InterviewSolution
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InterviewSolution
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If a, b, c are in A.P., prove that:(i) (a – c)2 = 4 (a – b) (b – c)(ii) a2 + c2 + 4ac = 2 (ab + bc + ca)(iii) a3 + c3 + 6abc = 8b3 |
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Answer» (i) (a – c)2 = 4 (a – b) (b – c) Let us expand the above expression a2 + c2 – 2ac = 4(ab – ac – b2 + bc) a2 + 4c2b2 + 2ac – 4ab – 4bc = 0 (a + c – 2b)2 = 0 a + c – 2b = 0 Since a, b, c are in AP b – a = c – b a + c – 2b = 0 a + c = 2b Hence, (a – c)2 = 4 (a – b) (b – c) (ii) a2 + c2 + 4ac = 2 (ab + bc + ca) Let us expand the above expression a2 + c2 + 4ac = 2 (ab + bc + ca) a2 + c2 + 2ac – 2ab – 2bc = 0 (a + c – b)2 – b2 = 0 a + c – b = b a + c – 2b = 0 2b = a + c b = (a + c)/2 Since a, b, c are in AP b – a = c – b b = (a + c)/2 Hence, a2 + c2 + 4ac = 2 (ab + bc + ca) (iii) a3 + c3 + 6abc = 8b3 Let us expand the above expression a3 + c3 + 6abc = 8b3 a3 + c3 – (2b)3 + 6abc = 0 a3 + (-2b)3 + c3 + 3a(-2b)c = 0 Since, if a + b + c = 0, a3 + b3 + c3 = 3abc (a – 2b + c)3 = 0 a – 2b + c = 0 a + c = 2b b = (a + c)/2 Since a, b, c are in AP a – b = c – b b = (a + c)/2 Hence, a3 + c3 + 6abc = 8b3 |
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