For any three positive real numbers a, b and c, 9(25a2 + b2 ) + 25(c2 – 3ac) = 15b(3a + c). Then :


For any three positive real numbers $a$, $b$ and $c$, $9\left(25 a^{2}+b^{2}\right)+25\left(c^{2}-3 a c\right)=15 b(3 a+c)$. Then $:$

  1. a,b and care 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.

Correct Option: 3

JEE Main Previous Year 1 Question of JEE Main from Mathematics Sequences and Series chapter.
JEE Main Previous Year 2017


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