Question -
Answer -
(i) 1/a, 1/b, 1/c are in AP
If 1/a, 1/b, 1/c arein 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 [bymultiplying with ‘c’ on both the numerator and denominator]
Let us consider RHS:
1/c – 1/b = (b-c)/bc
= a(b-c)/bc [bymultiplying with ‘a’ on both the numerator and denominator]
Since, (b+c)/a,(c+a)/b, (a+b)/c are in AP
(ii) bc, ca, ab are in AP
If bc, ca, ab are inAP then,
ca – bc = ab – ca
c (a-b) = a (b-c)
If 1/a, 1/b, 1/c arein AP then,
1/b – 1/a = 1/c – 1/b
c (a-b) = a (b-c)
Hence, the given termsare in AP