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AnswersIf the \[{p^{th}},{q^{th}}\,\& \,{r^{th}}\]terms of a H.P. area, b & c respectively, then the value of \[ab(p - q) + bc(q - r) + ca(r - p)\]is
Answer
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Hint: A harmonic progression (HP) is defined as a sequence of real numbers which is determined by taking the reciprocals of the arithmetic. Progression that does not contain O.In the HP, any term in the sequence is considered as the Harmonic mean of its two neighbors for example,
The sequence a, b, c, d is considered as an arithmetic progression, the harmonic progression can be written as \[\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},\dfrac{1}{d}\]
Complete step by step solution:
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
Note:
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
\[AM,GM,HM\,are\,in\,G.P\]
\[ \to AM \geqslant G.M \geqslant H.M\,\]
The sequence a, b, c, d is considered as an arithmetic progression, the harmonic progression can be written as \[\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},\dfrac{1}{d}\]
Complete step by step solution:
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
Note:
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
\[AM,GM,HM\,are\,in\,G.P\]
\[ \to AM \geqslant G.M \geqslant H.M\,\]