### Question**b) Given the sequence \( 6, 2, \frac{2}{3}, \frac{2}{9}, \ldots, \frac{2}{6561} \); how many terms are there?***(Show work Algebraically)*### ExplanationTo solve this problem, examine the given sequence and identify a pattern or common ratio. The sequence appears to be a geometric progression where each term after the first is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio.Let's break down the first few terms to find the common ratio \( r \):- \( \frac{2}{3} \div 2 = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3} \)- \( \frac{2}{9} \div \frac{2}{3} = \frac{2}{9} \cdot \frac{3}{2} = \frac{1}{3} \)This confirms that the common ratio \( r \) is indeed \( \frac{1}{3} \).Let's denote the first term of the sequence as \( a \). Here, \( a = 6 \).The general formula for the \( n \)-th term of a geometric sequence is:\[ a_n = a \cdot r^{(n-1)} \]In our case:\[ a_n = 6 \cdot \left( \frac{1}{3} \right)^{(n-1)} \]Set the \( n \)-th term equal to \( \frac{2}{6561} \) to find \( n \):\[ \frac{2}{6561} = 6 \cdot \left( \frac{1}{3} \right)^{(n-1)} \]To solve for \( n \), follow these algebraic steps:1. Divide both sides by 6:\[ \frac{2}{6561 \cdot 6} = \left( \frac{1}{3} \right)^{(n-1)} \]\[ \frac{2}{39366} = \left( \frac{1}{3} \right)^{(n-1)} \]2. Convert \( \frac{1}{6561} \) into a power of 3:\[

Name: ### Question**b) Given the sequence \( 6, 2, \frac{2}{3}, \frac{2}{9}
Uploaded: 2024-03-20T06:38:58.000Z
Channel: Bartleby
Description: ### Question**b) Given the sequence \( 6, 2, \frac{2}{3}, \frac{2}{9}, \ldots, \frac{2}{6561} \); how many terms are there?***(Show work Algebraically)*### ExplanationTo solve this problem, examine the given sequence and identify a pattern or common