,000 during the 12th year, 2023.

Decay

Geometric decay is found in real life instances such as depreciation and population decreases.

Geometric decay occurs when the common ratio is less than 1, that is $r<1$ .

The common ratio can be found by subtracting the percentage increase from 1, that is $r=1-%$ increase.

Say that the percentage decrease was 3%. The common ratio can be found by subtracting the percentage decrease (of 3%) from 1.

This would be the original amount minus 3% of the original amount.

The reasoning for this is as follows:

Original amount – 3% of the original amount
= original amount (1- 3%)
=original amount (1- 0.03)
=0.97 x original amount.

Example 2

A photocopier was purchased for $13,000 in 2014. The photocopier decreases in value by 20% of the previous year’s value.

a) What is an expression for the value of the photocopier, $V_n$ , after $n$ years?

We know that this is a geometric sequence as there is a 20% decrease on the previous year’s value.

Find $a$ and $r$ .

$a=13,000$
$r=1-20%$
$r=1-0.2$
$r=0.8$

Use the equation to find the expression. Take care to use $V_n$ instead of $t_n$ as this is what the question asked.

$V_n=ar^{n-1}$
$V_n=13,000\times 0.8^{n-1}$

Make sure that you write your answer as a full sentence.

The expression $V_n=13,000\times 0.8^{n-1}$ gives the value of the photocopier.

b) What is the value of the photocopier after three years?

Use the expression found in part a and substitute the value of $n$ .

$V_n=13,000\times 0.8^{n-1}$
$V_3=13,000\times 0.8^{3-1}$
$V_3=13,000\times 0.64$
$V_3=8,320$

Write the answer as a full sentence, make sure to include the dollar sign as we are referring to money.

The value of the photocopier after three years is $8,320.

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