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If $A_1, A_2$ be two arithmetic means between $\frac{1}{3}$ and $\frac{1}{24}$, then their values are
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The correct answer is:
$\frac{17}{72}, \frac{5}{36}$
Here $\frac{1}{3}, A_1, A_2, \frac{1}{24}$ will be in A.P.,
then $A_1-\frac{1}{3}=\frac{1}{24}-A_2 \Rightarrow A_1+A_2=\frac{3}{8}$\ldots(i)
Now, $A_1$ is a arithmetic mean of $\frac{1}{3}$ and $A_2$, we have
$2 A_1=\frac{1}{3}+A_2 \Rightarrow 2 A_1-A_2=\frac{1}{3}$\ldots(ii)
From (i) and (ii), we get,$A_1=\frac{17}{72}$ and $A_2=\frac{5}{36}$
Aliter: As we have formula $A_m=a+\frac{m(b-a)}{n+1}$
where, $n=2, a=\frac{1}{3}, b=\frac{1}{24}$
$\therefore \quad A_1=\frac{1}{3}+\frac{-7 / 24}{3}=\frac{17}{72}$
$A_2=\frac{1}{3}+\frac{-14 / 24}{3}=\frac{10}{72}=\frac{5}{36}$
then $A_1-\frac{1}{3}=\frac{1}{24}-A_2 \Rightarrow A_1+A_2=\frac{3}{8}$\ldots(i)
Now, $A_1$ is a arithmetic mean of $\frac{1}{3}$ and $A_2$, we have
$2 A_1=\frac{1}{3}+A_2 \Rightarrow 2 A_1-A_2=\frac{1}{3}$\ldots(ii)
From (i) and (ii), we get,$A_1=\frac{17}{72}$ and $A_2=\frac{5}{36}$
Aliter: As we have formula $A_m=a+\frac{m(b-a)}{n+1}$
where, $n=2, a=\frac{1}{3}, b=\frac{1}{24}$
$\therefore \quad A_1=\frac{1}{3}+\frac{-7 / 24}{3}=\frac{17}{72}$
$A_2=\frac{1}{3}+\frac{-14 / 24}{3}=\frac{10}{72}=\frac{5}{36}$
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