Geometric Sequences Revision Notes for Leaving Cert Mathematics

Geometric Sequences

Introduction

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A geometric sequence is a sequence where each term after the first term is obtained by multiplying the previous term by a constant factor, called the common ratio, denoted as $r$ .

The first term of any geometric sequence is denoted as $a$ , so in general, a geometric sequence is :

a,ar,ar^2, ar^3,ar^4, ...

The general term of any arithmetic sequence is

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General term of a geometric sequence : Page 22

T_n=ar^{n-1}

To derive the common ratio $r$ of any geometric sequence, take any term and divide by the term before it.

\begin{align*} r&=\frac{T_n}{T_{n-1}} \\ \end{align*}

Example

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Find the general term of the geometric sequence $1,5,25,...$

\begin{align*} r=\frac{25}{5}=5 \end{align*}

\begin{align*} T_n&=(1)(5)^{n-1} \\ &= 5^{n-1} \end{align*}

Example

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$x+9,x-6$ and $4$ are the first three terms in a geometric sequence, find $x$ .

We know that any term divided by the term before gives us the common ratio, given three terms we can form an equation :

\begin{align*} r&=\frac{T_2}{T_1}=\frac{T_3}{T_2} \\ \\ &=\frac{x-6}{x+9}=\frac{4}{x-6} \end{align*}

Cross multiply

\begin{align*} (x-6)(x-6)&=4(x+9) \\ x^2-12x+36 &= 4x+36 \\ x^2-16x &= 0 \\ x(x-16) &= 0 \end{align*}

Apply zero-product property :

\begin{align*} x=0, x=16 \\ \end{align*}

Test both values for $x$ ,

$x=16$

\begin{align*} (16)+9&,(16)-6,4 =25,10,4 \end{align*}

Test if the common ratio is constant for each division :

\begin{align*} \frac{4}{10}=\frac{10}{25}=r \end{align*}

$x=0$

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\begin{align*} (0)+9&,(0)-6,4 = 9,-6,4 \end{align*}

Test if the common ratio is constant for each division :

\begin{align*} \frac{4}{-6}=\frac{-6}{9}=r \end{align*}

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Example

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The amount of substance remaining in a solution reduces exponentially over time. An experiment measures the percentage of the substance remaining in the solution. The percentage is measured at the same time each day. The collected over the first $4$ days are given in the table below. Based on the data in the table, estimate which is the first day on which the percentage of the substance in the solution will be less than $0.01\%$ .

Day	1	2	3	4
Percentage of substance (%)	$95$	$42.75$	$19.2375$	$8.6569$

Determine the common ration of sequence

\begin{align*} \frac{T_n}{T_{n-1}}&=\frac{42.75}{95}=\frac{9}{20} \\ \end{align*}

We are looking for the day ( $n$ ) when the solution drops strictly less than $0.01\%$ .

\begin{align*} ar^{n-1} &< 0.01 & \text{} \\\\ (95)(\tfrac{9}{20})^{n-1} &< 0.01 & \text{\footnotesize\textcolor{gray}{(\( a = 95 \), \( r = \tfrac{9}{20} \))}} \\\\ (\tfrac{9}{20})^{n-1} &< \tfrac{1}{9500} & \text{\footnotesize\textcolor{gray}{(÷ 95)}} \\\\ \log_{\tfrac{9}{20}}(\tfrac{1}{9500}) &< n-1 & \text{\footnotesize\textcolor{gray}{(\( a^x<y \iff \log_a y < x \))}} \\\\ \log_{\tfrac{9}{20}}(\tfrac{1}{9500}) + 1 &< n & \text{\footnotesize\textcolor{gray}{(\(+n\))}} \\\\ 12.47020514 &< n & \text{\footnotesize\textcolor{gray}{(Evaluate)}} \end{align*}

On the day 13th day the solution will be less than $0.01\%$ .

It might seem intuitive to round down to $12$ , but observe that $n$ has to be strictly greater than $12.470...$ , and since $n$ needs to be a natural number, the answer is $13$ .

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Proving a Sequence is Geometric

To prove that a sequence is geometric, show that the common ratio is constant.

\begin{align*} \frac{T_n}{T_{n-1}}=r=c \end{align*}

where $c$ is a constant.

Example

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Determine if the sequence is geometric given the $n$ th term : $T_n=15n^2$

\begin{align*} T_n&=15n^2 \\ T_{n-1}&=15(n-1)^2 \end{align*}

Determine if the common ratio is constant :

\begin{align*} \frac{T_n}{T_{n-1}}&=\frac{15n^2}{15(n-1)^2} \\ \\ &=\frac{n^2}{(n-1)^2} \end{align*}

not a constant, so the sequence is not geometric.

Example

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Determine if the sequence is geometric given the $n$ th term : $T_n=x^ny^{n-1}$ , where $x,y$ are constant.

\begin{align*} T_n&=x^ny^{n-1} \\ T_{n-1}&=x^{(n-1)}y^{(n-1)-1} \\ &= x^{n-1}y^{n-2} \end{align*}

Determine if the common ratio is constant :

\begin{align*} \frac{T_n}{T_{n-1}}&=\frac{x^ny^{n-1}}{ x^{n-1}y^{n-2}} \\ \\ &=\frac{x^n}{ x^{n-1}}\cdot \frac{y^{n-1}}{y^{n-2}} \\ \end{align*}

Both fraction have the same base, you can apply indices rules to simplify :

\begin{align*} \\ &=(x^{(n)-(n-1)})\cdot (y^{(n-1)-(n-2)}) \\ &=(x^{n-n+1})\cdot (y^{n-1-n+2}) \\ &=(x^{1})\cdot (y^{1}) \\ &= xy \end{align*}

$xy$ is a constant, so the sequence is geometric.

Geometric Sequences (Leaving Cert Mathematics): Revision Notes