Harmonic Progression Solutions
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1. The sequence :math:`1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, ...` is an H. P.

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

   :math:`1, 3, 5, 7, ...` are in A. P. 100th term of whose is :math:`1 + 99*2 = 199`

   Thus, 100th term of the H. P. in question is :math:`\frac{1}{199}`.

2. Let :math:`a` be the first term and :math:`d` be the common difference of the
   corresponding A. P.

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

   For A. P. nth term = :math:`a + (n - 1)d`

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

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

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

   You can substitute :math:`d` in either for pth term or for qth term. Doing it for pth term
   we get,

   :math:`\frac{1}{qr} = a + \frac{p - 1}{pqr} \therefore a = \frac{1}{pqr}`

   Now it is trivial to find rth term.

3. :math:`\frac{1}{a}, \frac{1}{b}, \frac{1}{c}` are in A. P. and are pth, qth and rth term respectively.

   Let :math:`x` be the first term and :math:`d` be the common difference of this A. P.

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

   Multiplying with :math:`abc` we get

   :math:`bc = abc(a + (p - 1)d)`

   :math:`(q - r)bc = (q - r)abc(x + (p - 1)d)`

   Similarly, for qth and rth term we have

   :math:`ca = abc(x + (q - 1)d)`

   :math:`(r - p) = (r - p)abc(x + (q - 1)d)`

   :math:`ab = abc(x + (r - 1)d)`

   :math:`(p - q)ab = (p - q)abc(x + (r - 1)d)`

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

4. Because :math:`a, b, c` are in H. P. :math:`\frac{1}{a}, \frac{1}{b}, \frac{1}{c}` are in A. P.

   Thus, :math:`\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}`

   :math:`b = \frac{2ca}{c + a}`

   .. math::

      \frac{a - b}{b - c} = \frac{a - \frac{2ca}{c + a}}{\frac{2ca}{c + a} - c}

   .. math::

      \frac{a^2 - ac}{ac - c^2} = \frac{a}{c}

5. Since :math:`a, b, c, d` are in H. P. therefore :math:`\frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}`
   are in A. P.

   Let :math:`d` be the common difference.

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

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

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

   Adding last three equalities we get

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

   .. math::

      = \frac{1}{\frac{\frac{1}{d} - \frac{1}{a}}{4 - 1}}(a - d) = 3ad

6. This problem can be solved as previous problem has been.

7. Given :math:`a, b, c` are in H. P. which implies :math:`\frac{1}{a}, \frac{1}{b}, \frac{1}{c}`
   are in H. P.

   :math:`\Rightarrow \frac{a + b + c}{a}, \frac{a + b + c}{b}, \frac{a + b + c}{c}` are in A. P.

   :math:`\Rightarrow 1 + \frac{b + c}{a}, 1 + \frac{a + c}{b}, 1 + \frac{a + b}{c}` are in A. P.

   :math:`\Rightarrow \frac{b + c}{a}, \frac{a + c}{b} \frac{a + b}{c}` are in A. P.

   :math:`\Rightarrow \frac{a}{b + c}, \frac{b}{a + c}, \frac{c}{a + b}` are in H. P.

8. :math:`a^2, b^2, c^2` are in A. P.

   :math:`\Rightarrow b^2 - a^2 = c^2 - b^2`

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

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

   :math:`\Rightarrow \frac{1}{c + a} - \frac{1}{b + c} = \frac{1}{a + b} - \frac{1}{c + a}`

   :math:`\Rightarrow \frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}` are in A. P.

   :math:`\Rightarrow b + c, c + a, a + b` are in H. P.

9. :math:`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 :math:`3n - 2` which will constitute an A. P. as seen
   by the value of terms therefore this sequence is in A. P.

10. Reciprocal of the terms are :math:`\frac{11}{2}, 5, \frac{9}{2}, ...`

    Common difference = :math:`5 - \frac{11}{2} = -\frac{1}{2}`

    8th term of reciprocals = :math:`\frac{11}{2} + 7 * -\frac{1}{2} = 2`

    Thus, the term is :math:`\frac{1}{2}`

11. Reciprocals are :math:`3, \frac{23}{8}, \frac{11}{4}, ...`

    Common difference = :math:`\frac{23}{8} - 3 = -\frac{1}{8} = \frac{11}{4} - \frac{23}{8}`

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

    Thus the term is :math:`\frac{4}{9}`.

12. Reciprocals are :math:`t_7 = 20` and :math:`t_{13} = 38`

    Let :math:`a` be the first term and :math:`d` be the common difference.

    :math:`a + 6d = 20` and :math:`a + 12d =  38`

    Subtracting, :math:`6d = 18 \Rightarrow d = 3 \Rightarrow a = 2`

    :math:`t_4 = a + 3d = 11`, reciprocal of which is :math:`\frac{1}{11}`

13. Reciprocals would be :math:`t_m = \frac{1}{n}` and :math:`t_n = \frac{1}{m}` and in A. P.

    Let :math:`a` be the first term and :math:`d` be the common difference.

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

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

    Substituting :math:`d` in :math:`t_m` we have

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

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

    Reciprocal is :math:`\frac{mn}{m + n}` which is what we want.

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

14. Let the three numbers are :math:`\frac{1}{\alpha - \beta}, \frac{1}{\alpha}, \frac{1}{\alpha + \beta}`

    Thus, sum of reciprocals = :math:`3\alpha  = \frac{1}{4} \Rightarrow \alpha = \frac{1}{12}`

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

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

    Substitute for :math:`\alpha` and find :math:`\beta` and you will have the numbers.

15. Since :math:`a, b, c` are in H. P. :math:`\therefore \frac{2}{b} = \frac{1}{a} + \frac{1}{c}`.

    :math:`\Rightarrow b = \frac{2ac}{a + c}`

    :math:`\frac{1}{b - c} + \frac{1}{b - a} = \frac{1}{\frac{2ac}{a + c} - a} + \frac{1}{\frac{2ac}{a + c} - c}`

    :math:`= \frac{a + c}{a(c - a)} + \frac{a + c}{c(a - c)} = \frac{a + c}{a(c - a)} - \frac{a + c}{c(c - a)}`

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

16. From previous problem, we have

    :math:`b = \frac{2ac}{a + c}`

    Substituting for :math:`b`

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

    :math:`= \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.


19. Since :math:`b + c, c + a, a + b` are in H. P.

    :math:`\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}` are in A. P.

    :math:`\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

    :math:`\frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b}` are in A. P.

20. Since :math:`b + c, c + a, a + b` are in H. P.

    :math:`\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`_.


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

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

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

..  _Arithmetic Progression Problems Part1: /algebra/arithmetic_progressions_problems_1/