Harmonic Progressions Questions and Answers

Consider \frac{1}{6}, \frac{1}{4}, \frac{1}{3}, ...... ∞

Here T_{2} – T_{1} = T_{3} – T_{2} = \frac{1}{12} ⇒ \frac{1}{6}, \frac{1}{4}, \frac{1}{3} is an A.P.

6th term of this A.P. = \frac{1}{6} + 5 \times  \frac{1}{12} = \frac{1}{6} + \frac{5}{12} = \frac{7}{12},

And the 9th term = \frac{1}{6} + 8 \times \frac{1}{12} = \frac{5}{6}.

Hence the 9th term of the H.P. = \frac{6}{5} and the 6th term = \frac{12}{7}

Harmonic Mean = \frac{3abc}{ab +bc +ca}

= \frac{3\times 2\times 4\times 6}{2\times 4 + 4\times 6 + 6\times 2}

= \frac{144}{8 +24 +12} = 3.27

The given distribution is grouped data and the variable involved is the ages of first year students, while the number of students represents frequencies.

Calculations are as given below:

If x, y, z are in h.p, then \frac{1}{x} , \frac{1}{y} ,\frac{1}{z} will be in A.P. series and \frac{1}{y} will be the AM of x, z.

Let z be the harmonic mean,

z = \frac{2}{\frac{1}{x} + \frac{1}{y}} = \frac{2xy}{x+y}

Explanation:

Now from the above HP formula, it is clear the reciprocals of first 9 terms will make an AP.

The sum of first 9 terms of an AP =\frac{9}{2} [2a + (9 – 1) d]  = 81

⇒ 2a + 8d = 18

⇒ a + 4d = 9

Now there are 2 variables, but a + 4d = T_{5} in an AP series.

And reciprocal of 5th term of AP series will give the 5th term of corresponding HP series.

So, the 5th term of HP series is \frac{1}{9}

If a, a + d, a + 2d, a + 3d, ……. are in A.P.

Then \frac{1}{a}, \frac{1}{a+d},\frac{1}{a+2d}, \frac{1}{a+3d}, ....   are in H.P.

Now \frac{1}{(a+2d)} = 5  and \frac{1}{(a+5d)} = 8

Solving these two equations we get a = \frac{1}{4}

d =\frac{-1}{40}

Now a+10d = (\frac{1}{4}) + (\frac{-10}{40}) = 0

Hence the maximum terms that this H.P. can take is 10.

The terms of the HP are

\frac{1}{16}, \frac{1}{13} , \frac{1}{10}, \frac{1}{7}, \frac{1}{4},\frac{1}{1},\frac{1}{-2}\frac{1}{-5}

So the maximum partial sum is

\frac{1}{16} + \frac{1}{13} + \frac{1}{10} + \frac{1}{7} +\frac{1}{4} + \frac{1}{1} = 1.63