For any three positve real numbers a,b and c, `9(25a ^+b^2)+25(c^2-3ac)=15 b (3a+c)` then

Options

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

Answer

D

Explanation

We have,
`9(25a^(2)+b^(2))(c^(2)-3ac)=15(3a+c)`
`rArr225a^(2)+9b^(2)+25c^(2)-45ab-75ac=0`
`rArr(15a)^(2)+(3b)^(2)+(5c)^(2)-(15c)^(2)-(15a)(3b)-(3b)(5c)-(15a)(5c)=0`
`rArrx^(2)+y^(2)+z^(2)-xy-yz-zx=0`, where x=15a,y=3b,z=5c
`rArr2(x^(2)+y^(2)+z^(2)-xy-yz-zx)=0`
`rArr(x-y)^(2)+(y-z)^(2)+(z-x)^(2)=0`
`rArrx-y=0,y-z=0,z-x=0`
`rArrx=y=z`
`rArr15a=3b=5c`
`rArr(a)/(1)=(b)/(5)=(c)/(3)`
`rArr(a)/(1)=(b)/(5)=(c)/(3)=lamda("say")`
`rArra=lamda,b=5lamda,c=3lamda`
`rArr` a,c and b are in A.P. Also, b,c and a are in A.P.