A) \[5\frac{1}{5}\]
B) \[3\frac{1}{5}\]
C) 1/10
D) 10
Correct Answer: D
Solution :
Series, \[2,\ 2\frac{1}{2},\ 3\frac{1}{3},.......\]are in H.P. \[\Rightarrow \]\[\frac{1}{2},\ \frac{2}{5},\ \frac{3}{10},.......\] will be in A.P. Now first term \[a=\frac{1}{2}\] and common difference \[d=-\frac{1}{10}\] So, \[=\frac{2}{3}(10+{{10}^{2}}+{{10}^{3}}+..........+\] term of the A.P. \[=\frac{1}{2}+(5-1)\left( -\frac{1}{10} \right)=\frac{1}{10}\] Hence \[{{5}^{th}}\] term in H.P. is 10.
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