# What is the 8th term of the geometric sequence where a1 = 256 and a3 = 16?

Mar 9, 2018

$8$ th term of the geometric sequence is $\frac{1}{64}$

#### Explanation:

Geometric sequence is  {a, ar ,ar^2...}; a and r  being first term

and common ratio respectively. ${a}_{1} = 256 , {a}_{3} = 16$

$n$ th term is ${a}_{n} = {a}_{1} \cdot {r}^{n - 1} \therefore {a}_{3} = {a}_{1} \cdot {r}^{3 - 1} = 16$ or

$256 \cdot {r}^{2} = 16 \mathmr{and} {r}^{2} = \frac{16}{256} = \frac{1}{16} \therefore r = \frac{1}{4}$

$8$ th term is ${a}_{8} = {a}_{1} \cdot {r}^{n - 1} = 256 {\left(\frac{1}{4}\right)}^{8 - 1}$ or

${a}_{8} = 256 {\left(\frac{1}{4}\right)}^{7} = {4}^{4} / {4}^{7} = \frac{1}{4} ^ 3 = \frac{1}{64}$

$8$ th term of the geometric sequence is $\frac{1}{64}$ [Ans]