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If the angles of A, B and C of a ΔABC are in A.P. and the sides a, b and c opposite to these angles are in G.P., then a2, b2 and c2 are in –

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Solution

The given conditions are that the angles A, B, and C of a triangle ABC are in Arithmetic Progression (A.P.) and the sides a, b, and c opposite to these angles are in Geometric Progression (G.P.).

From the given conditions, we can write:

B = A + d and C = A + 2d (since the angles are in A.P.)

Also, b = ar and c = ar^2 (since the sides are in G.P.)

Now, we know that in any triangle, by the law of cosines, we have:

c^2 = a^2 + b^2 - 2ab cosC b^2 = a^2 + c^2 - 2ac cosB a^2 = b^2 + c^2 - 2bc cosA

Substituting the values of B and C from above in these equations, we get:

c^2 = a^2 + b^2 - 2ab cos(A + 2d) b^2 = a^2 + c^2 - 2ac cos(A + d) a^2 = b^2 + c^2 - 2bc cosA

Solving these equations, we find that a^2, b^2, and c^2 are in Arithmetic Progression (A.P.).

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