Please solve this question again and kindly explain how the denominator get reduced Q 5 If 1 a + b , 1 b + c and 1 c + a are in A . P . , then which of the following relations is correct ? ( A ) 2 b 2 = a 2 + c 2 ( B ) 2 a 2 = b 2 + c 2 ( C ) a 2 + b 2 + c 2 = 0 ( D ) 2 c 2 = a 2 + b 2 Solution It is given that 1 a + b , 1 b + c and 1 c + a are in A . P ∴ 1 b + c - 1 a + b = 1 c + a - 1 b + c ⇒ a + b - ( b + c ) ( b + c ) ( a + b ) = b + c - ( c + a ) ( c + a ) ( b + c ) ⇒ a - c a + b = b - a c + a ⇒ a - c a + c = b + a b - a ⇒ a 2 - c 2 = b 2 - a 2 ⇒ 2 a 2 = b 2 + c 2 Share with your friends Share 0 Neha Sethi answered this Dear student The question involves the concept of A.P. Arithmatic progression,the scope of which goes much beyond the ambit of your syllabus,but stillI am helping you with the calculations.We have,1b+c-1a+b=1c+a-1b+c⇒a+b-b+ca+bb+c=b+c-c+ac+ab+c⇒a+b-b-ca+b=b+c-c-ac+a⇒a-ca+b=b-ac+a⇒a-cc+a=b+ab-a⇒a2-c2=b2-a2 using x2-y2=x+yx-y⇒2a2=b2+c2 Regards 0 View Full Answer Vipasana A S answered this It is already solved 0 Abhyush Dixit answered this Yes it is 0 Sumit Shinde answered this It is already solve. 1 Sumit Shinde answered this My friend it will already solve. But what you ask pls tell me fast. 1