Question

If a, b, c are distinct positive real in H.P., then the value of the expression, $\frac{b+a}{b-a}+\frac{b+c}{b-c}$ is equal to

Answer 1

2

Answer 2

Understanding Harmonic Progression (H.P.):

If a, b, c are in H.P., then their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in Arithmetic Progression (A.P.).

The property of an A.P. states that the middle term is the average of the other two terms. So, for $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ in A.P.:

$$\frac{1}{b} = \frac{\frac{1}{a} + \frac{1}{c}}{2}$$

Multiplying both sides by 2, we get:

$$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$$

To simplify, find a common denominator on the right side:

$$\frac{2}{b} = \frac{c+a}{ac}$$

This is the fundamental relationship for a, b, c in H.P.

Manipulating the Expression:

The expression we need to evaluate is $E = \frac{b+a}{b-a}+\frac{b+c}{b-c}$.

Let $x = \frac{b}{a}$ and $y = \frac{b}{c}$.

Then, the relation becomes $x+y=2$.

Substitute $x$ and $y$ into the expression:

$$E = \frac{x+1}{x-1} + \frac{y+1}{y-1}$$

From $x+y=2$, we have $y = 2-x$. Substitute this into the second term:

$$\frac{y+1}{y-1} = \frac{(2-x)+1}{(2-x)-1} = \frac{3-x}{1-x}$$

Now substitute this back into the expression for E:

$$E = \frac{x+1}{x-1} + \frac{3-x}{1-x}$$

To combine these fractions, make the denominators the same. Note that $1-x = -(x-1)$:

$$E = \frac{x+1}{x-1} + \frac{3-x}{-(x-1)}$$ $$E = \frac{x+1}{x-1} - \frac{3-x}{x-1}$$

Now combine the numerators over the common denominator:

$$E = \frac{(x+1) - (3-x)}{x-1}$$ $$E = \frac{x+1-3+x}{x-1}$$ $$E = \frac{2x-2}{x-1}$$

Factor out 2 from the numerator:

$$E = \frac{2(x-1)}{x-1}$$

Since a, b, c are distinct positive real numbers, $a \neq b$. This means $\frac{b}{a} \neq 1$, so $x \neq 1$. Therefore, $x-1 \neq 0$, and we can cancel the $(x-1)$ terms.

$$E = 2$$

Thus, the value of the expression is 2.