A number is the reciprocal of the other. If t | vedclass

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Question

(d) Suppose that required numbers $a$ and $b$.

Therefore according to the conditions $a = \frac{1}{b}$

and $\frac{{a + b}}{2} = \frac{{13}}{{12}

Vedclass · Accepted Answer

$\frac{3}{2},\;\frac{2}{3}$

Vedclass · Answer

(d) Suppose that required numbers $a$ and $b$.

Therefore according to the conditions $a = \frac{1}{b}$

and $\frac{{a + b}}{2} = \frac{{13}}{{12}}$$ \Rightarrow $$a + b = \frac{{13}}{6}$

$ \Rightarrow $ $a + \frac{1}{a} = \frac{{13}}{6} \Rightarrow 6{a^2} - 13a + 6 = 0$

$ \Rightarrow $ $\left( {a - \frac{3}{2}} \right)\,\left( {a - \frac{2}{3}} \right) = 0$

$ \Rightarrow $$a = \frac{3}{2}$ and $b = \frac{2}{3}$

or $a = \frac{2}{3}$ and $b = \frac{3}{2}$.

Trick : Find the $A.M.$ of option $(a), (b), (c), (d)$ one by one.

A number is the reciprocal of the other. If the arithmetic mean of the two numbers be $\frac{{13}}{{12}}$, then the numbers are

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