# How do you write a rule for the nth term of the geometric term given the two terms a_2=2, a_6=512?

##### 1 Answer
Feb 3, 2018

${a}_{n} = a \cdot {r}^{n - 1} = \left(\frac{1}{2}\right) \cdot {4}^{n - 1}$

#### Explanation:

Let a be the first term and r the common ration.

First term ${a}_{1} = a = a {r}^{0}$

${a}_{2} = a \cdot r = a {r}^{2 - 1} = 2$

${a}_{6} = a \cdot r \cdot r \cdot r \cdot r \cdot r = a {r}^{5} = a {r}^{6 - 1} = 512$

${a}_{6} / {a}_{2} = \frac{a {r}^{5}}{a r} = {r}^{4} = \frac{512}{2} = 256$

$r = \sqrt[4]{256} = \sqrt[4]{{4}^{4}} = {\left({4}^{4}\right)}^{\frac{1}{4}} = 4$

${a}_{2} = a r = a \cdot 4 = 2$

$a = \frac{1}{2}$

Hence the ${n}^{t h}$ term of the geometric sequence is

${a}_{n} = a \cdot {r}^{n - 1} = \left(\frac{1}{2}\right) \cdot {4}^{n - 1}$