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Question

If a,b and c are positive real number such that 9(25a2+b2)+25(c23ac)=15b(3a+c), then which of the following is INCORRECT?


Correct Answer
A
a,b and c are in A.P.
Your Answer
B
b,c and a are in A.P.
Your Answer
C
a,2ca and 2ba are in A.P.
Your Answer
D
a+1,c+1 and b+1 are in A.P. 

Solution

The correct option is B a,b and c are in A.P.
9(25a2+b2)+25(c23ac)=15b(3a+c)225a2+9b2+25c275ac45ab15bc=012[(15a3b)2+(15a5c)2+(3b5c)2]=015a=3b=5c
a1=b5=c3=λa=λ, b=5λ, c=3λ2c=a+b
b,c and a are in A.P.
a,c and b are in A.P.
a+1,c+1 and b+1 are in A.P.
a,2ca and 2ba are in A.P.

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