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For any three positive real numbers a, b...

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

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

Text Solution

Verified by Experts

The correct Answer is:
C
`(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
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Knowledge Check

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    `1:2:3`
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    C
    `2:3:4`
    D
    `1:2:4`

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