Understand geometric sequences and series: concepts, formulas, and examples. This article explains them in detail with practical examples.
Geometric sequences and series are fundamental concepts in mathematics that have widespread applications in various fields, including economics, physics, and engineering. After previously studying arithmetic sequences and series, we will now continue with geometric sequences and series. We will discuss geometric sequences and series in detail, their associated formulas, and provide several examples to better understand these concepts.
What is a Geometric Series?
A geometric sequence is a series of numbers or terms formed in such a way that each term is obtained by multiplying the previous term by a fixed number called the ratio symbolized by the letter ($r$).
For example, the row is $$ U_1, U_2, U_3, U_4, U_5, U_6, U_7, …. $$ The way to calculate the ratio ($r$) is $$ r = \frac{u_2}{u_1} = \frac{u_3}{u_2} = \frac{u_4}{u_3} = … = \frac{u_n}{u_{n-1}}$$
Example
$2, 4, 8, 16, 32,…$ (Geometric Sequence with $r=2$)
$27, 9, 3, 1, \dfrac{1}{3}, \dfrac{1}{9}…$ (Geometric Line with $r=\dfrac{1}{3}$)
Formula for the th Term of a Geometric Sequence
If the first term = $a$ and the ratio = $r$, then in general the geometric sequence those are: $$\begin{matrix}U_1, & U_2, & U_3, & U_4, & & U_n \\ a, & ar, & ar^2, & ar^3, & \cdots, &ar^{n-1} \end{matrix} $$
So the formula for the nth term of a geometric sequence is $$U_n=ar^{n-1}$$ With :
- $U_n$ is the nth quarter
- $a$ is the first quarter
- $r$ is the geometric ratio
- $n$ is the index of the nth term
to make it easier to remember, the formula for the $n^{th}$ term can be read “arni”
Examples of Geometric sequences
Which of the following lines is a Geometric line?
- a. 1, 2, 4, 8, …..
- b. $\frac{1}{3} $, 1, 3, 9, 27, ….
- c. 1, 2, 6, 8, 16, ….
- d. 3, 4, 8, 2, 12, ….
- and. 16, 8, 4, 2, 1, ….
Solution βοΈ
A number line is called a geometric line if the ratio of two adjacent terms is equal. Let’s check each available row.- a. $\underbrace{1, 2}_{\times 2} \underbrace{, 4 }_{\times 2} \underbrace{, 8 }_{\times 2} , \cdots $
the ratio/comparison of the series above $r=\frac21=\frac42=2$ is the same, so it is a geometric series - b. $ \underbrace{\frac{1}{3}, 1}_{\times 3} \underbrace{, 3 }_{\times 3} \underbrace{, 9 }_{\times 3} \underbrace{, 27 }_{\times 3} , …. $
The ratio is the same, so it is included in the geometric series with the ratio 3. - c. $ \underbrace{1, 2}_{\times 2} \underbrace{, 6 }_{\times 3} \underbrace{, 8 }_{\times \frac{4}{3}} \underbrace{, 16 }_{\times 2} , …. $
The comparison is not the same, so it does not include the Geometry line. - d. $ \underbrace{3, 4}_{\times \frac{4}{3}} \underbrace{, 8 }_{\times 2} \underbrace{, 2 }_{\times \frac{1}{4}} \underbrace{, 12 }_{\times 6} , …. $
The comparison is not the same, so it does not include geometric lines. - e. $ \underbrace{16, 8}_{\times \frac{1}{2}} \underbrace{, 4 }_{\times \frac{1}{2}} \underbrace{, 2 }_{\times \frac{1}{2}} \underbrace{, 1 }_{\times \frac{1}{2}} , …. $
The ratio is the same, so it is a geometric sequence with the ratio $ \frac{1}{2}$.
How to find the ratio: $ r =\frac{u_2}{u_1} = \frac{8}{16} = \frac{1}{2} $ or $ r =\frac{u_3}{u_2} = \frac{4}{8} = \frac{1}{2} $ and so on.
Determine the 6th and 21st terms of the geometric sequence 3, 6, 12, 24, ….?
Solution βοΈ
- from the row obtained
$ a = 1 $ and
$ r = \frac{u_2}{u_1} = \frac{2}{1} = 2 $ - Determines the 6th quarter with $ U_n = a r^{n-1} $ $$\begin{align*} u_{6} &= a r^{6-1} \\&= 3 . 2^{5}\\&= 3\times 32= 94 \end{align*}$$
- Determines the 21st quarter with $ U_n = a r^{n-1} $ $$\begin{align*} u_{21} &= a r^{21-1} \\&= 3 . 2^{20} \end{align*}$$ So, the 6th quarter and the 21st quarter are 94 and $ 3\cdot2^{20} $.
- from the row obtained
The 3rd and 5th terms of a geometric sequence are 9 and 81, respectively, with a positive ratio. Determine the value of the 2nd term!
Solution βοΈ
known $ u_3 = 9 $ and $ u_5 = 81 $
To determine the value of a term in a sequence, we need the value of $ a $ and its ratio ($ r $) by explaining the known terms.
The formula for the $nth term : u_n = ar^{n-1} $
$ u_5 = ar^{5-1} = ar^4 \rightarrow a r^4 = 81 $ …. pers(i)
$ u_3 = ar^{3-1} = ar^2 \rightarrow a r^2 = 9 $ …. pers(ii)Determine the value of $ a $ and $ r $ by dividing pers(i) and pers(ii) $$ \begin{align*}\frac{U_5}{U_3}&=\frac{81}{9}\\ \frac{ar^4}{ar^2}&=\frac{81}{9}\\ r^2&=9\\ r&=\pm 3 \end{align*} $$ Since the ratio value is positive, then $ r = 3 $ is satisfied.
Substitute r=3 into Pers(ii): $$\begin{align*}a r^2 = 9 \\ \rightarrow a 3^2 = 9 \\ \rightarrow a = \frac{9}{9} = 1 \end{align*}$$
Determines the 2nd quarter $$ u_{2} = ar^{2-1} = 1.3^1 = 3 $$
So, the 2nd term is 3.
Practical Logic
If two terms in a geometric sequence are known, then the ratio of the geometric sequence can be determined by: $$r=\sqrt[p-q]{\frac{U_p}{U_q}}$$
Example
If $π_3 = 16$ and $π_7 = 256$ are known, determine the 9th quarter of the row!
Practical logic steps:
The 9th term is the 7th term multiplied by the power ratio of 2. $$\begin{align*} r&=\sqrt[p-q]{\frac{U_p}{U_q}} \\ &=\sqrt[7-3]{\frac{256}{16}} \\ &=\sqrt[4]{16} \\ &=2 \end{align*}$$ So the 9th term is $$\begin{align*} U_9&=U_7\times r^2\\ &=256\times 2^2\\ &=256\times 4\\ &=1024 \end{align*}$$
What is a Geometric Series?
A geometric series is the sum of the terms in a geometric sequence. The sum in question is the sum of a finite number of terms (the first n terms). The symbol used is $S_n$, which means the sum of the first n terms.
If we have a geometric sequence with the first term ($a$), geometric ratio ($r$), then:
$ S_1 = U_1 $ (sum of 1 first quarter)
$ S_2 = U_1 + U_2 $ (sum of the first 2 quarters)
$ S_3 = U_1 + U_2 + U_3 $ (sum of the first 3 quarters)
$ S_4 = U_1 + U_2 + U_3 + U_4 $ (sum of the first 4 quarters)
and so on.
$S_n=U_1 +U_2+U_3+U_4+\cdots +U_n$ (number of n first terms)
What if there are a lot of terms to be added, then we will use the formula directly. Here is the formula for the sum of the first $ n $ terms of a geometric series.
Formula for a Geometric Series with the First N Terms
Sum of the first $ n $ terms: $$ s_n = \frac{a(r^n - 1)}{r-1} \text{ for } |r| > 1 \\ s_n = \frac{a(1 - r^n)}{1-r} \text{ for } |r| < 1 $$
Where:
- $S_n$ is the sum of the geometric series
- $a$ is the first quarter
- $r$ is the geometric ratio
- $n$ is the number of terms in the series
Example of Geometric Series Problems
The ratio of the sequence $\dfrac{16}{27},\ \dfrac{8}{9},\ \dfrac{4}{3},\ 2,\ \cdots $ is…
Solution βοΈ
From the sequence $\dfrac{16}{27},\ \dfrac{8}{9},\ \dfrac{4}{3},\ 2,\ \cdots $ we can obtain the ratio, namely: $$\begin{align*} r &= \dfrac{U_{n}}{U_{n-1}} \\ &= \dfrac{U_{4}}{U_{3}} = \dfrac{2}{\frac{4}{3}} \\ &= 2 \cdot \dfrac{3}{4} = \dfrac{3}{2} \end{align*}$$So, the ratio is $\dfrac{3}{2}$
Known $9,\ 3,\ 1,\ \dfrac{1}{3},\ \cdots$ The $7$ term is…
Solution βοΈ
From the row $\dfrac{16}{27},\ \dfrac{8}{9},\ \dfrac{4}{3},\ 2,\ \cdots $ we can get: $$\begin{align*} r\ &= \dfrac{U_{n}}{U_{n-1}} \\ &= \dfrac{U_{3}}{U_{2}} = \dfrac{1}{3} \\ \hline U_{n}\ &= ar^{n-1} \\ U_{7}\ &= 9 \cdot \left( \dfrac{1}{3} \right)^{7-1} \\ &= 9 \cdot \left( \dfrac{1}{3} \right)^{6} = 9 \cdot \dfrac{1}{3^{6}} \\ &= \dfrac{9}{3^{6}} = \dfrac{1}{3^{4}} = \dfrac{1}{81} \end{align*}$$Given $3^{4},\ 3^{6},\ 3^{8},\ 3^{10},\ \cdots $ The $12th term is…
Solution βοΈ
From the row $3^{4},\ 3^{6},\ 3^{8},\ 3^{10},\ \cdots $ we can get: $$\begin{align*} r\ &= \dfrac{U_{n}}{U_{n-1}} \\ &= \dfrac{U_{2}}{U_{1}} = \dfrac{3^{6}}{3^{4}}=3^{6-4}=3^{2} \\ \hline U_{n}\ &= ar^{n-1} \\ U_{12}\ &= 3^{4} \cdot \left( 3^{2} \right)^{12-1} \\ &= 3^{4} \cdot \left( 3^{2} \right)^{11} \\ &= 3^{4} \cdot 3^{22} = 3^{26} \end{align*}$$
That’s the discussion about geometric sequences and series, next we will learn about infinite geometric series.