Question

find the recursive formula for the geometric sequence $-\frac{1}{9}, -\frac{1}{3}, -1, -3, -9, \dots$ (1 point)
\bigcirc $a_1 = -\frac{1}{9}, a_n = 3a_{n - 1}$
\bigcirc $a_1 = -\frac{1}{9}, a_n = -3a_{n - 1}$
\bigcirc $a_1 = -\frac{1}{9}, a_n = \frac{1}{3}a_{n - 1}$
\bigcirc $a_1 = -\frac{1}{9}, a_n = -\frac{1}{3}a_{n - 1}$

Explanation:

Step1: Identify the first term

The first term \(a_1\) of the geometric sequence \(-\frac{1}{9}, -\frac{1}{3}, -1, -3, -9, \dots\) is \(-\frac{1}{9}\).

Step2: Find the common ratio \(r\)

The common ratio \(r\) in a geometric sequence is given by \(\frac{a_n}{a_{n - 1}}\). Let's take the second term and the first term: \(r=\frac{-\frac{1}{3}}{-\frac{1}{9}}=\frac{-\frac{1}{3}\times(-9)}{1}= 3\). So the recursive formula is \(a_1 = -\frac{1}{9}\) and \(a_n=3a_{n - 1}\).

Answer:

A. \(a_1 = -\frac{1}{9}, a_n = 3a_{n-1}\)