A) \[1\frac{10}{17}\]
B) \[\frac{10}{17}\]
C) \[\frac{16}{27}\]
D) \[\frac{17}{27}\]
Correct Answer: A
Solution :
Given series \[27+9+5\ .\ \frac{2}{5}+3\ .\ \frac{6}{7}+......\] \[=27+\frac{27}{3}+\frac{27}{5}+\frac{27}{7}+.....+\frac{27}{2n-1}+......\] Hence \[{{n}^{th}}\] term of given series \[{{T}_{n}}=\frac{27}{2n-1}\] So, \[{{T}_{9}}=\frac{27}{2\times 9-1}=\frac{27}{17}=1\frac{10}{17}\].
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