A) 10th term
B) 8th term
C) 9th term
D) 11th term
Correct Answer: B
Solution :
Given, series is \[\sqrt{2},\frac{2}{3},\frac{2\sqrt{2}}{9},....\] or \[\frac{\sqrt{2}}{{{3}^{0}}},\frac{{{(\sqrt{2})}^{2}}}{3},\,\frac{{{(\sqrt{2})}^{3}}}{{{3}^{2}}},....\] which is an infinite GP whose, first term \[a=\frac{\sqrt{2}}{{{3}^{0}}}=\sqrt{2}\] and common ratio \[=\frac{{{(\sqrt{2})}^{2}}}{3}\times \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{3}\] Let nth term of GP \[=\frac{16}{2187}\] \[\therefore \] \[a{{t}^{n-1}}=\frac{16}{2187}\] \[\Rightarrow \] \[\sqrt{2}{{\left( \frac{\sqrt{2}}{3} \right)}^{n-1}}=\frac{16}{2187}\] \[\Rightarrow \] \[{{\left( \frac{\sqrt{2}}{3} \right)}^{n-1}}=\frac{8\sqrt{2}}{2187}={{\left( \frac{\sqrt{2}}{3} \right)}^{7}}\] On comparing the powers, we get \[n-1=7\] \[\Rightarrow \] \[n=8\]
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