A) \[\frac{7}{72},\,\frac{5}{36}\]
B) \[\frac{17}{72},\,\frac{5}{36}\]
C) \[\frac{7}{36},\,\frac{5}{72}\]
D) \[\frac{5}{72},\,\frac{17}{72}\]
Correct Answer: B
Solution :
Here \[\frac{1}{3},\ {{A}_{1}},\ {{A}_{2}},\ \frac{1}{24}\] will be in A.P., then \[{{A}_{1}}-\frac{1}{3}=\frac{1}{24}-{{A}_{2}}\]\[\Rightarrow \]\[{{A}_{1}}+{{A}_{2}}=\frac{3}{8}\] ......(i) Now, \[{{A}_{1}}\] is a arithmetic mean of \[\frac{1}{3}\] and \[{{A}_{2}}\], we have \[2{{A}_{1}}=\frac{1}{3}+{{A}_{2}}\Rightarrow 2{{A}_{1}}-{{A}_{2}}=\frac{1}{3}\] ......(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 \[n=2,\ a=\frac{1}{3},\ b=\frac{1}{24}\] \[\therefore \] \[{{A}_{1}}=\frac{1}{3}+\frac{-7/24}{3}=\frac{17}{72}\] \[{{A}_{2}}=\frac{1}{3}+\frac{-14/24}{3}=\frac{10}{72}=\frac{5}{36}\]
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