If A_1, A_2 are two arithmetic means between | vedclass

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(B) Let the sequence be $\frac{1}{3}, A_1, A_2, \frac{1}{24}$ in an $A.P.$Here,the first term $a = \frac{1}{3}$ and the fourth term $b = \frac{1}{24}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the sequence be $\frac{1}{3}, A_1, A_2, \frac{1}{24}$ in an $A.P.$Here,the first term $a = \frac{1}{3}$ and the fourth term $b = \frac{1}{24}$.In an $A.P.$ with $n$ arithmetic means,the common difference $d$ is given by $d = \frac{b - a}{n + 1}$.Here $n = 2$,so $d = \frac{\frac{1}{24} - \frac{1}{3}}{2 + 1} = \frac{\frac{1 - 8}{24}}{3} = \frac{-7/24}{3} = -\frac{7}{72}$.Now,$A_1 = a + d = \frac{1}{3} - \frac{7}{72} = \frac{24 - 7}{72} = \frac{17}{72}$.And $A_2 = a + 2d = \frac{1}{3} + 2\left(-\frac{7}{72}\right) = \frac{1}{3} - \frac{7}{36} = \frac{12 - 7}{36} = \frac{5}{36}$.Thus,the values are $\frac{17}{72}$ and $\frac{5}{36}$.

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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