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.

Harmonic mean
¯X=(∑f)/(∑(f/x)) = \frac{30}{1.7765} = 16.8871 ≈17 years

Calculations are as given below:

Harmonic Mean = \frac{100}{1.866} =53.59 ≈54

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:

HM = \frac{5}{0.4338} = 11.526

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

If  a, a+d, a+2d… are in A.P. then

\frac{1}{a}, \frac{1}{(a+d)}, \frac{1}{(a+2d)} … are in H.P.

Now \frac{1}{a+5d} = \frac{9}{106}  => a+5d = \frac{106}{9}

And \frac{1}{a+6d} = \frac{9}{122} => a+6d = \frac{122}{9}

Solving these two equations, we get

a= \frac{26}{9} and d=\frac{16}{9}

Now a+d = \frac{14}{3} and a+4d = \frac{10}{1}

So the sum of second and fifth term of H.P is \frac{3}{14} + \frac{1}{10} = \frac{11}{35}

As we know that 1/13, 1/a and 1/16 are in arithmetic progression.

or \frac{1}{a}-\frac{1}{13} = \frac{1}{16}-\frac{1}{a}

\frac{-1}{13}-\frac{1}{16} = \frac{-1}{a}-\frac{1}{a}

\frac{-16-13}{208} = \frac{-2}{a}

29a = 416

a = 14.34

G²=A.H=12²=13XH

144/13=H

Harmonic mean=144/13

(b+c), (c+a), (a+b) are in h.p.

\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} will be in A.P.

t₂-t₁=t₃-t₂

=\frac{1}{c+a}-\frac{1}{b+c}=\frac{1}{a+b}-\frac{1}{c+a}

By solving the equation we get:

(b-a)(a+b)=(c-b)(b+c)

b²-a²=c²-b²

a²,b²,c² are in A.P.

\frac{1}{a^{2}} , \frac{1}{b^{2}} , \frac{1}{c^{2}} are in H.P.

As we see the above series is in A.P.

So a_{n}=a+(n-1)d=2+(n-1)5=

5n-3

So nth term of h.p. = \frac{1}{(5n-3)}

Here a₁₀=3+(10-1)4

=39

So 10^th term will be 1/39.

H.P. Formula:

harmonic mean of a and b = \frac{2ab}{a+b} = \frac{2}{a+b}

As we know the above series is in AP thus

a_₇ = 2 + (7-1)3 = 20

\frac{1}{20} is h_₇ term of the series.

We have: {5, 9, 7, 10, 8, 5}

H.M = \frac{6}{\frac{1}{5} + \frac{1}{9} + \frac{1}{7} + \frac{1}{10} + \frac{1}{8} + \frac{1}{5}} = 6.82

GM^{2} = AM \times HM

HM = \frac{65^{2}}{100} = 42.25

GM^{2} = AM \times HM

HM = \frac{90^{2}}{75} = 108

Formula is

H.M = \frac{n}{\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}}+...\frac{1}{a_{n}}}

for number 17,22,13,26 harmonic mean is -
\frac{4}{( \frac{1}{17} + \frac{1}{22} + \frac{1}{13} + \frac{1}{26} )}
= \frac{4}{0.05+0.04+0.07+0.03}
harmonic mean = 21.05

The mean of the reciprocals of the terms = 1/150 and 1/132.

Thus, the total of the reciprocals of the terms in set i=1/150×25=1/6
and in set ii=1/132×12=1/11

The sum of the reciprocals of all 37 components is 1/6+1/11=17/66.

Hence their mean is 17/66÷37=9.53.

HM =2xy/(x+y)

So x=19 and y=57

H.M=(2*19*57)/(19+57)

H M=2166/76=28.50

HM =2xy/(x+y)

So x=18 and y=37

H.M=(2*18*37)/(18+37)

H.M=1332/55=24.21.

Reciprocal of this=1/24.21

G²=A.H=26²=27*H

676/27=H

Harmonic mean=25.03

Here  a_{11}=4+(11-1)4

=44

So 11^th term will be 1/44.

¹⁄_p, ¹⁄q, ¹⁄_r will be in a.m. series and ¹⁄q will be the A.M of p, r.

Let w be the harmonic mean, ²⁄_w = ¹⁄_u + ¹⁄_v,

W= \frac{2uv}{(u+v)}

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