If (b+c-a)/a , (c+a-b)/b & (a+b-c)/c are in A.P., then prove that 1/a, 1/b & 1/c are also in A.P. ...... answer urgently needed!!

add 2 to each term then divide each term by (a+b+c)

  • 9
adding 1 ,b+c-a /a  + 1 ,c+a-b /b +1 , a+b-c/c + 1 =b+c/a ,c+a/b ,a+b/c.
adding 1 again, we get a+b+c/a ,a+b+c/b ,a+b+c/c
cancelling the numerator we get 1/a,1/b,1/c which are thus  also in A.P.
  • -10
b+c-a/a, c+a-b/b, a+b-c/c in AP 








ac-bc = ab-ac 
1/b -1/a = 1/c-1/b 
Hence they are in AP 
  • 12
It's depend on you how you want to solve it
  • 0
What are you looking for?