For any three positive real numbers a, b and c, `9(25a^2+b^2)+25(c^2-3ac)=15b(3a+c)` Then: (1) b, c and a are in G.P. (2) b, c and a are in A.P. (3) a, b and c are in A.P (4) a, b and c are in G.P

Options

A. a,b and c are in G.P
B. b,c and a are in G.P
C. b,c and a are in A.P
D. a,b and c are in A.P

Answer

C

Explanation

`(15a)^2+(3b)^2+(5c)^2-(15a)(5c)-(15a)(3b)-(3b)(5c)=0`
`rArr 1/2 [ (15 a -3b)^2+(3b - 5c)^2+(5c-15a)^2+(5c -15 a)^2]=0`
`rArr 15a=3b =5c `
`rArr b=(5c)/3,a=c/3`
`rArr a+b =2c`
Thus, b,c and a are in A.P