Question

If m be a root of the equation $ {{x}^{2}}\left( 1-ab \right)-x\left( {{a}^{2}}+{{b}^{2}} \right)-\left( 1+ab \right)=0 $ and m harmonic means are inserted between a and b, then prove that $ {{H}_{1}}-{{H}_{m}}=ab\left( a-b \right) $ .

Answer 1

Hint:
 In this question, we are given m harmonic means between a and b and we need to find the difference between $ {{n}^{th}} $ harmonic mean and first harmonic mean. For this we will first suppose the given series and then use the formula of AP to find the common difference. Using that, we will find $ {{H}_{1}}\text{ and }{{H}_{m}} $ . At last, taking the difference between them, we can prove them as $ ab\left( a-b \right) $ . We will also use $ {{m}^{2}}\left( 1-ab \right)-m\left( {{a}^{2}}+{{b}^{2}} \right)-\left( 1+ab \right)=0 $ as m is root of the equation.

Complete step by step answer:
Here we are given that, there are m harmonic means inserted between a and b. Let these harmonic means be $ {{H}_{1}},{{H}_{2}},\ldots \ldots \ldots {{H}_{m}} $ . So our series looks like this,
 $ \dfrac{1}{a},\dfrac{1}{{{H}_{1}}},\dfrac{1}{{{H}_{2}}},\dfrac{1}{{{H}_{3}}},\cdots \cdots ,\dfrac{1}{{{H}_{m}}},\dfrac{1}{b}, $ .
Since $ {{H}_{1}},{{H}_{2}},\ldots \ldots \ldots {{H}_{m}} $ are harmonic means, so this series is in AP. It has (m+2) terms where first term is $ \dfrac{1}{a} $ . Let us calculate common difference in terms of a and b. Since $ \dfrac{1}{b} $ is the $ {{\left( m+2 \right)}^{th}} $ term of AP and we know $ {{n}^{th}} $ term of an AP is given by $ {{a}_{n}}=a+\left( n-1 \right)d $ . So we have,
\[\begin{align}
  & \dfrac{1}{b}=\dfrac{1}{a}+\left( m+2-1 \right)d \\
 & \Rightarrow \dfrac{1}{b}=\dfrac{1}{a}+\left( m+1 \right)d \\
 & \Rightarrow \left( \dfrac{1}{b}-\dfrac{1}{a} \right)=\left( m+1 \right)d \\
 & \Rightarrow d=\dfrac{1}{\left( m+1 \right)}\left( \dfrac{1}{b}-\dfrac{1}{a} \right) \\
\end{align}\].
So we have obtained a common difference.
Now the second term of AP is HP which will be given by $ {{a}_{2}}=a+d $ .
So $ \dfrac{1}{{{H}_{1}}}=\dfrac{1}{a}+\dfrac{1}{\left( m+1 \right)}\left( \dfrac{1}{b}-\dfrac{1}{a} \right) $ .
Taking LCM of ab on second term we have, $ \dfrac{1}{{{H}_{1}}}=\dfrac{1}{a}+\dfrac{1\left( a-b \right)}{\left( m+1 \right)ab} $ .
Taking LCM as (m+1)ab we get, \[\dfrac{1}{{{H}_{1}}}=\dfrac{\left( m+1 \right)b+\left( a-b \right)}{\left( m+1 \right)ab}\Rightarrow \dfrac{1}{{{H}_{1}}}=\dfrac{mb+b+a-b}{\left( m+1 \right)ab}\].
Simplifying we get, \[\dfrac{1}{{{H}_{1}}}=\dfrac{mb+a}{\left( m+1 \right)ab}\].
Taking reciprocal on both sides we get, \[\dfrac{1}{{{H}_{1}}}=\dfrac{\left( m+1 \right)ab}{mb+a}\].
Now let us calculate $ {{H}_{m}} $ .
For this, we see that $ \dfrac{1}{{{H}_{m}}} $ is the $ {{\left( m+1 \right)}^{th}} $ term of AP so \[{{a}_{n+1}}=a+nd\].
Hence we get $ \dfrac{1}{{{H}_{m}}}=\dfrac{1}{a}+m\left( \dfrac{1}{\left( m+1 \right)}\left( \dfrac{1}{b}-\dfrac{1}{a} \right) \right) $ .
Taking LCM of ab on second term, $ \dfrac{1}{{{H}_{m}}}=\dfrac{1}{a}+m\left( \dfrac{a-b}{ab\left( m+1 \right)} \right) $ .
Taking LCM as ab(m+1) we get $ \dfrac{1}{{{H}_{m}}}=\dfrac{\left( m+1 \right)b+m\left( a-b \right)}{ab\left( m+1 \right)}\Rightarrow \dfrac{1}{{{H}_{m}}}=\dfrac{mb+b+ma-mb}{ab\left( m+1 \right)} $ .
Simplifying we get, $ \dfrac{1}{{{H}_{m}}}=\dfrac{b+ma}{ab\left( m+1 \right)} $ .
Taking reciprocal on both sides we get, \[\dfrac{1}{{{H}_{m}}}=\dfrac{\left( m+1 \right)ab}{b+ma}\].
Now let us find the difference between $ {{H}_{1}}\text{ and }{{H}_{m}} $ we get,
 $ {{H}_{1}}-{{H}_{m}}=\dfrac{\left( m+1 \right)ab}{mb+a}-\dfrac{\left( m+1 \right)ab}{ma+b} $ .
Taking (m+1)ab common from both terms we get, $ \Rightarrow \left( m+1 \right)ab\left( \dfrac{1}{mb+a}-\dfrac{1}{ma+b} \right) $ .
Taking LCM as (mb+a)(ma+b) we get,
\[\begin{align}
  & \Rightarrow \left( m+1 \right)ab\left( \dfrac{ma+b-mb-a}{\left( mb+a \right)\left( ma+b \right)} \right) \\
 & \Rightarrow \left( m+1 \right)ab\left( \dfrac{ma-mb-a+b}{\left( mb+a \right)\left( ma+b \right)} \right) \\
 & \Rightarrow \left( m+1 \right)ab\left( \dfrac{m\left( a-b \right)-1\left( a-b \right)}{\left( mb+a \right)\left( ma+b \right)} \right) \\
 & \Rightarrow \dfrac{\left( m+1 \right)ab\left( m-1 \right)\left( a-b \right)}{\left( mb+a \right)\left( ma+b \right)} \\
\end{align}\].
Using $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ we get, $ \Rightarrow \dfrac{ab\left( {{m}^{2}}-1 \right)\left( a-b \right)}{\left( mb+a \right)\left( ma+b \right)} $ .
Simplifying the denominator we get,
 $ {{H}_{1}}-{{H}_{m}}=\dfrac{ab\left( {{m}^{2}}-1 \right)\left( a-b \right)}{{{m}^{2}}ab+am{{a}^{2}}+m{{b}^{2}}+ab}\cdots \cdots \cdots \left( 1 \right) $ .
Now we are given that m is the root of $ {{x}^{2}}\left( 1-ab \right)-x\left( {{a}^{2}}+{{b}^{2}} \right)-\left( 1+ab \right)=0 $ therefore,
 $ {{m}^{2}}\left( 1-ab \right)-m\left( {{a}^{2}}+{{b}^{2}} \right)-\left( 1+ab \right)=0 $ .
Simplifying we get,
 $ \begin{align}
  & {{m}^{2}}-{{m}^{2}}ab-m{{a}^{2}}-m{{b}^{2}}-1-ab=0 \\
 & \Rightarrow -{{m}^{2}}ab-m{{a}^{2}}-m{{b}^{2}}-ab=1-{{m}^{2}} \\
\end{align} $ .
Taking negative sign common from both sides we get,
 $ {{m}^{2}}ab+m{{a}^{2}}+m{{b}^{2}}+ab=1-{{m}^{2}} $ .
Putting this value in equation (1) we get,
 $ {{H}_{1}}-{{H}_{m}}=\dfrac{ab\left( {{m}^{2}}-1 \right)\left( a-b \right)}{\left( {{m}^{2}}-1 \right)} $ .
Cancelling $ \left( {{m}^{2}}-1 \right) $ from numerator and denominator we get,
 $ {{H}_{1}}-{{H}_{m}}=ab\left( a-b \right) $ .
Hence proved.

Note:
 This sum involves very complex calculations, so take care while solving. Solve it step by step taking care of signs. Harmonic progression is just reciprocal of the terms of an arithmetic progression. If a value is the root of the equation then it satisfies the equation.