Sequences and Series

NZ Level 7 (NZC) Level 2 (NCEA)

Introduction to Geometric Progressions

Lesson

Recall that an ordered list of numbers, separated by commas, is called a progression or sequence and when a pattern is detectable in a progression, a generating rule can often be established that allows us to determine any term in the sequence.

Geometric progressions 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`,`a``r`,`a``r`2,`a``r`3,... so that the $n$`n`th term is given by:

$t_n=ar^{n-1}$`t``n`=`a``r``n`−1

Having a formula for the $n$`n`th 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 $t_7=12\times\left(\frac{3}{2}\right)^6$`t`7=12×(32)6 or $136.6875$136.6875.

This applet will allow you to visualise the geometric side of geometric progressions. 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?

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

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

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

If $T_n$

`T``n` is the $n$`n`th term, evaluate $\frac{T_2}{T_1}$`T`2`T`1.Evaluate $\frac{T_3}{T_2}$

`T`3`T`2Evaluate $\frac{T_4}{T_3}$

`T`4`T`3Hence find the value of $T_5$

`T`5.

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

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

Use arithmetic and geometric sequences and series

Apply sequences and series in solving problems