Assume that the sequence converges and find its limit.
-$$a _ { 1 } = 3 , a _ { n + 1} = \frac { 72 } { 1 + a _ { n } }$$
A) 4
B) 8
C) 2
D) 3

Question

Assume that the sequence converges and find its limit.
-$$a _ { 1 } = 3 , a _ { n  + 1} = \frac { 72 } { 1 + a _ { n } }$$
A) 4
B) 8
C) 2
D) 3

Quizplus · Accepted Answer

To find the limit, we can set $a_{n+1}=a_n=a$, then solve for $a$:
\begin{align*}
a &= \frac{72}{1+a}\
a^2 + a -72 &= 0\
(a+9)(a-8) &= 0\
\end{align*}
Since the sequence is positive (starting with $a_1=3$) we choose $a=8$ as our limit. Therefore, the answer is B.

Assume That the Sequence Converges and Find Its Limit a1=3,an+1=721+ana _ { 1 } = 3 , a _ { n + 1} = \frac { 72 } { 1 + a _ { n } }a1​=3,an+1​=1+an​72​

Assume That the Sequence Converges and Find Its Limit $a _ { 1 } = 3 , a _ { n + 1} = \frac { 72 } { 1 + a _ { n } }$