If $a^{2} + b^{2} + c^{2} = 2 (a + 2 b - 2 c) - 9$ then find the value of a+b+c

2

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3

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1

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none of these

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Solution

The correct option is C $1$
$a^{2} + b^{2} + c^{2} = 2 (a + 2 b - 2 c) - 9$
$a^{2} - 2 a + b^{2} - 4 b + c^{2} + 4 c + 9 = 0$
$a^{2} - 2 (a) (1) + (1)^{2} + b^{2} - 2 (b) (2) + (2)^{2} + c^{2} + 2 (c) (2) + (2)^{2} = 0$
$(a - 1)^{2} + (b - 2)^{2} + (c + 2)^{2} = 0$
$⟹ a = 1, b = 2, c = - 2$
$a + b + c = 1 + 2 - 2 = 1$

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If a, b, c are in G.P., prove that:

(i) $a (b^{2} + c^{2}) = c (a^{2} + b^{2})$

(ii) $a^{2} b^{2} c^{2} (\frac{1}{a^{3}} + \frac{1}{b^{3}} + \frac{1}{c^{3}}) = a^{3} + b^{3} + c^{3}$

(iii) $\frac{(a + b + c)^{2}}{a^{2} + b^{2} + c^{2}} = \frac{a + b + c}{a - b + c}$

(iv) $\frac{1}{a^{2} - b^{2}} + \frac{1}{b^{2}} = \frac{1}{b^{2} - c^{2}}$

(v) $(a + 2 b + 2 c) (a - 2 b + 2 c) = a^{2} + 4 c^{2}$ .

\frac{(a + b + c)^{2}}{a^{2} + b^{2} + c^{2}} = \frac{c o t \frac{1}{2} A + c o t \frac{1}{2} B + c o t \frac{1}{2} C}{c o t A + c o t B + c o t C}

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