# (solved)Question 14 SSC-CGL 2020 March 3 Shift 1

If $\displaystyle x^{2a} = y^{2b} = z^{2c}$ and x² = yz, then (ab + bc + ca)/bc = ?

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### Question 14SSC-CGL 2020 Mar 3 Shift 1

If $\displaystyle x^{2a} = y^{2b} = z^{2c}$ and x² = yz, then (ab + bc + ca)/bc = ?

### Solution in Brief

Observe that if a =b = c = x = y = z = 1, all the given conditions are satisfied. So with a = b = c = 1, the required expression (ab + bc + ca)/bc becomes 3 ans!

### Solution in Detail

Let $\displaystyle x^{2a} = y^{2b} = z^{2c} = k$

Take $\displaystyle x^{2a} = k$

$\displaystyle \implies x = k^{1/2a}$

Likewise $\displaystyle y = k^{1/2b}$ and $\displaystyle z = k^{1/2c}$

Given $\displaystyle x^2 = yz$

$\displaystyle \implies (k^{1/2a})^2 = k^{1/2b} \cdot k^{1/2c}$

$\displaystyle \implies k^{2/2a} = k^{(1/2b + 1/2c)}$

Bases are same, so powers must also be same

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

Add $\displaystyle \frac{1}{2a}$ to both sides

$\displaystyle \frac{3}{2a} = \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c}$

$\displaystyle \therefore \frac{3}{2a} = \frac{bc + ca + ab}{2abc}$

Canceling $\displaystyle 2a$,

$\displaystyle 3 = \frac{bc + ca + ab}{bc}\:\underline{Ans}$

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