The harmonic progression is type of progression that is formed by taking the reciprocals of arithmetic progression. The sequence is the harmonic progression if each term is the harmonic mean of the neighboring terms. The infinite sequence it can be written in the form of a $$ \frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2d}, \frac{1}{a+3d},..., $$ where \(a\) is the non-zero and \(-a/d\) is a natural number. The finite sequence can be written as: $$ \frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2d}, \frac{1}{a+3d},..., \frac{1}{a+kd}$$ where \(a\neq 0\), \(k\) is the natural number and \(-a/d\) is not natural number or is greater than \(k\).
- Example 1 Determine the fourth, fifth and sixth member of harmonic progression defined as \(\frac{1}{5}, \frac{1}{10}, \frac{1}{15},... \)
Solution The first step is to determine \(d\). We know that the second member of the harmonic progression in general form is \(\frac{1}{a+d}\) From the general form and the real value of the second member we can calculate d. The parameter \(a\) is equal to 5 and this can be found by equalizing the general formula of the first member and the real value of the first member. \begin{eqnarray} \frac{1}{a+d} &=& \frac{1}{10}\\ a+d &=& 10 \\ d &=& 10 - a \Rightarrow d = 10 - 5 = 5 \end{eqnarray} The \(k\) for the fourth member is equal to 3, for the fifth 4, and for the sixth 5. \begin{eqnarray} k &=& 3, a = 5, d = 5 \\ \frac{1}{a+3d} &=& \frac{1}{5+3\cdot 5} = \frac{1}{20}\\ k &=& 4\\ \frac{1}{a+4d} &=& \frac{1}{5+4\cdot 5} = \frac{1}{25}\\ k &=& 5 \\ \frac{1}{a+5d} &=& \frac{1}{5+5\cdot 5} = \frac{1}{30} \end{eqnarray} - Example 2 Calculate the third member of the harmonic sequence: \(\frac{1}{4}, \frac{1}{7},...\)
SolutionThe procedure is to calculate the parameter \(d\) and then to calculate the third member. \begin{eqnarray} a &=& 4\\ \frac{1}{a+d} &=& \frac{1}{7}\\ 4+d &=& 7\\ d &=& 3 \\ k &=& 2 \\ \frac{1}{a + kd } &=& \frac{1}{4+2\cdot 3} = \frac{1}{10} \end{eqnarray} - Example 3 Calculate the thousandth member of a harmonic progression which can be obtained from series \(6, 13, 20, 27, 34,...\). Test if the previous series is arithmetic progression or not.
Solution The first step is to investigate if this is arithmetic progression. If this is in fact arithmetic progression the harmonic progression can be created by taking the reciprocals of the arithmetic progression. First step - Investigate if this is arithmetic progression i.e calculate the difference between third and second member, and second and first member. By calculating the difference we will check if the difference is constant. \begin{eqnarray} a_3 - a_2 &=& 20 - 13 = 7 \\ a_2 - a_1 &=& 12 - 6 = 7\\ d &=& 7 \end{eqnarray} From the previous calculation we have concluded that the difference \(d\) is constant. Second step is to create the harmonic progression from arithmetic progression i.e. create reciprocal value of each member. $$ \frac{1}{6}, \frac{1}{13}, \frac{1}{20},...$$ The general formula of nth member of harmonic progression can be written as $$ \frac{1}{a+kd} $$ where $k$ is $n-1$. So for thousand member (\(n=1000\)) the \(k = n-1 = 999\). The thousand member is equal \begin{eqnarray} a&=& 6, d= 7, k = 999\\ \frac{1}{a+kd} &=& \frac{1}{6+999\cdot 7} = \frac{1}{6999} \end{eqnarray} - Example 4Calculate the fifteenth, and twentieth member of harmonic progression \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4},...\)
SolutionDetermine the parameters \(a\) and \(d\) \begin{eqnarray} \frac{1}{a} &=& \frac{1}{2} \Rightarrow a = 2 \\ \frac{1}{a+d} &=& \frac{1}{3}\\ a+d &=& 3 \Rightarrow d = 3-2 = 1 \\ \end{eqnarray} The general formula for determining each member of harmonic progression can be written as: $$ \frac{1}{a+kd} = \frac{1}{a+(n-1)d}$$ Since we are looking for the fifteenth member the \(n = 15\) and the \(k\) is determine from formula \(k= n-1\). $$ k = n-1 = 15 -1 = 14 $$ Then the fifteenth member is equal to: \begin{eqnarray} n&=& 15, k = 14, a = 2, d=1\\ \frac{1}{a+14d} &=& \frac{1}{2+14\cdot 1} = \frac{1}{16} \end{eqnarray} For the twentieth member the \(n= 20\) the \(k=19\) and the value of the member is: \begin{eqnarray} n&=& 20, k = 19, a = 2, d=1\\ \frac{1}{a+kd} &=& \frac{1}{2+19\cdot 1} = \frac{1}{21} \end{eqnarray} - Example 5Calculate the sum of first five members of harmonic progression \(\frac{1}{4}, \frac{1}{7}, \frac{1}{10}...\)
Solution In this example we have only three members of the harmonic progression. To determine the remaining 2 which are needed for calculating the sum we need to determine the parameters \(a\), and \(d\). \begin{eqnarray} \frac{1}{a} &=& \frac{1}{4} \Rightarrow a = 4 \\ \frac{1}{a+d} &=& \frac{1}{7}\\ a+d &=& 7 \Rightarrow d = 7-4 = 3 \end{eqnarray} The parameters of the forth member are \(n=4, k = n-1 = 4-1 = 3\). The value of the forth member can be calculated as: \begin{eqnarray} \frac{1}{a+kd} &=& \frac{1}{4 + 3\cdot 3} = \frac{1}{13} \end{eqnarray} The parameters of the fifth member are \(n=5, k = n-1 = 5-1 = 4\). the value of the fifth member can be calculated as: \begin{eqnarray} \frac{1}{a+kd} &=& \frac{1}{4+4\cdot 3} = \frac{1}{16} \end{eqnarray} The sum of the first five members of the harmonic progression can is equal to. \begin{eqnarray} \frac{1}{4} + \frac{1}{7} + \frac{1}{10} + \frac{1}{13} + \frac{1}{16} = \frac{4603}{7280} \end{eqnarray} - Example 6 Calculate the tenth member of the harmonic progression \(1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, ...\)
Solution \begin{eqnarray} a &=& 1 \\ \frac{1}{a+d} &=& \frac{1}{3}\\ a+d &=& 3 \Rightarrow d = 2\\ n &=& 10, k = 10-1 = 9\\ \frac{1}{a+kd} &=& \frac{1}{1 + 9\cdot 2} = \frac{1}{19} \end{eqnarray} - Example 7 Calculate the parameters \(d\) of harmonic progression\(\frac{1}{4}, \frac{1}{8}, \frac{1}{12},...\)
Solution \begin{eqnarray} \frac{1}{a} &=& \frac{1}{4} \Rightarrow a = 4\\ \frac{1}{a+d} &=& \frac{1}{8}\\ d &=& 8-4 = 4 \end{eqnarray} - Example 8 Calculate the fortyfifth member of the harmonic progression \(\frac{1}{10}, \frac{1}{12}, \frac{1}{14}, \frac{1}{16},...\)
Solution The first step is to find \(a\) and \(d\). \begin{eqnarray} \frac{1}{a} &=& \frac{1}{10}\Rightarrow a = 10 \\ \frac{1}{a+d} &=& \frac{1}{12}\\ a+d &=& 12 \Rightarrow d = 2 \end{eqnarray} The formula for determining nth member of harmonic progression can be written as: $$\frac{1}{a+nd} = \frac{1}{a+(n-1)d} $$ The \(n\) and \(k\) parameters of fortyfifth member are \(n=45\), and \(k = n-1 = 45-1 = 44\). \begin{eqnarray} n &=& 45, k = 44, a = 10, d = 2 \\ \frac{1}{a+kd} &=& \frac{1}{10 + 44\cdot 2} = \frac{1}{98} \end{eqnarray} - Example 9 Calculate the product of the first 5 members of harmonic progression \(\frac{1}{10}, \frac{1}{13}, \frac{1}{16},...\)
Solution In this example we have only three members of harmonic progression and we have to find remaining 2 to calculate the product of the first 5 members. To do this first we have to determine \(a\) and \(d\) parameters. \begin{eqnarray} \frac{1}{a} &=& \frac{1}{10}\Rightarrow a = 10\\ \frac{1}{a+d} &=& \frac{1}{13}\\ a+d &=& 13 \Rightarrow d = 3 \end{eqnarray} The forth member can be calculated as: \begin{eqnarray} n&=& 4, k = 4-1 = 3 \\ a &=& 10, d = 3 \\ \frac{1}{a+kd} &=& \frac{1}{10+3\cdot 3} = \frac{1}{19} \end{eqnarray} The fifth member can be clacualted as: \begin{eqnarray} n&=& 5, k = 5-1 = 4 \\ a &=& 10, d = 3 \\ \frac{1}{a+kd} &=& \frac{1}{10+4\cdot 3} = \frac{1}{22} \end{eqnarray} The product of first five members is equal to: \begin{eqnarray} \frac{1}{10}\cdot \frac{1}{13}\cdot\frac{1}{16}\cdot \frac{1}{19}\cdot \frac{1}{22} &=& \frac{1}{869400} \end{eqnarray} - Example 10 Determine the fivethousandth member of the harmonic progression \(\frac{1}{10}, \frac{1}{14}, \frac{1}{18},...\)
SolutionThe first step is to find parameters \(a\) and \(d\) \begin{eqnarray} \frac{1}{a} &=& \frac{1}{10} \Rightarrow a = 10 \\ \frac{1}{a+d} &=& \frac{1}{14} \\ a+d &=& 14 \Rightarrow d = 4 \end{eqnarray} The fivethousand member is equal to: \begin{eqnarray} n &=& 5000, k = n-1 = 4999\\ \frac{1}{a+kd} &=& \frac{1}{10+4999\cdot 4} = \frac{1}{20006} \end{eqnarray}
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