If $A_{1}, A_{2}$ are two arithmetic means between $\frac{1}{3} and \frac{1}{24}$ , then their values are

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$\frac{7}{72}, \frac{25}{36}$

$\frac{17}{72}, \frac{5}{36}$

$\frac{7}{36}, \frac{5}{72}$

$\frac{5}{72}, \frac{17}{72}$

answer is B.

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Detailed Solution

Given, $A_{1}, A_{2}$ are two arithmetic means between $\frac{1}{3} and \frac{1}{24}$ then

$\frac{1}{3}, A_{1}, A_{2}, \frac{1}{24} are in AP.$

Here, $a = \frac{1}{3}, l = \frac{1}{24}, n = 4$

Let d be the common difference.

Then, $l = a + (n - 1) d$

$\begin{aligned} \Rightarrow \frac{1}{24} = \frac{1}{3} + (4 - 1) d \\ \Rightarrow \frac{1}{24} - \frac{1}{3} = 3 d \\ \Rightarrow \frac{1 - 8}{24} = 3 d \\ \Rightarrow - \frac{7}{24} = 3 d \Rightarrow d = - \frac{7}{72} \\ ∴ A_{1} = a + d^{'} = \frac{1}{3} - \frac{7}{72} = \frac{24 - 7}{72} = \frac{17}{72} \\ and A_{2} = a + 2 d = \frac{1}{3} - \frac{14}{72} = \frac{24 - 14}{72} = \frac{10}{72} = \frac{5}{36} \end{aligned}$