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



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