# 42. Harmonic Progression Solutions

The sequence \(1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, ...\) is an H. P.

Thus, reciprocals would form an A. P. i.e.

\(1, 3, 5, 7, ...\) are in A. P. 100th term of whose is \(1 + 99*2 = 199\)

Thus, 100th term of the H. P. in question is \(\frac{1}{199}\).

Let \(a\) be the first term and \(d\) be the common difference of the corresponding A. P.

The pth and qth term of the A. P. will be \(\frac{1}{qr}\) and \(\frac{1}{pr}\) respectively.

For A. P. nth term = \(a + (n - 1)d\)

\(\frac{1}{qr} = a + (p - 1)d\)

\(\frac{1}{pr} = a + (q - 1)d\)

Subtracting we get, \(\frac{q - p}{pqr} = (q - p)d \therefore d = \frac{1}{pqr}\)

You can substitute \(d\) in either for pth term or for qth term. Doing it for pth term we get,

\(\frac{1}{qr} = a + \frac{p - 1}{pqr} \therefore a = \frac{1}{pqr}\)

Now it is trivial to find rth term.

\(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A. P. and are pth, qth and rth term respectively.

Let \(x\) be the first term and \(d\) be the common difference of this A. P.

\(\therefore \frac{1}{a} = x + (p - 1)d\)

Multiplying with \(abc\) we get

\(bc = abc(a + (p - 1)d)\)

\((q - r)bc = (q - r)abc(x + (p - 1)d)\)

Similarly, for qth and rth term we have

\(ca = abc(x + (q - 1)d)\)

\((r - p) = (r - p)abc(x + (q - 1)d)\)

\(ab = abc(x + (r - 1)d)\)

\((p - q)ab = (p - q)abc(x + (r - 1)d)\)

Now it is trivial to prove the equality in the question.

Because \(a, b, c\) are in H. P. \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A. P.

Thus, \(\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}\)

\(b = \frac{2ca}{c + a}\)

\[\frac{a - b}{b - c} = \frac{a - \frac{2ca}{c + a}}{\frac{2ca}{c + a} - c}\]\[\frac{a^2 - ac}{ac - c^2} = \frac{a}{c}\]Since \(a, b, c, d\) are in H. P. therefore \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}\) are in A. P.

Let \(d\) be the common difference.

\(\frac{1}{b} - \frac{1}{a} = d \Rightarrow ab = \frac{1}{d}(a - b)\)

Similarly, \(bc = \frac{1}{d}(b - c)\)

and \(cd = \frac{1}{d}(c - d)\)

Adding last three equalities we get

\(ab + bc + cd = \frac{1}{c}(a - d)\)

\[= \frac{1}{\frac{\frac{1}{d} - \frac{1}{a}}{4 - 1}}(a - d) = 3ad\]This problem can be solved as previous problem has been.

Given \(a, b, c\) are in H. P. which implies \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in H. P.

\(\Rightarrow \frac{a + b + c}{a}, \frac{a + b + c}{b}, \frac{a + b + c}{c}\) are in A. P.

\(\Rightarrow 1 + \frac{b + c}{a}, 1 + \frac{a + c}{b}, 1 + \frac{a + b}{c}\) are in A. P.

\(\Rightarrow \frac{b + c}{a}, \frac{a + c}{b} \frac{a + b}{c}\) are in A. P.

\(\Rightarrow \frac{a}{b + c}, \frac{b}{a + c}, \frac{c}{a + b}\) are in H. P.

\(a^2, b^2, c^2\) are in A. P.

\(\Rightarrow b^2 - a^2 = c^2 - b^2\)

\(\Rightarrow \frac{b - a}{(c + a)(b + c)} = \frac{c - b}{(a + b)(c + a)}\)

\(\Rightarrow \frac{b + c - c - a}{(c + a)(b + c)} = \frac{c + a - a - b}{(a + b)(c + a)}\)

\(\Rightarrow \frac{1}{c + a} - \frac{1}{b + c} = \frac{1}{a + b} - \frac{1}{c + a}\)

\(\Rightarrow \frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}\) are in A. P.

\(\Rightarrow b + c, c + a, a + b\) are in H. P.

\(t_1 = 1, t_2 = \frac{1}{3*2 - 1} = \frac{1}{4}, t_3 = \frac{1}{7}, t_4, \frac{1}{10}\)

Since the reciprocal of nth term is \(3n - 2\) which will constitute an A. P. as seen by the value of terms therefore this sequence is in A. P.

Reciprocal of the terms are \(\frac{11}{2}, 5, \frac{9}{2}, ...\)

Common difference = \(5 - \frac{11}{2} = -\frac{1}{2}\)

8th term of reciprocals = \(\frac{11}{2} + 7 * -\frac{1}{2} = 2\)

Thus, the term is \(\frac{1}{2}\)

Reciprocals are \(3, \frac{23}{8}, \frac{11}{4}, ...\)

Common difference = \(\frac{23}{8} - 3 = -\frac{1}{8} = \frac{11}{4} - \frac{23}{8}\)

7th term = \(3 + (7 - 1) * -\frac{1}{8} = \frac{18}{8} = \frac{9}{4}\)

Thus the term is \(\frac{4}{9}\).

Reciprocals are \(t_7 = 20\) and \(t_{13} = 38\)

Let \(a\) be the first term and \(d\) be the common difference.

\(a + 6d = 20\) and \(a + 12d = 38\)

Subtracting, \(6d = 18 \Rightarrow d = 3 \Rightarrow a = 2\)

\(t_4 = a + 3d = 11\), reciprocal of which is \(\frac{1}{11}\)

Reciprocals would be \(t_m = \frac{1}{n}\) and \(t_n = \frac{1}{m}\) and in A. P.

Let \(a\) be the first term and \(d\) be the common difference.

\(t_m = a + (m - 1)d = \frac{1}{n}\) and \(t_n = a + (n - 1)d = \frac{1}{m}\)

Subtracting \((m - n)d = \frac{m - n}{mn}\) i.e. \(d = \frac{1}{mn}\)

Substituting \(d\) in \(t_m\) we have

\(a = \frac{1}{n} - \frac{(m - 1)}{mn} = \frac{1}{mn}\)

Now, \(t_{m + n} = \frac{1}{mn} + (m + n - 1)\frac{1}{mn} = \frac{m + n}{mn}\)

Reciprocal is \(\frac{mn}{m + n}\) which is what we want.

Similarly, \(t_{mn} = a + (mn - 1)d = \frac{1}{mn} + (mn - 1)\frac{1}{mn} = 1\)

Let the three numbers are \(\frac{1}{\alpha - \beta}, \frac{1}{\alpha}, \frac{1}{\alpha + \beta}\)

Thus, sum of reciprocals = \(3\alpha = \frac{1}{4} \Rightarrow \alpha = \frac{1}{12}\)

Sum of three terms = \(\frac{\alpha(\alpha + \beta) + (\alpha - \beta)(\alpha + \beta) + \alpha(\alpha - \beta)}{\alpha(\alpha^2 - \beta^2)} = 37\)

\(\Rightarrow \frac{3\alpha^2 - \beta^2}{\alpha(\alpha(\alpha^2 - \beta^2))} = 37\)

Substitute for \(\alpha\) and find \(\beta\) and you will have the numbers.

Since \(a, b, c\) are in H. P. \(\therefore \frac{2}{b} = \frac{1}{a} + \frac{1}{c}\).

\(\Rightarrow b = \frac{2ac}{a + c}\)

\(\frac{1}{b - c} + \frac{1}{b - a} = \frac{1}{\frac{2ac}{a + c} - a} + \frac{1}{\frac{2ac}{a + c} - c}\)

\(= \frac{a + c}{a(c - a)} + \frac{a + c}{c(a - c)} = \frac{a + c}{a(c - a)} - \frac{a + c}{c(c - a)}\)

\(= \frac{ac + c^2 - a^2 - ac}{ac(c - a)} = \frac{a + c}{ac} = \frac{1}{a} + \frac{1}{c}\).

From previous problem, we have

\(b = \frac{2ac}{a + c}\)

Substituting for \(b\)

\(\frac{b + a}{b - a} + \frac{b + c}{b - c} = \frac{3ac + a^2}{a(c - a)} + \frac{3ac + c^2}{c(a - c)}\)

\(= \frac{3c + a}{c - a} - \frac{3a + c}{c - a} = \frac{2c - 2a}{c - a} = 2\).

17 and 18 are similar to 6 and has been left as an exercise to the reader.

Since \(b + c, c + a, a + b\) are in H. P.

\(\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}\) are in A. P.

\(\frac{a + b + c}{b + c}, \frac{a + b + c}{c + a}, \frac{a + b + c}{a + b}\) are in A. P.

Subtracting 1 from each term

\(\frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b}\) are in A. P.

Since \(b + c, c + a, a + b\) are in H. P.

\(\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}\) are in A. P.

Now, this problem is same as 23 in Arithmetic Progression Problems Part1.

This problem is similar to 27.3 in Arithmetic Progression Problems Part1.

This problem is similar to 28 in Arithmetic Progression Problems Part1.

This is a similar problem and has been left as an exercise.