A) \[\frac{18}{11}\]
B) \[\frac{22}{13}\]
C) \[\frac{20}{11}\]
D) \[\frac{16}{9}\]
Correct Answer: C
Solution :
nth term of the series is, \[{{T}_{n}}=\frac{1}{\frac{n(n+1)}{2}}=\frac{2}{n(n+1)}\] \[\Rightarrow \] \[{{T}_{n}}=2\left\{ \frac{1}{n}-\frac{1}{n+1} \right\}\] \[\Rightarrow \] \[{{T}_{1}}=2\left( \frac{1}{1}-\frac{1}{2} \right),\,\,{{T}_{2}}=2\,\left( \frac{1}{2}-\frac{1}{3} \right)\] \[{{T}_{3}}=2\,\left( \frac{1}{3}-\frac{1}{4} \right).....{{T}_{10}}=2\,\left( \frac{1}{10}-\frac{1}{11} \right)\] \[\therefore \] \[{{S}_{10}}={{T}_{1}}+{{T}_{2}}+.....+{{T}_{20}}\] \[{{S}_{10}}=2\left[ 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11} \right]\] \[=2\left( 1-\frac{1}{11} \right)=2\cdot \,\frac{10}{11}=\frac{20}{11}\].
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