Child Labor 14

Algebra Level 4

$\begin{cases} \dfrac{1}{a^2} + \dfrac{1}{ab} + \dfrac{1}{b^2} = 3 \\ \dfrac{1}{b^2} + \dfrac{1}{bc} + \dfrac{1}{c^2} = 1 \\ \dfrac{1}{c^2} + \dfrac{1}{ca} + \dfrac{1}{a^2} = 2 \end{cases}$

Given that $a,b$ and $c$ satisfy the system of equations above, then $a,b$ and $c$ follows a/an:

Harmonic Progression Geometric Progression Arithmetic Progression None of these choices

1 solution

Rajen Kapur
Oct 21, 2015

Multiply the first equation by $(\dfrac{1}{a} - \dfrac{1}{b})$ to get $(\dfrac{1}{a^3} - \dfrac{1}{b^3}) =( \dfrac {3}{a} - \dfrac{3}{b})$ . Similarly, the next two by $(\dfrac{1}{b} - \dfrac{1}{c})$ and $(\dfrac{1}{c} - \dfrac{1}{a})$ , respectively. Adding all the three equations, $0 = (\dfrac {3}{a} - \dfrac{3}{b}) +( \dfrac {1}{b} - \dfrac{1}{c})+ ( \dfrac {2}{c} - \dfrac{2}{a})$ which simplifies to $\dfrac {1}{a} + \dfrac{1}{c} = \dfrac {2}{b}$ . Hence a, b, c are in H.P.

Child Labor 14

1 solution

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