(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. .

Options

A. `a^(n)+c^(n)gtb^(n)`
B. `a^(n)+c^(n)gt2b^(n)`
C. `a^(n)+b^(n)gt2c^(n)`
D. `b^(n)+c^(n)gt2a^(n)`

Answer

B

Explanation

From (i) in illustration 1 we have
`(a+c)/(2)gtb` . . . (i)
We know that `(a^(n)+c^(n))/(2)gt((a+c)/(2))^(n)` . . . (ii)
From (i) and (ii), we get
`(a^(n)+c^(n))/(2)gtb^(n)rArra^(n)+c^(n)gt2b^(n)`