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If a, b, c are in G.P., prove that : (a + 2b + 2c) (a – 2b + 2c) = a2 + 4c2. |
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Answer» 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 + 4a 2r4 = a2(1 + 4r4) Clearly, LHS = RHS Hence proved |
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