If a, b, c are in A.P., show that \(\frac{1}{\sqrt{b} +\sqrt{c}}\),

1.	If a, b, c are in A.P., show that \(\frac{1}{\sqrt{b} +\sqrt{c}}\), \(\frac{1}{\sqrt{c} +\sqrt{a}}\), \(\frac{1}{\sqrt{a} +\sqrt{b}}\) are in A.P.
Answer» Given a, b, c are in A.P. ⇒ 2b = a + c Now , \(\frac{1}{\sqrt{b} +\sqrt{c}}\), \(\frac{1}{\sqrt{c} +\sqrt{a}}\), \(\frac{1}{\sqrt{a} +\sqrt{b}}\) are in A.P., if \(\frac{1}{\sqrt{c} +\sqrt{a}}\)- \(\frac{1}{\sqrt{b} +\sqrt{c}}\) = \(\frac{1}{\sqrt{a} +\sqrt{b}}\)- \(\frac{1}{\sqrt{c} +\sqrt{a}}\) ⇒ \(\frac{2}{\sqrt{c} +\sqrt{a}}\) = \(\frac{1}{\sqrt{b} +\sqrt{c}}\)+ \(\frac{1}{\sqrt{a} +\sqrt{b}}\) ⇒ \(\frac{2}{\sqrt{c} +\sqrt{a}}\) = \(\frac{\sqrt{a} + \sqrt{b} + \sqrt{b} + \sqrt{c}}{(\sqrt{b} + \sqrt{c}) (\sqrt{a} + \sqrt{b})}\) ⇒ 2(√b + √c)(√a +√b) = (√c +√a)(√a +2√b +√c) ⇒ 2(√ab + b + √ac + √bc) = (√ac + 2√bc + c + a + 2√ba + √ac) ⇒ 2√ab +2b +2√ac +2√bc = 2√ac + 2√bc + c + a + 2√ba ⇒ 2b = a + c, which is true as a, b, c are in A.P. ⇒ \(\frac{1}{\sqrt{b} +\sqrt{c}}\), \(\frac{1}{\sqrt{c} +\sqrt{a}}\), \(\frac{1}{\sqrt{a} +\sqrt{b}}\) are in A.P

If a, b, c are in A.P., show that \(\frac{1}{\sqrt{b} +\sqrt{c}}\), \(\frac{1}{\sqrt{c} +\sqrt{a}}\), \(\frac{1}{\sqrt{a} +\sqrt{b}}\) are in A.P.

Answer»

Given

a, b, c are in A.P. ⇒ 2b = a + c

Now ,

\(\frac{1}{\sqrt{b} +\sqrt{c}}\), \(\frac{1}{\sqrt{c} +\sqrt{a}}\), \(\frac{1}{\sqrt{a} +\sqrt{b}}\) are in A.P., if

\(\frac{1}{\sqrt{c} +\sqrt{a}}\)- \(\frac{1}{\sqrt{b} +\sqrt{c}}\) = \(\frac{1}{\sqrt{a} +\sqrt{b}}\)- \(\frac{1}{\sqrt{c} +\sqrt{a}}\)

⇒ \(\frac{2}{\sqrt{c} +\sqrt{a}}\) = \(\frac{1}{\sqrt{b} +\sqrt{c}}\)+ \(\frac{1}{\sqrt{a} +\sqrt{b}}\)

⇒ \(\frac{2}{\sqrt{c} +\sqrt{a}}\) = \(\frac{\sqrt{a} + \sqrt{b} + \sqrt{b} + \sqrt{c}}{(\sqrt{b} + \sqrt{c}) (\sqrt{a} + \sqrt{b})}\)

⇒ 2(√b + √c)(√a +√b) = (√c +√a)(√a +2√b +√c)

⇒ 2(√ab + b + √ac + √bc) = (√ac + 2√bc + c + a + 2√ba + √ac)

⇒ 2√ab +2b +2√ac +2√bc = 2√ac + 2√bc + c + a + 2√ba

⇒ 2b = a + c, which is true as a, b, c are in A.P.

⇒ \(\frac{1}{\sqrt{b} +\sqrt{c}}\), \(\frac{1}{\sqrt{c} +\sqrt{a}}\), \(\frac{1}{\sqrt{a} +\sqrt{b}}\) are in A.P