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The sequence a, b, c, d is considered as an arithmetic progression, the harmonic progression can be written as \[\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},\dfrac{1}{d}\]

Given that a, b, c is the \[{p^{th}},{q^{th}},{r^{th}}\] terms of a H.P.

We know that H.P is the reciprocal of the arithmetic progression so,

\[\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},\dfrac{1}{d}\]are the \[{p^{th}},{q^{th}},{r^{th}}\] term of A.P.

Let the A.P have \[x\] as the first term and as the common difference

\[\dfrac{1}{a} = x + (p - 1)d......(1)\]

\[\dfrac{1}{b} = \,x + \,(q - 1)d\].......(2)

\[\dfrac{1}{c} = x + (r - 1)d......(3)\]

From the question we need \[(p - q)\], so we subtract \[e{q^n}(2)\] from \[e{q^n}(1)\], so we get

\[\dfrac{1}{a} - \dfrac{1}{b} = x + (p - 1)d - (x + (q - 1)d\]

\[ = \not x + (p - 1)d - \not x(q - 1)d\]

\[ = \not x + (p - 1)d - \not x(q - 1)d\]

We took d common from the equation we have

\[\dfrac{1}{a} - \dfrac{1}{b} = (p - q)d\]

\[e{q^n}(2)\]

\[d = \dfrac{{b - a}}{{ab(p - q)}}\]

Similarly,

We subtract \[e{q^n}\] from \[e{q^n}(2)\]so we get

\[\dfrac{1}{b} - \dfrac{1}{c} = (q - r)d\]

\[d = \dfrac{{c - b}}{{cb(q - r)}}\]

\[d = \dfrac{{c - b}}{{cb(q - r)}}\]

Similarly

We subtract \[e{q^n}(1)\]from \[e{q^n}(3)\] we get

\[\dfrac{1}{c} - \dfrac{1}{a} = \,(r - p)d\]

\[d = \dfrac{{a - c}}{{ac(r - p)}}\]

Now we have 3 values of d

i.e.

\[d = \dfrac{{b - a}}{{ab(p - q)}} = \dfrac{{c - b}}{{bc(q - r)}} = \dfrac{{a - v}}{{ac(r - p)}}\]

We know that if

\[x = \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f}\] then

then\[x = \dfrac{{a + c + e}}{{b + d + f}}\]

Now,\[d = \dfrac{{\not b - \not a + \not c - \not b + \not a - \not c}}{{ab(p - q) + cb(q - r) + ac(r - p)}}\]

\[d = \dfrac{O}{{ab(p - q) + cb(q - r) + ac(r - p)}}\]

We took the denominator in opposite side,

\[ab(p - q) + cb(q - r) + ac(r - p)d = O\]

\[ab(p - q) + cb(q - r) + ac(r - p) = O\]

Hence,

\[ab(p - q) + cb(q - r) + ac(r - p)\] is equal to Zero

Harmonic mean is calculated as the reciprocal of the arithmetic mean of the reciprocals

The formula to calculate the harmonic mean is given by

\[Harmonic\,Mean = \dfrac{h}{{\left[ {\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} + \dfrac{1}{d}....} \right]}}\]

a, b, c, d are the values and n are the number of value present

relation between AM, GM and HM

\[ \to AM \geqslant G.M \geqslant H.M\,\] where

\[AM,GM,HM\,are\,in\,G.P\]

\[ \to AM \geqslant G.M \geqslant H.M\,\]