Question

if a,b,c are in A.P., show that a(1/b +1/c), b(1/c + 1/a), c(1/a + 1/b) are also in A.P.​

Answer 1

proof

Step-by-step explanation:

Given If abc are in ap;show that a(1÷b+1÷c),b(1÷c+1÷a),c(1÷a+1÷b)are also in ap

If abc are in A.P we need to show that a, b, c are in A.P. So we need to prove b-a = c - b

We have a(1/b + 1/c), b(1/c + 1/a), c(1/a + 1/b) are in A.P

So b(1/c + 1/a) - a(1/b + 1/c)

   b/c + b/a - a/b - a/c

  ab^2 + b^2c - a^2c - ab^2 / abc

  ab(b - a) + c(b^2 - a^2) /abc

  ab(b -a) + c((b + a)(b - a) / abc (a^2  - b^2 = (a + b)(a - b))

 (b - a)(ab + bc + ca ) / abc----------(1)

  c(1/a + 1/b) - b(1/c + 1/a)

 c/a + c/b - b/c - b/a

  bc^2 + ac^2 - ab^2 - b^2c / abc

  bc(c - b) + a(c^2 - b^2) / abc

  bc(c - b) + a(c + b)(c - b) / abc

  (c - b)(bc + ac + ab) / abc

 Now (c - b) = (b - a)

 So (b - a) (ab + bc + ca) / abc----------(2)

From 1 and 2 we get

difference between the terms is same. So the terms a(1÷b+1÷c),b(1÷c+1÷a),c(1÷a+1÷b) are ina.p.