Given that \( a, b, c \) are in Arithmetic Progression (AP), we have to prove that \(\displaystyle \frac{1}{\sqrt{b}+\sqrt{c}}, \frac{1}{\sqrt{c}+\sqrt{a}}, \frac{1}{\sqrt{a}+\sqrt{b}} \) are also in AP.
Solution:
Since \( a, b, c \) are in Arithmetic Progression (AP), we have:
\( 2b = a + c \dots \text{(1)} \) and \(a-b=b-c \dots \text{(2)}\)
Let the given terms be:
\(\displaystyle x = \frac{1}{\sqrt{b} + \sqrt{c}}, \quad y = \frac{1}{\sqrt{c} + \sqrt{a}}, \quad z = \frac{1}{\sqrt{a} + \sqrt{b}} \)
To prove that \( x, y, z \) are in AP, we have to show that \( 2y = x + z \).
First find \( x+z \):
\( \begin{aligned} x + z &= \frac{1}{\sqrt{b} + \sqrt{c}} + \frac{1}{\sqrt{a} + \sqrt{b}} \\ &= \frac{\sqrt{b}\, – \sqrt{c}}{(\sqrt{b} + \sqrt{c})(\sqrt{b}\, – \sqrt{c})} + \frac{\sqrt{a}\, – \sqrt{b}}{(\sqrt{a} + \sqrt{b})(\sqrt{a}\, – \sqrt{b})} \\ &= \frac{\sqrt{b}\, – \sqrt{c}}{b\, – c} + \frac{\sqrt{a}\, – \sqrt{b}}{a\, – b} \\ &= \frac{\sqrt{b}\, – \sqrt{c}}{a\, – b} + \frac{\sqrt{a}\, – \sqrt{b}}{a\, – b} \quad \because a-b=b-c \text{ from (2)} \\ &= \frac{\sqrt{b}\, – \sqrt{c} + \sqrt{a}\, – \sqrt{b}}{a\, – b} \\ &= \frac{\sqrt{a}\, – \sqrt{c}}{a\, – b} \\ \end{aligned} \)Now find \( 2y \):
\( \begin{aligned} 2y &= 2 \frac{1}{\sqrt{c} + \sqrt{a}} \\ &= 2 \frac{\sqrt{c}\, – \sqrt{a}}{(\sqrt{c} + \sqrt{a})(\sqrt{c}\, – \sqrt{a})} \\ &= 2 \frac{\sqrt{c}\, – \sqrt{a}}{c\, – a} = 2 \frac{\sqrt{c}\, – \sqrt{a}}{2b -a -a} \quad \because c=2b-a \text{ from (1)}\\ &= \frac{2(\sqrt{c}\, – \sqrt{a})}{2(b-a)} \\ &= \frac{\sqrt{a}\, – \sqrt{c}}{a-b} \\ \end{aligned} \)Thus \(x + z = 2y \). Hence, the given expressions are also in AP.