10) The geometric and harmonic means of two numbers x₁ and x₂ are 18 and 16$\frac{8}{13}$ respectively. The value of $\left|x_1-x_2\right| is$
(a) 5
(b) 10
(c) 15
(d) 20

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$$\begin{gathered} Given : Two numbers x_{1 }and x_2 ,Geometric mean \left(GM\right) of the numbers is 18 and Harmonic mean\left(HM\right) of the numbers is 16\frac{8}{13}. \\ To Calculate : \left|x_1-x_2\right| \\ Solution : GM of x_1 and x_2=\sqrt{x_1 x_2} \\ \Rightarrow 18=\sqrt{x_1 x_2} \\ \Rightarrow x_1 x_2={18}^2 \\ \Rightarrow x_1 x_2=324..................................................................................\overline{)1} \\ \Rightarrow x_2=\frac{324}{x_1}..................................................................................\overline{)2} \\ Now, \\ HM of x_1 and x_2=\frac{2 x_1 x_2}{x_1+x_2} \\ \Rightarrow 16\frac{8}{13}=\frac{2 x_1 x_2}{x_1+x_2} \\ \Rightarrow \frac{2 x_1 x_2}{x_1+x_2}=\frac{216}{13} \\ \Rightarrow \frac{2 \times 324}{x_1+x_2}=\frac{216}{13} [from \overline{)1}] \\ \Rightarrow \frac{3}{x_1+x_2}=\frac{1}{13} \\ Cross multiplying \\ \Rightarrow x_1+x_2=39 \\ \Rightarrow x_1+\frac{324}{x_1}=39 \\ \Rightarrow \frac{{x_1}^2+324}{x_1}=39 \\ \Rightarrow {x_1}^2+324=39 x_1 \\ \Rightarrow {x_1}^2-39x_1+324=0 \\ \Rightarrow {x_1}^2-27x_1-12x_1+324=0 \\ \Rightarrow x_1(x_1-27)-12(x_1-27)=0 \\ \Rightarrow (x_1-27)(x_1-12)=0 \\ either x_1=27 or x_1=12 \\ putting these values in \overline{)2} \\ either x_2=\frac{324}{27} or x_2=\frac{324}{12} \\ either x_2=12 or x_2=27 \\ So, (x_1, x_2) be either (27, 12) or (12, 27) \\ So, \left|x_1-x_2\right|=\left|27-12\right| or \left|12-27\right| \\ So, \left|x_1-x_2\right|=\left|15\right| or \left|-15\right| \\ Hence, \left|x_1-x_2\right|=15 \end{gathered}$$

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