<h3><u>Question:</u></h3>
A geometric sequence is defined by the equation an = (3)^3 − n. What are the first three terms of the sequence and what is the common ratio, r?
<h3><u>Answer:</u></h3>
The first three terms of sequence is 9, 3 , 1
The common ratio is ![\frac{1}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D)
<h3><u>Solution:</u></h3>
The geometric sequence is defined by the equation:
![a_n = (3)^{3-n}](https://tex.z-dn.net/?f=a_n%20%3D%20%283%29%5E%7B3-n%7D)
To find the first three terms of sequence, substitute n = 1, n = 2, n = 3
<em><u>First term:</u></em>
Put n = 1 in given equation
![a_1=(3)^{3-1}\\\\a_1 = 3^2\\\\a_1 = 9](https://tex.z-dn.net/?f=a_1%3D%283%29%5E%7B3-1%7D%5C%5C%5C%5Ca_1%20%3D%203%5E2%5C%5C%5C%5Ca_1%20%3D%209)
Thus first term of sequence is 9
<em><u>Second term:</u></em>
Put n = 2 in given equation
![a_2 = (3)^{3-2}\\\\a_2 = 3^1\\\\a_2 = 3](https://tex.z-dn.net/?f=a_2%20%3D%20%283%29%5E%7B3-2%7D%5C%5C%5C%5Ca_2%20%3D%203%5E1%5C%5C%5C%5Ca_2%20%3D%203)
Thus second term of sequence is 3
<em><u>Third term:</u></em>
Put n = 3 in given sequence
![a_3 = (3)^{3-3}\\\\a_3=3^0\\\\a_3 = 1](https://tex.z-dn.net/?f=a_3%20%3D%20%283%29%5E%7B3-3%7D%5C%5C%5C%5Ca_3%3D3%5E0%5C%5C%5C%5Ca_3%20%3D%201)
Thus third term of sequence is 1
Thus the first three terms of sequence is 9, 3 , 1
<em><u>To find common ratio:</u></em>
Common ratio is found by dividing the two consecutive terms
![a_1 = 9\\\\a_2 = 3\\\\a_3 = 1](https://tex.z-dn.net/?f=a_1%20%3D%209%5C%5C%5C%5Ca_2%20%3D%203%5C%5C%5C%5Ca_3%20%3D%201)
Thus common ratio is obtained as:
![r = \frac{a_2}{a_1} = \frac{3}{9} = \frac{1}{3}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7Ba_2%7D%7Ba_1%7D%20%3D%20%5Cfrac%7B3%7D%7B9%7D%20%3D%20%5Cfrac%7B1%7D%7B3%7D)
![r = \frac{a_3}{a_2}=\frac{1}{3}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7Ba_3%7D%7Ba_2%7D%3D%5Cfrac%7B1%7D%7B3%7D)
Thus common ratio is ![\frac{1}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D)