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If demotes the sum of first n terms of an A.P., prove that S12 = 3 (s8 – S4) . |
Answer» S8 = \(\frac{8}{2}\)[2a + 7d] = 4 (2a + 7d)..... (i) S4 = \(\frac{4}{2}\)[2a + 7d] = 2 (2a + 7d)....... (ii) L.H.S = S12 = \(\frac{12}{2}\)[2a + 11d] = 6 [2a + 11d] R.H.S = 3 (S8 – S4) = 3 [4 (2a + 7d) – 2 (2a + 3d)] [From (i) and (ii)] = 6 [4a + 14d – 2a – 3d] = 6 [2a + 11d] Since, L.H.S = R.H.S Hence, proved |
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