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Question
If the angles A, B, C of ΔABC are in A.P. and its sides a, b, c are in G.P., then show that a2, b2, c2 are in A.P.
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Solution
A, B, C are in A.P.
∴ A + C = 2B
We know that A + B + C = 180°
∴ 2B + B = 180°
∴ 3B = 180°
∴ ∠B = 60° .......(i)
Also, it is given that sides a, b, c are in G.P.
∴ ac = b2 .......(ii)
Consider, cos B = `("a"^2 + "c"^2 - "b"^2)/(2"ac")` .......[By cosine rule]
∴ cos(60°) = `("a"^2 + "c"^2 - "b"^2)/(2"b"^2)` .......[From (i) and (ii)]
∴ `1/2 = ("a"^2 + "c"^2 - "b"^2)/(2"b"^2)`
∴ b2 = a2 + c2 – b2
∴ a2 + c2 = 2b2
∴ a2, b2, c2 are in A.P.
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