If b + c a , c + a b , a + b c are in A.P., prove that: 1 a , 1 b , 1 c are in A.P.

Given that, b + c a , c + a b , a + b c are in A.P., To prove : 1 a , 1 b , 1 c are in A.P. Therfore, b + c a , c + a b , a + b c are in A.P. ⇒ b + c a + 1 , c + a b + 1 , a + b c + 1 are also in A.P. (adding 1 to each term) ⇒ a + b + c a , a + b + c b , a + b + c c are in A.P. ⇒ 1 a , 1 b , 1 c are in A.P. (dividing each term by a + b + c ) Hence, proved.

If b + c a , c + a b , a + b c are in A.P., prove that: bc , ca , ab are in A.P.

Given that, b + c a , c + a b , a + b c are in A.P. To prove : bc , ca , ab are in A.P. Now, b + c a + 1 , c + a b + 1 , a + b c + 1 also in A.P. (adding 1 in each term) ⇒ a + b + c a , a + b + c b , a + b + c c are in A.P. ⇒ 1 a , 1 b , 1 c are in A.P. (dividing each term by a + b + c ) ⇒ abc a , abc b , abc c are in A.P. (multiplying each term by abc ) ⇒ bc , ca , ab are in A.P. ⇒ ab , bc , ca are in A.P. Hence, Proved.

If a 1 b + 1 c , b 1 c + 1 a , c 1 a + 1 b are in A.P., prove that a , b , c are in A.P.

Given that, a 1 b + 1 c , b 1 c + 1 a , c 1 a + 1 b are in A.P. ⇒ ac + ba bc , ba + bc ac , ca + cb ab are in A.P. ⇒ ac + ba bc + 1 , ba + bc ac + 1 , ca + cb ab + 1 are in A.P. (Adding 1 to each term) ⇒ ab + bc + ca bc , ab + bc + ca ac , ab + bc + ca ab ⇒ 1 bc , 1 ac , 1 ab are in A.P. (Dividing each term by ab + bc + ca ) ⇒ abc bc , abc ac , abc ab are in A.P. (Multiplying each term by abc ) ⇒ a , b , c are in A.P. Hence, Proved.