Question

find the common ratio and write out the first four terms of the geometric sequence \\(\frac{g^{n - 3}}{6}\\)

common ratio is \\(square\\)

\\(a_1 = \square\\), \\(a_2 = \square\\), \\(a_3 = \square\\), \\(a_4 = \square\\)

question help: \\(\boldsymbol{\text{message instructor}}\\)

Explanation:

Step1: Recall geometric sequence formula

A geometric sequence has the form \( a_n = a_1 r^{n - 1} \), where \( r \) is the common ratio. The given sequence is \( a_n=\frac{g^{n - 3}}{6} \). We can rewrite this as \( a_n=\frac{1}{6}g^{n - 3}=\frac{g^{-2}}{6}\cdot g^{n - 1} \) (using exponent rules: \( g^{n-3}=g^{(n - 1)-2}=g^{-2}\cdot g^{n - 1} \)).

Step2: Find the common ratio

In the form \( a_n=a_1r^{n - 1} \), the base of the exponential term with \( n \) is the common ratio. Here, the term with \( n \) is \( g^{n - 1} \) (after re - writing), so the common ratio \( r = g \).

Step3: Find the first term (\( a_1 \))

To find \( a_1 \), substitute \( n = 1 \) into the formula \( a_n=\frac{g^{n - 3}}{6} \). So \( a_1=\frac{g^{1-3}}{6}=\frac{g^{-2}}{6}=\frac{1}{6g^{2}} \).

Step4: Find the second term (\( a_2 \))

Substitute \( n = 2 \) into the formula \( a_n=\frac{g^{n - 3}}{6} \). Then \( a_2=\frac{g^{2 - 3}}{6}=\frac{g^{-1}}{6}=\frac{1}{6g} \).

Step5: Find the third term (\( a_3 \))

Substitute \( n = 3 \) into the formula \( a_n=\frac{g^{n - 3}}{6} \). Then \( a_3=\frac{g^{3 - 3}}{6}=\frac{g^{0}}{6}=\frac{1}{6} \) (since \( g^{0}=1 \) for \( g
eq0 \)).

Step6: Find the fourth term (\( a_4 \))

Substitute \( n = 4 \) into the formula \( a_n=\frac{g^{n - 3}}{6} \). Then \( a_4=\frac{g^{4 - 3}}{6}=\frac{g^{1}}{6}=\frac{g}{6} \).

Answer:

Common ratio is \( g \)
\( a_1=\frac{1}{6g^{2}} \), \( a_2=\frac{1}{6g} \), \( a_3=\frac{1}{6} \), \( a_4=\frac{g}{6} \)