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

A

b, c and a are in GP

B

b, c and a are in AP

C

a, b and c are in AP

D

a, b and c are in GP

Text Solution

Verified by Experts

The correct Answer is:
B
We have,
`225a^(2) + 9b^(2) + 25c^(2) -- 75ac - 45ab - 15 bc = 0`
`rArr (15a)^92) + (3b)^(2) + (5c)^(2) - (15a) (5c) - (15a) (3b) - (3b) (5c) = 0`
`rArr (1)/(2) [(15a - 3b)^(2) + (3b -5c)^(2) + (5c - 15a)^(2)] = 0`
`rArr 15a = 3b, 3b = 5c and 5c = 15a`
`:. 15a = 3b = 5c`
`rArr (a)/(1) = (b)/(5) = (c)/(3) = lamda` (say)
`rArr a = lamda, b = 5 lamda, c = 3 lamda`
`:. b, c, a` are in AP.
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