Question

Let the sequence $< b_{n} >$ of real numbers satisfies the recurrence relation $b_{n+1}=\frac{1}{3}\left(2 b_{n}+\frac{125}{b_{n}^{2}}\right), b_{n} \neq 0 .$ Then find $\displaystyle\lim _{n \rightarrow \infty} b_{n}$

• A. 10
• B. 15
• C. 5
• D. 25

Answer 1

Answer: (C)

Let $\displaystyle\lim _{n \rightarrow \infty} b_{n}=b$ Now, $b_{n+1}=\frac{1}{3}\left(2 b_{n}+\frac{125}{b_{n}^{2}}\right)$
or $\displaystyle\lim _{n \rightarrow \infty} b_{n+1}=\frac{1}{3}\left(2 \displaystyle\lim _{n \rightarrow \infty} b_{n}+\frac{125}{\displaystyle\lim _{n \rightarrow \infty} b_{n}^{2}}\right)$
or $b=\frac{1}{3}\left(2 b+\frac{125}{b^{2}}\right)$
$\Rightarrow \frac{b}{3}=\frac{125}{3 b^{2}}$
$\Rightarrow b^{3}=125$ or $b=5.$