Question

What is the 7th term of a geometric sequence where a1=256 and a3=16?

Answer 1

I fiddled around with the problem for a while but no, just find the formula:

$a_n=256\left(\frac{1}{4}\right)^{n-1}$

The ratio is 1/4 because:
4^4 = 256 = $a_1$
4^2 = 16 = $a_3$

First element of sequence: 256
So with these two numbers you can get the formula.

Now plug in 7 for n and solve.

$a_7 = 256( \frac{1}{4})^{7-1} = 256( \frac{1}{4})^{6} = \frac{256}{4096} = \frac{1}{16}$

Answer 2

The answer is 0.0625