Q44 / 90 · JEE Main 2017 (02 Apr) · Mathematics

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

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

Solution

225 a^{2} + 9 b^{2} + 25 c^{2} - 75 a c - 45 a b - 15 b c = 0

\Rightarrow 2 ({(15 a)}^{2} + {(3 b)}^{2} + {(5 c)}^{2} - (15 a) (5 c) - (15 a) (3 b) - (3 b) (5 c) = 0)

\Rightarrow 2 {(15 a)}^{2} + 2 {(3 b)}^{2} + 2 {(5 c)}^{2} - 2 (15 a) (5 c) - 2 (15 a) (3 b) - 2 (3 b) (5 c) = 0

\Rightarrow {(15 a - 3 b)}^{2} + {(3 b - 5 c)}^{2} + {(5 c - 15 a)}^{2} = 0

\Rightarrow 15 a - 3 b = 3 b - 5 c = 5 c - 15 a = 0

\Rightarrow 15 a = 3 b = 5 c

\Rightarrow \frac{a}{1} = \frac{b}{5} = \frac{c}{3} = k

\Rightarrow a = k, b = 5 k, c = 3 k

\Rightarrow a, c, b a r e i n A . P .

\Rightarrow b, c, a a r e i n A . P .

Hence, the correct answer is option: 'B'.

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