Question

(a-b),(b-c),(c-a) in g.P prove that (a+b+c)^2=3(ab+bc+ca)

Answer 1

Ratios of any two consecutive numbers are same in GP

Answer 2

(a - b), (b - c) , (c - a) are in G.P

so, (b - c)/(a - b) = (c - a)/(b - c) [ common ratio remains same ]

⇒(b - c)² = (c - a)(a - b)

⇒b² + c² - 2bc = ca - bc - a² + ab

⇒a² + b² + c² = ab + bc + ca

⇒a² + b² + c² + 2(ab + bc + ca) = ab + bc + ca + 2(ab + bc + ca)

= 3(ab + bc + ca)

⇒(a + b + c)² = 3(ab + bc + ca) [ hence proved]

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