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| 1. | If (b + c)/a, (c + a)/b, (a + b)/c are in AP., prove that:(i) 1/a, 1/b, 1/c are in AP(ii) bc, ca, ab are in AP | 
| Answer» (i) 1/a, 1/b, 1/c are in AP If 1/a, 1/b, 1/c are in AP then, 1/b – 1/a = 1/c – 1/b Let us consider LHS: 1/b – 1/a = (a - b)/ab = c(a - b)/abc [by multiplying with ‘c’ on both the numerator and denominator] Let us consider RHS: 1/c – 1/b = (b - c)/bc = a(b - c)/bc [by multiplying with ‘a’ on both the numerator and denominator] Since, (b + c)/a, (c + a)/b, (a + b)/c are in AP a(b - c) = c(a - b) LHS = RHS Hence, the given terms are in A.P (ii) bc, ca, ab are in AP If bc, ca, ab are in AP then, ca – bc = ab – ca c (a - b) = a (b - c) If 1/a, 1/b, 1/c are in AP then, 1/b – 1/a = 1/c – 1/b c (a - b) = a (b - c) Hence, the given terms are in AP | |