If a, b, c is a arithmetic progression, prove that the series:?

Question

1/ rtsq b + rtsq c , 1/rtsq c +rtsq a, 1/ rtsq a + rtsq b
is an arithmetical progression.

Answer 1

It will greatly help to "rationalize" the terms.

1/(√b + √c) = (√c - √b)/[(√c + √b)( √c - √b)] = (√c - √b)/(c-b) = (√c - √b)/d

where d is the common difference. Similarly,

1/(√c + √a) = (√c - √a)/(c-a) = (√c - √a)/(2d) and

1/(√a + √b) = (√b - √a)/(b-a) = (√b - √a)/d

I leave it to you to show that

(√b - √a)/d - (√c - √a)/(2d) = (√c - √a)/(2d) - (√c - √b)/d

Answer 2

Let a = 1, b =2 , c = 3. Certainly is arithmetic progression. Try these in your "prove" sequence.