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Question: If b is the harmonic mean between a and c, prove that \(\dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{a}+\...

If b is the harmonic mean between a and c, prove that 1b−a+1b−c=1a+1c\dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{a}+\dfrac{1}{c}.

Explanation

Solution

We first use the condition of HP for the given numbers where 1a+1c=2b\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{2}{b}. We simplify the condition and multiply with abcabc. Then we form the equation taking difference of two numbers to find the given form of 1b−a+1b−c=1a+1c\dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{a}+\dfrac{1}{c}.

Complete step by step solution:
As b is the harmonic mean between a and c, we can write 1a+1c=2b\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{2}{b}.
We simplify the expression by multiplying with abcabc and get
abc(1a+1c)=2abcb ⇒bc+ab=2ac \begin{aligned} & abc\left( \dfrac{1}{a}+\dfrac{1}{c} \right)=\dfrac{2abc}{b} \\\ & \Rightarrow bc+ab=2ac \\\ \end{aligned}
We can write the equation as
bc+ab=2ac ⇒ac−bc=ab−ac \begin{aligned} & bc+ab=2ac \\\ & \Rightarrow ac-bc=ab-ac \\\ \end{aligned}
We take the common terms and get
ac−bc=ab−ac ⇒−c(b−a)=a(b−c) \begin{aligned} & ac-bc=ab-ac \\\ & \Rightarrow -c\left( b-a \right)=a\left( b-c \right) \\\ \end{aligned}
We now write them as dfraction form and get
−c(b−a)=a(b−c) ⇒ab−a=−cb−c \begin{aligned} & -c\left( b-a \right)=a\left( b-c \right) \\\ & \Rightarrow \dfrac{a}{b-a}=\dfrac{-c}{b-c} \\\ \end{aligned}
We multiply 1b\dfrac{1}{b} both sides to get
ab−a=−cb−c ⇒ab(b−a)=−cb(b−c) \begin{aligned} & \dfrac{a}{b-a}=\dfrac{-c}{b-c} \\\ & \Rightarrow \dfrac{a}{b\left( b-a \right)}=\dfrac{-c}{b\left( b-c \right)} \\\ \end{aligned}
We form the numerators with respect to the denominators and get
ab(b−a)=−cb(b−c) ⇒b−(b−a)b(b−a)=(b−c)−bb(b−c) \begin{aligned} & \dfrac{a}{b\left( b-a \right)}=\dfrac{-c}{b\left( b-c \right)} \\\ & \Rightarrow \dfrac{b-\left( b-a \right)}{b\left( b-a \right)}=\dfrac{\left( b-c \right)-b}{b\left( b-c \right)} \\\ \end{aligned}
Simplifying we get
b−(b−a)b(b−a)=(b−c)−bb(b−c) ⇒1(b−a)−1b=1b−1(b−c) ⇒1b−a+1b−c=2b ⇒1b−a+1b−c=1a+1c \begin{aligned} & \dfrac{b-\left( b-a \right)}{b\left( b-a \right)}=\dfrac{\left( b-c \right)-b}{b\left( b-c \right)} \\\ & \Rightarrow \dfrac{1}{\left( b-a \right)}-\dfrac{1}{b}=\dfrac{1}{b}-\dfrac{1}{\left( b-c \right)} \\\ & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{2}{b} \\\ & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{a}+\dfrac{1}{c} \\\ \end{aligned}
Thus proved the given expression.

Note: A sequence of numbers is said to be a harmonic progression if the reciprocal of those numbers is in AP. We actually convert the given numbers into their AP form to find the given expression of 1b−a+1b−c=1a+1c\dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{a}+\dfrac{1}{c} as 1a,1b,1c\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c} are in AP. The easier way to solve the problem is to go for back process and get the HP relation of the given numbers from 1b−a+1b−c=1a+1c\dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{a}+\dfrac{1}{c} and then solve the problem the way it is done in the solution part.