*The progressions that we will see next are sequences of numbers of an ordered set of numbers formed according to a given law.*

The only requirement for a set of numbers to be a sequence is that there is a formula or law with which it is possible to obtain any number from said sequence.

Let’s quickly look at an example. If we have that u_{n} represents the nth term of a sequence, then there must be an expression for u_{n} in terms of n, this means that said nth term must be a function of n. Taking the following as a succession:

3, \ 5, \ 7, \ \dots \ , 2n +1

u_{n} is a formula that allows us to obtain any term of the sequence.

u_{n}= 2n+1

## Arithmetic progressions

*An arithmetic progression is a sequence of numbers such that each of the terms following the first are obtained by adding a fixed number to the previous number, which is called the difference of the progression.*

The example where our first term is a_{1}, our second term is a_{1} + d, the third term is a_{1} + 2d, etc… which can be represented as that the nth term is:

a_{n} = a_{1} +(n-1)d

Thanks to algebra and all the studies that exist of arithmetic progressions, we have the following theorem that says:

### Arithmetic progression theorem

If in an arithmetic sequence a_{1} is the first term, a_{n} is the nth term, d is the difference and s_{n} is the sum of the first n terms , then we have the following two independent relationships:

a_{n} = a_{1} + \left(n - 1 \right)d

s_{n} = \cfrac{n}{2} \left(a_{1} + a_{n} \right)

In simpler terms, the expression a_{n} will help us to calculate the value in a position and the expression s_{n} will help us to calculate the sum of the consecutive numbers from the first value to the nth term that we have wanted to calculate.

Now, playing with these two relations, we can obtain another formula that all it does is substitute the a_{n} of the s_{n} formula for all the equality of the first relation:

s_{n} = \cfrac{n}{2} \left( a_{1} + \left[ a_{1} + \left(n - 1 \right)d \right]\right)

We simplify to get our second formula for s_{n}:

s_{n} = \cfrac{n}{2} \left[2a_{1} + \left(n - 1 \right)d \right]

### Arithmetic progression example 1

*In the arithmetic sequence 4, 6, 8, 10, 12, \dots, calculate the term in place 14 and calculate the sum of the first 14 terms.*

Using the two relations that we have from the arithmetic progression theorem, we have that a_{1} = 4, the jumps in which it grows is two by two (d = 2) and we want the 14th place (n = 14):

a_{14} = a_{1} + (n-1)d = 4 + \left(14-1\right)\cdot 2 = 4 + 13\cdot 2 = 30

s_{14}=\cfrac{n}{2}\left(a_{1} + a_{n}\right) = \cfrac{14}{2} \left(4 + 30 \right) = 238

So our term 14 is a_{14}=30 and the sum of the first 14 terms is s_{14}=238.

### Arithmetic progression example 2

*In an arithmetic sequence where a_{1} = 2 and d = 4, how many terms must be taken for the sum to be 512?*

Since we know that the sum of the terms must give us 512, we will take the formula obtained from the union of the first two formulas of our arithmetic progression theorem.

s_{n} = \cfrac{n}{2} \left[2a_{1} + \left(n - 1 \right)d \right]

We have a_{1}=2, d = 4 and s_{n} = 512:

512 = \cfrac{n}{2}\left[ 2\cdot 2 + \left( n - 1 \right) 4 \right]

512 = \cfrac{n}{2} \left[4 + 4n - 4\right]

512 = \cfrac{4n}{2} + \frac{4n^{2}}{2} - \cfrac{4n}{2} = 2n + 2n^{2} - 2n

512 = 2n^{2}

In this example, it was easy for us to find the result, since nothing else is to pass dividing the 2 and then apply the square root:

\cfrac{512}{2} = n^{2}

256 = n^{2}

And so we have our two results

16, \quad -16

And since n must be a positive integer, our result is 16.

### Arithmetic progression example 3

*Interpolate seven numbers between 10 and -14.*

Since we have to find seven numebers, it means that our progression is going to be 9 terms, since they are the seven arithmetic terms and our extremes are 10 and -14. So we have a_{1} = 10 and a_{9} = -14. As we have our first relation of the arithmetic progression theorem:

a_{n} = a_{1} + \left(n-1\right)d

We substitute a_{1} and a_{n} as our last term a_{9} = -14:

-14 = 10 + \left(9 - 1 \right)d \quad \rightarrow \quad -24 = 8d

We spend dividing the 8 so that we have our difference:

d = -3

And since we already have the difference, we simply take our first term and we are going to subtract it to obtain our seven arithmetic terms between 10 and -14, which are the following:

10, \ 7, \ 4, \ 1, \ -2, \ -5, \ -8, \ -11, \ -14

## Geometric progressions

*A geometric progression is a sequence of numbers such that any term following the first one is obtained by multiplying the previous term by a non-zero term which we will call the ratio of the progression.*

An example would be the following:

1, \ 2, \ 4, \ 8, \ \dots

Where a_{1} is the first term and r is the ratio. Therefore, we have that a_{n} represents the nth term, which would have our following formula:

a_{n} = a_{1}r^{n-1}

Since we have the expression to find the term in a certain position, I am going to write you the formula for the sum of the terms:

s_{n} = \cfrac{a_{1}\left(1 - r^{n} \right)}{1-r}, \qquad r \ne 1

### Geometric progression theorem

If in a geometric progression a_{1} is the first term, a_{n} is the nth term, r is the ratio and s_{n} is the sum of the first n terms , then we have the two independent relationships mentioned above:

a_{n} = a_{1}r^{n-1}

s_{n} = \cfrac{a_{1}\left(1-r^{n}\right)}{1-r}, \qquad r=\ne 1

The first equality what we will do is multiply it by r to obtain the following:

ra_{n} = a_{1}r^{n-1}r = a_{1}r^{n-1+1} = a_{1}r^{n}

ra_{n} = a_{1}r^{n}

So all we do is substitute a_{1}r^{n} from our second equality of our theorem:

s_{n} = \cfrac{a_{1} - a_{1}r^{n}}{1-r} = \cfrac{a_{1} - ra_{n}}{1-r}, \qquad r\ne 1

### Geometric progression example 1

*In our first simple example we will simply substitute in the formulas. We have our geometric progression 1, \ 2, \ 4, \ 8, \ \dots, in which we want to find the ninth term and the sum of the first nine terms:*

We have that a_{1} = 1, r = 2, n = 9, therefore, we will directly apply the formulas:

a_{n} = a_{1}r^{n-1}

a_{9} = 1\dot 2^{8} = 256

a_{9} = 256

And now we will calculate the sum of the first nine terms:

s_{n} = \cfrac{a_{1}\left(1-r^{n}\right)}{1-r}

s_{9} = \cfrac{1\left(1-2^{9}\right)}{1-2} = \cfrac{1\left(1-512\right)}{-1}=\cfrac{1\left(-511\right)}{-1} = \cfrac{-511}{-1}

s_{9} = 511

### Geometric progression example 2

*Our first term is 4, the last term is \frac{15625}{16}, and the sum of the terms is \frac{25999}{16}. Find the ratio and the number of terms.*

We have a_{1} = 4, a_{n} = \frac{15625}{16} and s_{n} = \frac{25999}{16}, so what we want to calculate is r and n. For this progression exercise we will use one of our formulas that are mentioned in the theorem:

s_{n} = \cfrac{a_{1} - ra_{n}}{1-r}

We substitute all the data we have to find r:

\cfrac{25999}{16} = \cfrac{4 - \left( \frac{15625}{16}\right)r}{1 - r}

And now that we have our expression, all we have to do is solve it to get the result of r:

\cfrac{25999}{16} - \cfrac{25999}{16}r=4 - \cfrac{15625}{16}r

\cfrac{25999}{16} - 4 = \cfrac{25999}{16}r - \cfrac{15625}{16}r

\cfrac{25935}{16} = \cfrac{5187}{8}r

\cfrac{8}{5187}\cdot \cfrac{25935}{16} = r

r = \cfrac{5}{2}

And now to find the value of n, we will use the formula a_{n} = a_{1}r^{n-1}, since we already have r, we will only need to solve the equation to find n:

\cfrac{15625}{16} = 4\left(\cfrac{5}{2}\right)^{n-1}

\cfrac{1}{4}\cdot \cfrac{15625}{16} = \left(\cfrac{5}{2}\right)^{n-1}

\cfrac{15625}{64} = \left(\cfrac{5}{2}\right)^{n-1}

In this part comes an algebraic trick, what we will do is look for a number that raised to the power leaves us the left part of the equation with the same base as the right part of the equation, what I mean is that we will look for a number (if we can do it by trial and error), that raising \frac{5}{2} to the x power, can give us \frac{15625}{64}. We already took the trouble to look for it and it is 6:

\left(\cfrac{5}{2}\right)^{6} = \left(\cfrac{5}{2} \right)^{n-1}

And since we have that in our equality we have \frac{5}{2} of both sides, we can now equal our exponents:

6 = n-1

n=7

Giving us as a result that r = \cfrac{5}{2} and n = 7.

### Geometric progression example 3

*Interpolate seven geometric terms between \frac{1}{18} and \frac{6561}{18}.*

What we have to do is find seven numbers that make up a geometric progression, those numbers start with the first number \frac{1}{18} and end with the number \frac{6561}{18}. Which means that our geometric progression has nine terms.

So we have a_{1} = \frac{1}{18} and a_{n} = a_{9} = \frac{6561}{18}. We will use the following equation and find the value of the ratio:

a_{n} = a_{1}r^{n-1}

We substitute values and solve:

\cfrac{6561}{18} = \cfrac{1}{18}r^{9-1}

\cfrac{6561}{\cancel{18}} \cdot \cfrac{\cancel{18}}{1} = r^{8}

6561= r^{8}

r = 3

And now that we are right, we simply apply the formula to determine each of the values:

n | Equation |

n | a_{n}=a_{1}r^{n-1} |

1 | a_{1} = \left(\frac{1}{18}\right)\cdot 3^{0} = \frac{1}{18} |

2 | a_{2} = \left(\frac{1}{18}\right)\cdot 3^{1} = \frac{3}{18} |

3 | a_{3} = \left(\frac{1}{18}\right)\cdot 3^{2} = \frac{9}{18} |

4 | a_{4} = \left(\frac{1}{18}\right)\cdot 3^{3} = \frac{27}{18} |

5 | a_{5} = \left(\frac{1}{18}\right)\cdot 3^{4} = \frac{81}{18} |

6 | a_{6} = \left(\frac{1}{18}\right)\cdot 3^{5} = \frac{243}{18} |

7 | a_{7} = \left(\frac{1}{18}\right)\cdot 3^{6} = \frac{729}{18} |

8 | a_{8} = \left(\frac{1}{18}\right)\cdot 3^{7} = \frac{2187}{18} |

9 | a_{9} = \left(\frac{1}{18}\right)\cdot 3^{8} = \frac{6561}{18} |

So we have that as an answer we have that the ratio is r = 3 and our seven values are: \frac{3}{18}, \frac{9}{18}, \frac{27}{18}, \frac{81}{18}, \frac{243}{18}, \frac{729}{18} and \frac{2187}{18}.

## Harmonic progressions

The harmonic progression is a succession of numbers whose reciprocals form an arithmetic progression.

A simple example would be: \frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \dots, \frac{1}{n}, \dots since 2, 4, 6, 8, \dots, 2n, \dots is an arithmetic progression . Remember that they are only reciprocals, so the numerator always has to be the same.

We will not see examples of the harmonic progression since they are solved the same as arithmetic progressions. What we will see are several formulas that will be used for all the progressions.

If we interpolate a single harmonic mean between two numbers, we will obtain the harmonic mean. What we mean is that a and b are given numbers, so H is our harmonic mean:

\cfrac{1}{a}, \quad \cfrac{1}{H}, \quad \cfrac{1}{b}

So with this we can obtain another theorem that will be very useful to us

### Theorem

If A, G, and H are the arithmetic mean, geometric mean, and harmonic mean, respectively, of two different positive numbers a and b, the following formulas hold:

A = \cfrac{a + b}{2}, \qquad G=\sqrt{ab}, \qquad H = \cfrac{2ab}{a+b}

Being then related in the following way:

G^{2} = AH \qquad A > G > H

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