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
common ratio = 8/2 = 32/8 = 4
10th term = a1r^(n - 1) where a1 = 2 , r = 4 and n = 10
= 2 4^9
= 524,288
Answer:
\(524,288\)
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
\(a_n=a(r)^{n-1}\)
Where 'a' is the first term and 'r' is the common ratio
To find common ratio we divide the second term by first term
\(\frac{8}{2} =4\)
\(\frac{32}{8} =4\)
\(\frac{128}{32} =4\)
\(r=4\) and \(a=2\)
Plug in the values in the formula, n=10
\(a_n=a(r)^{n-1}\)
\(a_{10}=2(4)^{10-1}\)
\(a_{10}=2(4)^{9}\)
\(a_{10}= 524,288\)