Assertion :If a, b, c are three positive real numbers such that a+c≠0 and 1a+1a−b+1c+1c−b=0 then a,b,c are in H.P Reason: If a,b,c are distinct positive real numbers such that a(b−c)x2+b(c−a)xy+c(a−b)y2 is a perfect square, then a,b,c are in H.P.
A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Assertion is correct but Reason is incorrect
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Assertion is incorrect but Reason is correct
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C Assertion is correct but Reason is incorrect Rewrite the given expression as 1a+1c−b+1c+1a−b=0 ⇒a+c−ba(c−b)+a+c−bc(a−b)=0 ⇒a(c−b)=c(b−a)⇒b=2aca+c For statement-2, put t=x/y, and write the expression as y2[a(b−c)t2+b(c−a)t+c(a−b)]