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## For any three positive real numbers a,b and c 9(25a^2+b^2)+25(c^2-3ac)=15b(3a+c).Then-JEE MAINS-Mathematics-2017

(1)a, b and c are in G.P.

(2) b, c and a are in G.P.

(3) b, c and a are in A.P.

(4) a, b and c are in A.P.

By:Purnima

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|akialwayz

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## Answers

(15a)^2 + (3b)^2 + (5c)^2 – (15a)(5c) – (15a)(3b)– (3b)(5c) = 0

1[(15a – 3b)2 + (3b – 5c)2 + (5c – 15a)2]/2=0

it is possible when 15a = 3b = 5c

b =5c/3 , a =c/3

a + b = 2c

Thus, b, c, a in A.P.

leo

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