If A_1, A_2 are two arithmetic means between | vedclass

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(B) Let the sequence be $\frac{1}{3}, A_1, A_2, \frac{1}{24}$. This is an arithmetic progression $(A.P.)$ with $n=4$ terms.The common difference $d$ i

Vedclass · Accepted Answer

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

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(B) Let the sequence be $\frac{1}{3}, A_1, A_2, \frac{1}{24}$. This is an arithmetic progression $(A.P.)$ with $n=4$ terms.The common difference $d$ is given by $d = \frac{b-a}{n+1}$,where $a = \frac{1}{3}$,$b = \frac{1}{24}$,and $n=2$ (number of means).$d = \frac{\frac{1}{24} - \frac{1}{3}}{2+1} = \frac{\frac{1-8}{24}}{3} = \frac{-7/24}{3} = -\frac{7}{72}$.Now,$A_1 = a + d = \frac{1}{3} - \frac{7}{72} = \frac{24-7}{72} = \frac{17}{72}$.$A_2 = A_1 + d = \frac{17}{72} - \frac{7}{72} = \frac{10}{72} = \frac{5}{36}$.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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