Question -
Answer -
(i) (b+c)/a, (c+a)/b,(a+b)/c are in A.P.
We know that, if a, b,c are in AP, then b – a = c – b
If, 1/a, 1/b, 1/c arein AP
Then, 1/b – 1/a = 1/c– 1/b
If (b+c)/a, (c+a)/b,(a+b)/c are in AP
Then, (c+a)/b –(b+c)/a = (a+b)/c – (c+a)/b
Let us take LCM
Since, 1/a, 1/b, 1/c arein AP
1/b – 1/a = 1/c – 1/b
C (b – a) = a (b-c)
Hence, the given termsare in AP.
(ii) a(b + c), b(c + a),c(a + b) are in A.P.
We know that if, b(c +a) – a(b+c) = c(a+b) – b(c+a)
Consider LHS:
b(c + a) – a(b+c)
Upon simplification weget,
b(c + a) – a(b+c) = bc+ ba – ab – ac
= c (b-a)
Now,
c(a+b) – b(c+a) = ca +cb – bc – ba
= a (c-b)
We know,
1/a, 1/b, 1/c are inAP
So, 1/a – 1/b = 1/b –1/c
Or c(b-a) = a(c-b)
Hence, given terms arein AP.