A) 10
B) 15
C) 5
D) 25
Correct Answer: C
Solution :
[c] Let \[\underset{n\to \infty }{\mathop{\lim }}\,{{b}_{n}}=b\] now, \[{{b}_{n+1}}=\frac{1}{3}\left( 2{{b}_{n}}+\frac{125}{b_{n}^{2}} \right)\] or \[\underset{n\to \infty }{\mathop{\lim }}\,{{b}_{n+1}}=\frac{1}{3}\left( 2\underset{n\to \infty }{\mathop{\lim }}\,{{b}_{n}}+\frac{125}{\underset{n\to \infty }{\mathop{\lim }}\,b_{n}^{2}} \right)\] or \[b=\frac{1}{3}\left( 2b+\frac{125}{{{b}^{2}}} \right)\] \[\Rightarrow \frac{b}{3}=\frac{125}{3{{b}^{2}}}\Rightarrow {{b}^{3}}=125\,\,or\,\,b=5.\]
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