Geometric Progression Definition

A sequence of numbers \({{a}_{1}},~{{a}_{2}},{{a}_{3}},\ldots \); It is called a geometric progression if, starting from the second, each element is obtained from the multiplication of the previous one by a number \(r\ne 0\), that is, if:
\({{a}_{n+1}}={{a}_{n}}r\)
Where:
- The number \(r\) is called the ratio of the geometric progression.
- The element \({{a}_{1}}\) is called the first element of the arithmetic progression.

The elements of the geometric progression can be expressed in terms of the first element and its ratio, that is:
\({{a}_{1}},{{a}_{1}}r,{{a}_{1}}{{r}^{2}},{{a}_{1} }{{r}^{3}}\)

They are the first four elements of the arithmetic progression; in general, the \(k-\)th element is expressed as follows:
\({{a}_{k}}={{a}_{1}}{{r}^{k-1}}\)

When \({{a}_{1}}\ne 0,~\)of the previous expression we obtain:

\(\frac{{{a}_{k}}}{{{a}_{l}}}=\frac{{{a}_{1}}{{r}^{k-1}} }{{{a}_{1}}{{r}^{l-1}}}\)

\(\frac{{{a}_{k}}}{{{a}_{l}}}={{r}^{k-l}}\)

The above expression is equivalent to:

\({{a}_{k}}={{a}_{l}}{{r}^{k-l}}\)

Example/exercise 1. Find the difference of the arithmetic progression: \(2,6,18,54,\ldots \) ​​and find the elements \({{a}_{20}},~{{a}_{91}}\)

Solution

Since \(\frac{6}{2}=\frac{18}{6}=\frac{54}{18}=3\) we can conclude that the ratio is:

\(r=3\)

\({{a}_{20}}=2\left( {{3}^{20-1}} \right)=2{{\left( 3 \right)}^{19}}\)

\({{a}_{91}}=2\left( {{3}^{91-1}} \right)=2{{\left( 3 \right)}^{90}}\)

Example/exercise 2. In an arithmetic progression we have: \({{a}_{17}}=20~\)y \({{a}_{20}}=-1280\), determine the ratio of the geometric progression and write the first 5 elements.

Solution

Wearing

\(\frac{{{a}_{k}}}{{{a}_{l}}}={{r}^{k-l}}\)

\(\frac{{{y}_{20}}}{{{y}_{17}}}={{r}^{20-17}}\)

\(\frac{-1280}{20}={{r}^{3}}\)

\(-64={{r}^{3}}\)

\(\sqrt[3]{-64}=\sqrt[3]{{{r}^{3}}}\)

\(-4=r\)

To find the first 5 elements of the arithmetic progression; we will calculate \({{a}_{1}}\):

\({{a}_{k}}={{a}_{1}}{{r}^{k-1}}\)

\({{a}_{17}}={{a}_{1}}{{\left( r \right)}^{17-1}}\)

\(20={{a}_{1}}{{\left( -4 \right)}^{16}}\)

\(\frac{20}{{{4}^{16}}}={{a}_{1}}\)

\(\frac{5\left( 4 \right)}{{{4}^{16}}}={{a}_{1}}\)

\(\frac{5}{{{4}^{15}}}={{a}_{1}}\)

The first 5 elements of the geometric progression are:

\(\frac{5}{{{4}^{15}}},~\frac{5}{{{4}^{15}}}\left( -4 \right),\frac{5} {{{4}^{15}}}{{\left( -4 \right)}^{2}},\frac{5}{{{4}^{15}}}{{\left( -4 \right)}^{3}},\frac{5}{{ {4}^{15}}}{{\left( -4 \right)}^{4}}\)

\(\frac{5}{{{4}^{15}}},-~\frac{5}{{{4}^{14}}},\frac{5}{{{4}^{ 13}}},-\frac{5}{{{4}^{12}}},\frac{5}{{{4}^{11}}}\)

Example/exercise 3. A thin glass absorbs 2% of the sunlight that passes through it.

to. What percentage of light will pass through 10 of those thin glasses?

b. What percentage of light will pass through 20 of those thin glasses?

c. Determine the percentage of light that passes through \(n\) thin glasses with the same characteristics, placed consecutively.

Solution

We will represent with 1 the total light; by absorbing 2% of the light, then 98% of the light goes through the glass.

We will represent with \({{a}_{n}}\) the percentage of light that passes through the glass \(n\) .

\({{a}_{1}}=0.98,~{{a}_{2}}=0.98\left( 0.98 \right),~{{a}_{3}}={{\left( 0.98 \right)}^{2}}\left( 0.98 \right),\)

In general \({{a}_{n}}={{\left( 0.98 \right)}^{n}}\)

to. \({{a}_{10}}={{\left( 0.98 \right)}^{10}}=0.81707\); which tells us that after glass 10 passes 81.707% of light

b. \({{a}_{20}}={{\left( 0.98 \right)}^{20}}=~0.66761\); which tells us that after glass 20 passes 66.761%

The sum of the first \(n\) elements of a geometric progression

Given the geometric progression \({{a}_{1}},{{a}_{1}}r,{{a}_{1}}{{r}^{2}},{{a} 1}}{{r}^{3}}\)….

When \(r\ne 1\) is the sum of the first \(n\) elements, the sum:

\({{S}_{n}}={{a}_{1}}+{{a}_{1}}r+{{a}_{1}}{{r}^{2}} +{{a}_{1}}{{r}^{3}}+\ldots +{{a}_{1}}{{r}^{n-1}}\)

It can be calculated with

\({{S}_{n}}={{a}_{1}}\frac{\left( 1-{{r}^{n}} \right)}{1-r},~r \n1\)

Example/exercise 4. From example 2 calculate \({{S}_{33}}\).

Solution

In this case \({{a}_{1}}=\frac{5}{{{4}^{15}}}\) and \(r=-4\)

applying

\({{S}_{n}}={{a}_{1}}\frac{\left( 1-{{r}^{n}} \right)}{1-r}\)

\({{S}_{22}}=\frac{5}{{{4}^{15}}}\frac{1-{{\left( -4 \right)}^{22}}} {1-\left( -4 \right)}\)

\({{S}_{22}}=\frac{5}{{{4}^{15}}}\frac{1-{{\left( -4 \right)}^{22}}} {5}\)

\({{S}_{22}}=\frac{1-{{\left( 4 \right)}^{22}}}{{{4}^{15}}}\)

\({{S}_{22}}=\frac{1}{{{4}^{15}}}-\frac{{{\left( 4 \right)}^{22}}}{{ {4}^{15}}}\)

\({{S}_{22}}=\frac{1}{{{4}^{15}}}-{{4}^{7}}\)

Example/exercise 5. Suppose a person uploads a photo of her pet and shares it with 3 of her friends on an internet social network, and in one hour each of them, shares the photograph with three other people and then the latter, in one more hour, each of them shares the photograph with 3 others people; And so it goes on; each person who receives the photograph shares it with 3 other people within an hour. In 15 hours, how many people already have the photograph?

Solution

The following table shows the first calculations
Time People who receive the photograph People who have the photograph
1 3 1+3=4
2 (3)(3)=32=9 4+9=13
3 32(3)= 33=27 13+27=40

The number of people who receive the photograph in hour \(n\) is equal to: \({{3}^{n}}\)

The number of people who already have the photograph in the hour is equal to:

\(3+{{3}^{2}}+{{3}^{3}}+\ldots +{{3}^{n}}\)

applying

\({{S}_{n}}={{a}_{1}}\frac{\left( 1-{{r}^{n}} \right)}{1-r}\)

With \({{a}_{1}}=3,\) \(r=3\) and \(n=15\)

Whereby:

\({{S}_{n}}=\frac{\left( 1-{{3}^{15}} \right)}{1-3}=7174453\)

geometric means

Given two numbers \(a~\) and \(b,\) the numbers \({{a}_{2}},{{a}_{3}},\ldots ,{{a}_{k +1}}\) are called \(k\) geometric means of the numbers \(a~\) and \(b\); if the sequence \(a,{{a}_{2}},{{a}_{3}},\ldots ,{{a}_{k+1}},b\) is a geometric progression.

To know the values ​​of \(k\) geometric means of the numbers \(a~\) and \(b\), it is enough to know the ratio of the arithmetic progression, for this the following must be considered:

\(a={{a}_{1}},{{a}_{2}},{{a}_{3}},\ldots ,{{a}_{k+1}},{ {a}_{k+2}}=b,\)

From the above we establish the relationship:

\(b=a{{r}^{k+1}}\)

Solving for \(d\), we obtain:

\(b=a{{r}^{k+1}}\)

\(\frac{b}{a}={{r}^{k+1}}\)

\(r=\sqrt[k+1]{\frac{b}{a}}\)

Example/exercise 6. Find 2 geometric means between the numbers -15 and 1875.

Solution

When applying

\(r=\sqrt[k+1]{\frac{b}{a}}\)

with \(b=375,~a=-15\) and \(k=2~\):

\(r=\sqrt[2+1]{\frac{1875}{-15}}\)

\(r=\sqrt[3]{-125}=-5\)

The 3 geometric means are:

\(75,-375\)

Example/exercise 7. A person invested money and received interest every month for 6 months and his capital increased by 10%. Assuming the rate did not change, what was the monthly interest rate?

Solution

Let \(C\) be the invested capital; the final capital is \(1.1C\); To solve the problem we must place 5 geometric means, by applying the formula:

\(r=\sqrt[k+1]{\frac{b}{a}}\)

With \(k=5,~b=1.1C\) and \(a=C.\)

\(r=\sqrt[5+1]{\frac{1.1C}{C}}=\sqrt[6]{1.1}=1.016\)

The monthly rate received was \(1.6%\)