Question

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  

Answer 1

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

$a_n=a(r)^{n-1}$

$a_{10}=2(4)^{10-1}$

$a_{10}=2(4)^{9}$

$a_{10}= 524,288$

Answer 2

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