Find the 10th term of the following geometric sequence: 2, 8, 32, 128. A. 164,357 B. 621,325 C. 524,288 D. 248,221
2 answers:
Answer:
![524,288](https://tex.z-dn.net/?f=524%2C288)
Step-by-step explanation:
Find the 10th term of the following geometric sequence
2, 8, 32, 128.......
To find the nth term of any geometric sequence we use formula
![a_n=a(r)^{n-1}](https://tex.z-dn.net/?f=a_n%3Da%28r%29%5E%7Bn-1%7D)
Where 'a' is the first term and 'r' is the common ratio
To find common ratio we divide the second term by first term
![\frac{8}{2} =4](https://tex.z-dn.net/?f=%5Cfrac%7B8%7D%7B2%7D%20%3D4)
![\frac{32}{8} =4](https://tex.z-dn.net/?f=%5Cfrac%7B32%7D%7B8%7D%20%3D4)
![\frac{128}{32} =4](https://tex.z-dn.net/?f=%5Cfrac%7B128%7D%7B32%7D%20%3D4)
and ![a=2](https://tex.z-dn.net/?f=a%3D2)
Plug in the values in the formula, n=10
![a_n=a(r)^{n-1}](https://tex.z-dn.net/?f=a_n%3Da%28r%29%5E%7Bn-1%7D)
![a_{10}=2(4)^{10-1}](https://tex.z-dn.net/?f=a_%7B10%7D%3D2%284%29%5E%7B10-1%7D)
![a_{10}=2(4)^{9}](https://tex.z-dn.net/?f=a_%7B10%7D%3D2%284%29%5E%7B9%7D)
![a_{10}= 524,288](https://tex.z-dn.net/?f=a_%7B10%7D%3D%20524%2C288)
Common ratio = 8/2 = 32/8 = 4
10th term = a1*r^(n - 1) where a1 = 2 , r = 4 and n = 10
= 2 * 4^9
= 524,288
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