If a, b, c are in G.P., prove that :

(a + 2b + 2c) (a – 2b + 2c) = a2 + 4c2.

As,


a, b, c are in G.P, let r be the common ratio.


Therefore,


b = ar … (1)


c = ar2 … (2)


To prove: (ab + bc + cd)2 = (a + 2b + 2c) (a – 2b + 2c) = a2 + 4c2


As, LHS = (a + 2b + 2c) (a – 2b + 2c)


LHS = (a + 2ar + 2ar2)(a – 2ar + 2ar2)


LHS = a2(1 + 2r + 2r2)(1 – 2r + 2r2)


LHS = a2 (1 + 4r2 + 4r4 – 4r2)


LHS = a2(1 + 4r4)


And RHS = a2 + 4a2r4 = a2(1 + 4r4)


Clearly, LHS = RHS


Hence proved


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