The sequence converges to 0.
What is geometric sequence?
"It is a type of sequence where the ratio of every two consecutive terms is a constant."
Formula to find the nth term of geometric sequence:
"[tex]a_n=a_1\times r^{n-1}[/tex]
where [tex]a_1[/tex] is the first term, and r is the common ratio."
What is the convergent sequence?
"If [tex]\lim_{n \to \infty} a_n[/tex] exists and is finite we say that the sequence is convergent.
where [tex]a_n[/tex] is the n-th term of the sequence."
What is the divergent sequence?
" If [tex]\lim_{n \to \infty} a_n[/tex]doesn't exist or is infinite we say the sequence diverges.
where [tex]a_n[/tex] is the n-th term of the sequence."
For given question,
We have been given a sequence.
36, - 6, 1, - , . . .
The first term of the sequence is [tex]a_1=36[/tex]
[tex]\frac{-6}{36}=-\frac{1}{6}\\\\ \frac{1}{-6}=-\frac{1}{6}[/tex]
The ratio of every two consecutive terms is a constant.
⇒ [tex]r=-\frac{1}{6}[/tex]
So, the n-th term of given geometric sequence would be,
[tex]a_n=36\times (-\frac{1}{6})^{n-1}[/tex]
Now we determine whether the sequence is convergent or divergent.
[tex]\lim_{n \to \infty} a_n \\\\= \lim_{n \to \infty}36\times (-\frac{1}{6} )^{n-1}\\\\=36\times \lim_{n \to \infty} (-\frac{1}{6} )^{n-1}\\\\=36\times 0\\\\=0[/tex]
Since [tex]\lim_{n \to \infty} a_n[/tex] exists and is finite we say that the sequence is convergent.
Therefore, the sequence converges to 0.
Learn more about the convergent sequence here:
https://brainly.com/question/21961097
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