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(D) Let the two numbers be $a$ and $b$.Given that one number is the reciprocal of the other,so $a = \frac{1}{b}$ or $ab = 1$.The arithmetic mean of th

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

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(D) Let the two numbers be $a$ and $b$.Given that one number is the reciprocal of the other,so $a = \frac{1}{b}$ or $ab = 1$.The arithmetic mean of the two numbers is given as $\frac{a + b}{2} = \frac{13}{12}$.This implies $a + b = \frac{13}{6}$.Substituting $b = \frac{1}{a}$ into the equation,we get $a + \frac{1}{a} = \frac{13}{6}$.Multiplying by $6a$,we get $6a^2 - 13a + 6 = 0$.Factoring the quadratic equation: $6a^2 - 9a - 4a + 6 = 0 \Rightarrow 3a(2a - 3) - 2(2a - 3) = 0$.$(3a - 2)(2a - 3) = 0$.Thus,$a = \frac{2}{3}$ or $a = \frac{3}{2}$.If $a = \frac{3}{2}$,then $b = \frac{2}{3}$. If $a = \frac{2}{3}$,then $b = \frac{3}{2}$.Therefore,the numbers are $\frac{3}{2}$ and $\frac{2}{3}$.

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

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