Question

If $\dfrac{1}{a}$ , $\dfrac{1}{H}$, $\dfrac{1}{b}$ are in AP, then $\dfrac{{\left( {H + a} \right)}}{{\left( {H - a} \right)}} + \dfrac{{\left( {H + b} \right)}}{{\left( {H - b} \right)}} = $
A) $2$
B) $4$
C) $0$
D) $1$

Answer 1

Hint: In the given question we use the concept of Arithmetic Progression. Here, we will use the common difference formula to find the terms of the final expressions that are to be found. To do so equate the common difference between ${1^{st}}$ and ${2^{nd}}$ terms with ${2^{nd}}$ and ${3^{rd}}$ terms. Then add and subtract $a$ and $b$ from both the sides of equation one by one to get the values of $\left( {H + a} \right)$ , $\left( {H - a} \right)$, $\left( {H + b} \right)$ and $\left( {H - b} \right)$ respectively and substitute in the final expression to get the answer.

Complete step by step Solution:
 Given AP: $\dfrac{1}{a}$ , $\dfrac{1}{H}$, $\dfrac{1}{b}$
Common difference: $\dfrac{1}{H} - \dfrac{1}{a} = \dfrac{1}{b} - \dfrac{1}{H}$
 $\dfrac{2}{H} = \dfrac{1}{a} + \dfrac{1}{b}$
Taking LCM on the right side of the equation, we get
 $\dfrac{2}{H} = \dfrac{{a + b}}{{ab}}$
Inverting the equations, we get
 $\dfrac{H}{2} = \dfrac{{ab}}{{a + b}}$
 $H = \dfrac{{2ab}}{{a + b}}$ ...(1)
Adding $a$ to both the sides of the equation (1), we get
 $H + a = \dfrac{{2ab}}{{a + b}} + a$
 $H + a = \dfrac{{2ab + a\left( {a + b} \right)}}{{a + b}}$
Taking $a$ common from the numerator, we get
 $H + a = \dfrac{{a(a + 3b)}}{{a + b}}$ ...(2)
Subtracting $a$ from both sides of the equation (1), we get
 $H - a = \dfrac{{2ab}}{{a + b}} - a$
 $H - a = \dfrac{{2ab - a\left( {a + b} \right)}}{{a + b}}$
Again taking $a$ common from the numerator, we get
 $H - a = \dfrac{{a(b - a)}}{{a + b}}$ ...(3)
Similarly, by adding $b$ to both sides of the equation (1), we get
 $H + b = \dfrac{{2ab}}{{a + b}} + b$
 $H + b = \dfrac{{2ab + b\left( {a + b} \right)}}{{a + b}}$
Taking $b$ common from the numerator, we get
 $H + b = \dfrac{{b(3a + b)}}{{a + b}}$ ...(4)
Now subtracting $b$ from both sides of the equation (1), we get
 $H - b = \dfrac{{2ab}}{{a + b}} - b$
 $H - b = \dfrac{{2ab - b\left( {a + b} \right)}}{{a + b}}$
Again taking $a$ common from the numerator, we get
 $H - b = \dfrac{{b(a - b)}}{{a + b}}$ ...(5)
Now we substitute equations (2), (3), (4), and (5) in equation (1), and we get
 $\dfrac{{\left( {H + a} \right)}}{{\left( {H - a} \right)}} + \dfrac{{\left( {H + b} \right)}}{{\left( {H - b} \right)}} = \dfrac{{\dfrac{{a(a + 3b)}}{{a + b}}}}{{\dfrac{{a(b - a)}}{{a + b}}}} + \dfrac{{\dfrac{{b(3a + b)}}{{a + b}}}}{{\dfrac{{b(a - b)}}{{a + b}}}}$
 $\dfrac{{\left( {H + a} \right)}}{{\left( {H - a} \right)}} + \dfrac{{\left( {H + b} \right)}}{{\left( {H - b} \right)}} = \dfrac{{(a + 3b)}}{{(b - a)}} + \dfrac{{(3a + b)}}{{(a - b)}}$
Taking a negative sign out from the denominator of the second term, we get
 $\dfrac{{\left( {H + a} \right)}}{{\left( {H - a} \right)}} + \dfrac{{\left( {H + b} \right)}}{{\left( {H - b} \right)}} = \dfrac{{(a + 3b)}}{{(b - a)}} - \dfrac{{(3a + b)}}{{(b - a)}}$
 $\dfrac{{\left( {H + a} \right)}}{{\left( {H - a} \right)}} + \dfrac{{\left( {H + b} \right)}}{{\left( {H - b} \right)}} = \dfrac{{a + 3b - 3a - b}}{{(b - a)}}$
Solving further, we get
 $\dfrac{{\left( {H + a} \right)}}{{\left( {H - a} \right)}} + \dfrac{{\left( {H + b} \right)}}{{\left( {H - b} \right)}} = \dfrac{{2b - 2a}}{{(b - a)}}$
Taking 2 commons from the numerator of the right-hand side of the equation, we get
 $\dfrac{{\left( {H + a} \right)}}{{\left( {H - a} \right)}} + \dfrac{{\left( {H + b} \right)}}{{\left( {H - b} \right)}} = \dfrac{{2(b - a)}}{{(b - a)}}$
Thus we get, $\dfrac{{\left( {H + a} \right)}}{{\left( {H - a} \right)}} + \dfrac{{\left( {H + b} \right)}}{{\left( {H - b} \right)}} = 2$

Therefore, the correct option is A.

Note:More such questions can be given in Geometric Progression too where the formula of common difference is not the same as that of Arithmetic Progression. Common Difference for Geometric Progression (GP) $x,y,z$ is $\dfrac{y}{x}$ or $\dfrac{z}{y}$ i.e., $\dfrac{{{n^{th}}term}}{{{{(n + 1)}^{th}}term}}$ .