If a-b-c - 2 - a2 - b2 - c2 - 1 and abc -3 then 1 a - 1 b - 1 c is equal to

The correct option is A 1 2 Given that, #160; a+b+c=2 , a^2+b^2+c^2=1 abc=3 Considering #160; a+b+c=2 and squaring both sides #1 ...

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Standard X

Mathematics

Sum of Infinite Terms of a GP

If a+b+c = ...

Question

If $a + b + c = 2$ , $a^{2} + b^{2} + c^{2} = 1$ and $a b c = 3$ then $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is equal to

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\frac{1}{2}

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none of these

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Solution

The correct option is A $\frac{1}{2}$
Given that,

$a + b + c = 2, a^{2} + b^{2} + c^{2} = 1 a b c = 3$

Considering $a + b + c = 2$ and squaring both sides
$⟹ a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a = 4$

$⟹ 2 (a b + b c + c a) = 4 - 1$

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

$⟹ 2 \times 3 \times (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = 3$

$⟹ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2}$

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If a+b+c=2 , a2+b2+c2=1 and abc=3 then 1a+1b+1c is equal to

The correct option is A 12Given that, a+b+c=2,a2+b2+c2=1abc=3Considering a+b+c=2 and squaring both sides⟹a2+b2+c2+2ab+2bc+2ca=4⟹2(ab+bc+ca)=4−1⟹2(a)(b)(c)(1a+1b+1c)=3⟹2×3×(1a+1b+1c)=3⟹1a+1b+1c=12

If $a + b + c = 2$ , $a^{2} + b^{2} + c^{2} = 1$ and $a b c = 3$ then $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is equal to