Progression

In mathematics, progression/sequence/series, referes to the numbers that are arranged in particular, predictable order. The predictable order means that for a given sequence of numbers the next number in the sequence can easily be predicted/calculated. The simple example of progression is the progression of even numbers. If the part of the progression of even numbers is given let's say 2,4,6,8,... we can easily find the next number in the progression. In this case it's the number 10 since this number is the next number in the progression of even numbers.
The general definition of progression would be: The progression in the set \(S\) is every function \(a:\mathbf{N} \rightarrow S\) which associates the natural element \(n\) with the element \(a(n) = a_n \) of the set \(S\). The number \(a_n\) is called general or n-th number of progression/sequence/series, and the progression itself is denoted by \(a_n.\)

Progression examples

There are four different types of progression definition and these are

• Progression defined with general number \(a_n:\)
  • the sequence of even natural numbers \(a:\mathbf{N} \rightarrow \mathbf{N}, a(n) = a_n = 2n\) - 2,4,6,8,10,...
  • a series of multiples of number 5 \(a:\mathbf{N} \rightarrow \mathbf{N}, a(n) = a_n = 5n\) - 5, 10, 15, 20, 25,...
  • a series of numbers that when divided by 4 give the remainder 3. In general form this progression can be written as \(a:\mathbf{N} \rightarrow \mathbf{N}, a(n) = a_n = 4n-1\), 3,7,11,15,...
  • a sequence of negative integers \(a:\mathbf{N} \rightarrow \mathbf{N}, a(n) = a_n = -n \) - -1,-2,-3,-4,...
  • a sequece
• Progression defined with recrusive formulas
• \(a:\mathbf{N} \rightarrow \mathbf{N}, a(n) = a_n = a_{n-} + a_{n-2}, a_1 = 1, a_2 = 3\) The progression example: 1,3,4,7,11,18,...,
• \(a:\mathbf{N} \rightarrow \mathbf{N}, a(n) = a_n = a_{n-1} - a_{n-2}, a_1 = 1, a_2 =2 \) The progression example: 1,2,1,-1, -2, -1, 1,...
• \(a;\mathbf{N} \rightarrow \mathbf{N}, a(n) = a_n = a_{n-1}- a_{n-2}^2, a_1 = 1, a_2 = 2\) The progression example: 1, 2, 1, -3, -4, -13,...
• The progression defined graphically
• The progression defined with table

  \(\mathbf{n}\)	1	2	3	4	5
  \(a_n\)	32	35	39	44	50

• progression defined descriptivly

In mathematics there are three types of progression and these are arithmetic progression, geometric progression and harmonic progression. Arithmetic progression is the sequence of numbers such that the difference of any two successive members of the sequence is a constant. The geometric progression is the sequence of numbers such that the quotient of any two successive members of the sequence is a constant. The harmonic progression is the sequence of numbers such that their reciprocals form an arithmetic progression. As stated earlier the arithemtic progression or arithmetic sequence is the sequence of numbers ushc that the difference between the consecutive ters is constant. Example of arithmetic sequence or artithemtic progression is 1,3,5,7,9,11,13,15,... with a common difference of 2. The initial term of the arithmetic progression is indicated by \(a_1\) and the difference between two successive members in the sequence by term $d$, then the n-th term of the sequence \(a_n\) can be written in the following form: $$ a_n = a_1 + (n-1)d,$$ whic can be generalized to: $$ a_n = a_m + (n-m)d.$$ The general form of the arithmetic progression with constant distance between sequential members can be written as: $$ a, a+d, a+2d,..., a+(n-1)d. $$ If the last term is denoted as \(l = a +(n-1)d\), then the sum is equal to: $$ s = \frac{n}{2}(a+l) = \frac{n}{2}(a + a+(n-1)d)$$
A geometric prgoression or geometric sequence is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. Example of the geometric progression is 2,6,18,54,... with common ratio of 3. Another example of geometric progression with common ration of 0.5 is the sequence of 10, 5, 2.5, 1.25,... The goemetric progression can simbolicly be written in the following form $$ a,ar, ar^2,...,ar^{n-1}$$ and the sum is $$S = \frac{a-ar^n}{1-r}$$ The harmonic progression or harmonic sequence is the type of progression formed by taking the reciprocals of an arithmetic progression. A sequence is a harmonic progression when each term is the harmonic mean of the neighboring terms. The general from of harmonic sequence/progression can be written as $$\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2d}, \frac{1}{a+3d},...,\frac{1}{a+kd}, $$ where \(a\) is not zero, \(k\) is a natural number and \(-a/d\) is not a natural number or is greater than \(k\). Some examples of harmonic progression/sequence are

• 1, 1/2, 1/3, 1/4, 1/5, 1/6, ...
• 12, 6, 4, 3, \(\frac{12}{5}\), 2, ..., \(\frac{12}{n}\),...

• Example 1 Determine the general recursive formula using which the sequence members are calculated. \begin{eqnarray}a_1 &=& 1\\ a_2 &=& a_1 + 2 = 3\\ a_3 &=& a_2 + 4 = 7 \\ a_4 &=& a_3 + 8 = 15\end{eqnarray}
  Solution Using previously shown numbers sequence the general recursive formula can be written in following form \begin{eqnarray}a_n &=& a_{n-1} + 2^{n-1}\end{eqnarray} for \(a_1 = 1.\) The next number in sequence can be caluclated using general recursive formula \begin{eqnarray} a_n &=& a_{n-1} + 2^{n-1}\\ \mathrm{for}\quad n = 5 \Rightarrow a_5 &=& a_4 + 2^4 \\ a_5 &=& 15 + 16 = 31.\end{eqnarray}
• Example 2 The sequence is defined with recursive formula \(a_n = a_{n-2} + a_{n-1}\) where \(a_1 = 1\) and \(a_2= 4.\) Calculate the fifth and sixth member of the sequence.
  Solution\begin{eqnarray}a_1 &=& 1, a_2 = 4\\ a_3 &=& a_1 + a_2 = 1+ 4 = 5 \\ a_4 &=& a_2 + a_3 = 4+5 = 9 \\ a_5 &=& a_3 + a_4 = 5+9 = 14 \\ a_6 &=& a_4 + a_5 = 9 + 14 = 23\end{eqnarray}
• Example 3 The first four members of the sequence are 12, 18, 30 and 54. Determine the recursive formula and calcualte the fifth member of the sequence.
  Solution\begin{eqnarray}a_1 &=& 12, a_2 = 18, a_3 = 30, a_4= 54\\ a_2 &=& 2\cdot 12 - 6 = 18\\ a_3 &=& 2\cdot 18 - 6 = 30 \\ a_4 &=& 2\cdot 30 - 6 = 54\end{eqnarray} The general recursive fromula of the sequence is \(a_{n+1} = 2a_n - 6\). The fifth member of the sequence can be determine from recursive formula \begin{eqnarray}a_5 &=& 2a_4 - 6 = 2\cdot 54 - 6 = 102\end{eqnarray}
• Example 4 The equation \(t_{n+1} = a\cdot t + 6\) is recursive formula that defines the sequence 5, 21, 69, 213,... Determine the value of variable a.
  Solution\begin{eqnarray}t_2 &=& a\cdot t_4 + 6 \\ 21 &=& 5a + 6\\ 5a &=& 21 - 6 \Rightarrow a = 3 \\ t_3 &=& a\cdot t_2 + 6 \\ 69 &=& 21a + 6 \\ 21a &=& 63 /: 3 \Rightarrow a = 3\end{eqnarray} The value of \(a\) is 3.
• Example 5 Determine the fifteenth member of sequence \(a_n = (-1)^{n-1}\cdot n^2 \)
  Solution \begin{eqnarray} n &=& 15\\ a_{15} &=& (-1)^{15-1} \cdot 15^2 \\ a_{15} &=& 225 \end{eqnarray}
• Example 6 Determine the thirty fifth, fortieth, fiftieth number of the sequence \(a_n = (-1)^n\cdot 2^{n-25}\)
  Solution \begin{eqnarray} n &=& 35\\ a_{35} &=& (-1)^{35} \cdot 2^{35-25}\\ a_{35} &=& -2^{10} \\ a_{35} &=& -1024\\ n &=& 40 \\ a_{40} &=& (-1)^{40} \cdot 2^{40-25}\\ a_{40} &=& 2^{15} \\ a_{40} &=& 2^{10}\cdot 2^5\\ a_{40} &=& 1024 \cdot 32 \\ a_{40} &=& 32768\\ n &=& 50\\ a_{50} &=& (-1)^{50} \cdot 2^{50-25}\\ a_{50} &=& 2^{25} \\ a_{50} &=& 2^{10} \cdot 2^{10} \cdot 2^5\\ a_{50} &=& 1024 \cdot 1024 \cdot 32 \\ a_{50} &=& 33554432 \end{eqnarray}
• Example 7 The longer the FIFA World Cup lasts, the more people come to the matches. The equation that predicts the weekly attendance of various football matches can be written in the following form \(W_{n+1} = 1.5 W_n + 50000, W_1 = 100000,\) where \(W_n\) is the weekly attendance for nth week. Using the equation, calculate the number of people possibly attending football matches in 5th week.
  Solution \begin{eqnarray} n&=& 0\\ W_{1} &=& 100000\\ n &=& 1 \\ W_2 &=& 1.5\cdot W_1 + 50000\\ W_2 &=& 200000\\ n &=& 2 \\ W_3 &=& 1.5 \cdot W_2 + 50000\\ W_3 &=& 350000\\ n &=& 3\\ W_4 &=& 1.5\cdot W_3 + 50000\\ W_4 &=& 575000\\ n &=& 4\\ W_5 &=& 1.5 \cdot W_4 + 50000\\ W_5 &=& 912500 \end{eqnarray}
• Example 8 Calculate the third member of the sequence which is defined by equation \(t_{n+1} = 2t_n+40\)
  Solution \begin{eqnarray} n &=& 0 \\ t_1 &=& 2\cdot 0 + 40 \\ t_1 &=& 40 \\ n &=& 1 \\ t_2 &=& 2\cdot t_1 + 40 \\ t_2 &=& 2 \cdot 40 + 40 \\ t_2 &=& 120\\ n &=& 2 \\ t_3 &=& 2 \cdot t_2 + 40 \\ t_3 &=& 2 \cdot 120 + 40 \\ t_3 &=& 280 \end{eqnarray}
• Example 9 The progression is defined with recursive formula \(a_{n+1} = 3a_n + 2, a_1 = 10\)Calculate the fifth member of the progression.
  Solution \begin{eqnarray} a_1 &=& 10\\ n&=& 1\\ a_2 &=& 3\cdot a_1 + 2 \\ a_2 &=& 3\cdot 10 + 2 \Rightarrow a_2 = 32 \\ n&=& 2\\ a_3 &=& 3\cdot a_2 + 2 \\ a_3 &=& 3\cdot 32 + 2 \Rightarrow a_3 = 98\\ n&=& 3\\ a_4 &=& 3\cdot a_3 + 2 \\ a_4 &=& 3\cdot 98 + 2 \Rightarrow a_4 = 296\\ n&=& 4\\ a_5 &=& 3\cdot a_4 + 2\\ a_5 &=& 3\cdot 296 + 2 \Rightarrow a_5 = 890 \end{eqnarray}
• Example 10 Calculate the second, third and fourth member of the sequence defined with equation \(a_{n+1} = a_n + 20, a_1 = 1\\
  Solution \begin{eqnarray} a_1 &=& 1 \\ n &=& 1 \\ a_2 &=& a_1 + 20 \\ a_2 &=& 1 + 20 = 21 n &=& 2 \\ a_3 &=& a_2 + 20 \\ a_3 &=& 21 + 20 = 41 \\ n &=& 3 \\ a_4 &=& a_3 + 20 \\ a_4 &=& 41 + 20 = 61 \end{eqnarray}