Question

If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then
(a) a, b, c are in AP
(b) a², b², c² are in AP
(c) 1/1, 1/b, 1/c are in AP
(d) None of these​

Answer 1

Answer:

If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then

(a) a, b, c are in AP

$\rm\underline\bold{b)\:a², b², c² are in AP \red{\huge{\checkmark}}}$

(c) 1/1, 1/b, 1/c are in AP

(d) None of these

Explanation

Given, 1/(b + c), 1/(c + a), 1/(a + b)

⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)

⇒ 2b² = a² + c²

⇒ a², b², c² are in AP