SATHEE: Sequences and Series 1 Question 3

Sequences and Series 1 Question 3

3.

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(2017 Main)

(a) $b, c$ and $a$ are in GP

(b) $b, c$ and $a$ are in AP

(c) $a, b$ and $c$ are in $A P$

(d) $a, b$ and $c$ are in GP

Show Answer

Answer:

Correct Answer: 3. (b)

Solution:

We have,

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

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

$\Rightarrow \frac{1}{2} [(15 a - 3 b)^{2} + (3 b - 5 c)^{2} + (5 c - 15 a)^{2}] = 0$

$\Rightarrow 15 a = 3 b, 3 b = 5 c$ and $5 c = 15 a$

$∴ 15 a = 3 b = 5 c$

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

$\Rightarrow a = λ, b = 5 λ, c = 3 λ$

$∴ b, c, a$ are in AP.

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