(i) a , b, c are in H.P. , show that ` (b + a)/(b -a) + (b + c)/(b - c) = 2 `
(ii) If ` a^(2), b^(2), c^(2) ` are A.P. then b + c , c + a , a + b are in H.P. .

Explanation

(i) As a, b, c are in H.P.
` rArr b =(2ac)/(a+c) ` …(A)
`rArr (b)/(a) = (2c) /(a+c)`
` rArr (b + a)/(b-a) = (3c + a)/(c + a)" "` …(B) (From componendo and dividendo )
Rather then finding ` (b+c)/(b-c)` from (A) , observe that as a functioin of a and c, d is symmetric
in a and c. So changing a to c in (B) we get
` (b + c)/(b -a) =(3 a + C)/(a-c)` ...(C)
Adding (B) and (C) ,
`(b +a)/(b-a) +(b + c)/(b-c) = (3c+a)/(a-c) = ((3c+a)-(3a+c))/(c-a) = (2c -2c)/(c-a) = 2. (because c ne a) `
(ii) ` a^(2) b^(2) , c^(2)` are in A.P.
` rArr a^(2) + ab + bc + ca , b^(2) + ab + bc + ca, c^(2) + ab + bc + ca ` are in A.P. ...(A)
[Adding ` ab + bc + ca ` to each term of the A.P. ]
Now, ` a^(2) + ab + bc + ca = a^(2) + ab + ac + bc `
`= a^(2) + a(b + c) +be `
= ` (a + b ) (a + c) ,` etc .
So (A) reads
` (a + b ) (a + c), (b + a) (b + c), (c + a) (c + b) 1` are in A.P.
` rArr (1)/(b +c), (1)/(c +a), (1)/(a + b)` are in A.P. [ Dividing each term by `( a + b ) (b + c) (c + a)]`
`rArr b + c , c + a , a + b ` are in H.P.`