Math in Python: What is arithmetic progression ?

The sequence of numbers can be categorized as arithmetic progression if the difference of every sequence member and the previous member is constant i.e. $a_n - a_{n-1} = d, n>1.$ The first member of the arithmetic progression is $a_1 = 1$, the second member $a_2 = a_1 + d$ the third $a_3 = a_2 + d$ or generally $$ a_n = a_{n+1} + d = a_1 + (n+1)d $$ The sum of the first $n$ members can be written as: $$ S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}(2a_1 + (n-1)d).$$ For every three consecutive members of arithmetic progression, it holds $$ a_n - a_{n-1} = a_{n+1} - a_n \Rightarrow a_n = \frac{a_{n-1} + a_{n+1}}{2}.$$ The aritmetic mean of $n$ numbers: $$ A= \frac{a_1 + a_2 + ... + a_n}{n} $$ The interpolated arithmetic progression that consist of $r$ members between numbers $a$ and $b$ means to determine the $r$ numbers that together with $a$ and $b$ form artihmetic progression. The number $a$ represents the first member, while $b$ represents the last member of the sequence. The difference $d$ between the members of the arithmetic progressions can be calculated as: \begin{eqnarray} a_1 &=& a, \\ a_n &=& b,\\ n &=& r + 2,\\ a_n &=& a_1 + (n-1) d \\ b &=& a + (r+1) d \Rightarrow d = \frac{b-a}{r+1} \end{eqnarray}

Example 1 - The member of arithmetic progression is 3, and the difference $d$ is 5. Determine the eight member and sum of the first 8 members of arithmetic progression.
Solution \begin{eqnarray}a_1 &=& 3, d = 5\\ a_8 &=& a_1 + 7d = 3+ 7\cdot 5 = 38\\S_8 &=& \frac{8}{2}(a_1 + a_8) = 4(3+38) = 4\cdot 41 = 164\end{eqnarray}
Example 2 The first member of arithmetic progression is 3 while the third is 29. Calculate the forth member of arithmetic progression.
Solution\begin{eqnarray}a_1 &=& 15\\a_3 &=& a_1 + 2d \\29 &=& 15 + 2d \Rightarrow 2d = 29 - 15 \\2d &=& 14/: 2 \Rightarrow d = 7\end{eqnarray} Now that the difference d is deterimed the fourth member of arithmetic progression can be determine from formula. \begin{eqnarray} a_4 &=& a_1 + 3d = 15+ 21 = 36\end{eqnarray}
Example 3 How many members of arithmetic progression -3, 2, 7, ... we have to sum up to get 116?
Solution \begin{eqnarray} S_n &=& \frac{n}{2}(2a_1 + (n-1)d)\\ 116 &=& \frac{n}{2}(-6+(n-1) 5)/\cdot 2 \\ 232 &=& n(-6 + 5n - 5) \\ 5n^2 - 11 n &=& 232 \\ 5n^2 - 11n - 232 &=& 0\\ n_{1,2} &=& \frac{11 \pm \sqrt{121 + 4\cdot 232 \cdot 5}}{10}\\ n_{1,2} &=& \frac{11 \pm 69}{10}\\ n_1 &=& 8 \\ n_2 &=& -\frac{58}{10} \end{eqnarray} The second solution is omitted due to the fact that n can only be positive integer value. So the solution is to obtain the sum of arithmetic progression equal to 116 the first 8 members of the aritmetic progression must be summed up.
Example 4 The sum of first $n$ members of arithmetic progression is equal to $S_n = 2n^2 + 3n$. Calculate the twelfth member of arithmetic progression.
Solution To find the twelfth member of the arithmetic progression we have to find the first member $a_1$, the difference between members of arithmetic progression $d$. When those to parameters are found than the following formula is utilized $a_{12} = a_1 + (12-1)d = a_1 + 11d$. To find the first member of the arithmetic progression simply utilize the sum formula given in the description of the problem by setting the $n$ value to 1. \begin{eqnarray} n &=& 1 \Rightarrow S_1 = 2\cdot 1^2 + 3\cdot 1 = 2 + 3 = 5️ \\ S_1 &=& 1 \Rightarrow a_1 = 5 \end{eqnarray} The second member of the arithmetic progression is found using the same formula and by setting the value $n$ to 2. \begin{eqnarray} n &=& 2 \Rightarrow S_2 = 2\cdot 2^2 + 3\cdot 2\\ S_2 &=& 8+6 = 14 \Rightarrow S_2 = a_1 + a_2 \\ a_2 &=& 14 - a_1 = 14 - 5 = 9 \end{eqnarray}
Example 5 The sum of 5 consecutive natural numbers is equal to 50. Find those numbers.
Solution The sequence of natural numbers is arithmetic progression. The first member is $a_1$ and the fifth member is $a_5 = a_1 + 4$. From the formula of a sum for first 5 members we can determine the first member of arithmetic progression: \begin{eqnarray} S_5 &=& \frac{5}{2}(a_1 + a_5) \\ S_5 &=& \frac{5}{2}(a_1 + a_1 + 4)\\ S_5 &=& \frac{5}{2} 2 (a_1 + 2) \\ S_5 &=& 5(a_1 + 2) \\ 5a_1 + 10 &=& 50 \\ 5a_1 &=& 40/:5 \\ a_1 &=& 8 \end{eqnarray} So the first member of the arithmetic progression is 8 and since the first 5 consecutive natural numbers is equal 50 the members of this progression are $8,9,10,11$ and $12$.
Example 6 Between 1000 and 50000 there are 9 numbers so that all 11 numbers form arithmetic progression. Determine these numbers and the sum of all 11 numbers.
Solution \begin{eqnarray} d &=& \frac{b-a}{r+1} = \frac{50000 - 1000}{9+1}\\ d &=& \frac{49000}{10} = 4900 \end{eqnarray} The members of the artimetic progression are: \begin{eqnarray} a_1 &=& 1000 \\ a_2 &=& a_1 + d = 1000 + 4900 = 5900\\ a_3 &=& a_1 + 2d = 1000 + 9800 = 10800\\ a_4 &=& a_1 + 3d = 1000 + 14700 = 15700\\ a_5 &=& a_1 + 4d = 1000 + 19600 = 20600\\ a_6 &=& a_1 + 5d = 1000 + 24500 = 25500\\ a_7 &=& a_1 + 6d = 1000 + 29400 = 30400\\ a_8 &=& a_1 + 7d = 1000 + 34300 = 35300\\ a_9 &=& a_1 + 8d = 1000 + 39200 = 40200\\ a_10 &=& a_1 + 9d = 1000 + 44100 = 45100\\ a_11 &=& a_1 + 10d = 1000 + 49000 = 50000 \end{eqnarray} The sum of all 11 members of arithmetic progression is: \begin{eqnarray} S_{11} &=& \frac{11}{2} (a_1 + a_{11}) \\ S_{11} &=& \frac{11}{2} (1000 + 50000)\\ S_{11} &=& 280500 \end{eqnarray}
Example 7 The sum of first $n$ members is equal to $S_n = n^2$. Determine the fifteenth member.
Solution \begin{eqnarray} S_1 &=& 1^2 = 1 \Rightarrow a_1 = 1\\ S_2 &=& 2^2 = 4 \Rightarrow a_1 + a_2 = 4 \\ a_2 &=& 4-a_1 = 4-1 = 3 \\ d &=& a_2 - a_1 = 3 - 1 = 2 \\ a_{15} &=& a_1 + (15-1) d \\ a_{15} &=& 1 + 14\cdot 2 \\ a_{15} &=& 29 \end{eqnarray}
Example 8 Determine if the sequence of numbers 1,3,9,27,81,.. is the arithmetic sequence.
Solution The sequence is arithmetic if the difference between every member and the previous member is constant. To check if the difference is constant will calculate the difference between third and second member and the difference between second and first member of the sequence. \begin{eqnarray} a_3 &=& 9, a_2 = 3\\ d &=& a_3 - a_2 = 9 -3 = 6\\ a_2 &=& 3, a_1 = 1 \\ d &=& a_2 - a_1 = 3-1 = 2 \end{eqnarray} From calculated difference between third and second member and the difference between second and the first member it can be concluded that this is not arithmetic seuquence due to the fact that the difference between sequence members is not constant.
Example 9 Determine if the sequence of numbers -4, -1, 2, 5, 8,... is the arithmetic sequence.
Solution The sequence is arithmetic if the difference between every member and the previous member is constant. To test if the difference is constant for a current sequence of numbers the difference between third and second as well as second and first member will be calculated. \begin{eqnarray} a_3 &=& 2, a_2 = -1 \\ d_2 &=& 2-(-1) = 2+1 =3 \\ a_2 &=& -1, a_1 = -4 \\ d_1 &=& -1 -(-4) = -1 +4 = 3\\ d_2 &=& d_1 = d \end{eqnarray} From obtained results it can be seen that the diffrence between third and second member and second and first member are equal which means that this is arithmetic sequence of numbers.
Example 10 The total of 8600 papers must be printed. How many more sheets of paper are left to copy after one hour of photocopying if 25 sheets of paper are photocopied per minute?
Solution This is classic arithmetic sequence since the difference is constant. The difference in this case is the number of copied papers in a minute which is 25. After 1 minute the total of 25 papers is copied. After 60 minutes (1 hour) the number of papers copied is: $$ a_60 = 25+ 59\cdot 25 = 1500$$ The remaining number of papers that must be copied is \begin{eqnarray} 8600 - a_{60} &=& 8600 - 1500 = 7100 \end{eqnarray}

Math in Python

What is arithmetic progression ?

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