# How to write a formula for the nth term of the geometric sequence below and how to use that formula for a_n to find a_8, which is the 8th term of the sequence? (1) 4, -4, 4, -4, ..... (2) 16, 8, 4, .....

Jul 5, 2018

(1) $- 4$

(2) $\frac{1}{8}$

#### Explanation:

A geometric series ${a}_{n}$ starts from a certain number ${a}_{1} = x$ and computes the next term ${a}_{n + 1}$ by multiplying the previous one by a fixed ratio $r$. So, you have

${a}_{1} = x$
${a}_{2} = {a}_{1} \cdot r = x \cdot r$
${a}_{3} = {a}_{2} \cdot r = \left(x \cdot r\right) \cdot r = x \cdot {r}^{2}$
${a}_{4} = {a}_{3} \cdot r = \left(x \cdot {r}^{2}\right) \cdot r = x \cdot {r}^{3}$

So, it should be clear that the general rule is

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

So, to find the generi expression, we need both $x$ and $r$. Note that, since we have the first term in each series, we immediately know $x$: you just need to look at the first term, since ${a}_{1} = x$.

So, if we call the first series ${a}_{n}$ and the second ${b}_{n}$, we have ${a}_{1} = 4$ and ${b}_{1} = 16$.

To compute $r$, we can divide two consecutive terms, since you have

${a}_{n + 1} / {a}_{n} = \frac{x \cdot {r}^{n}}{x \cdot {r}^{n - 1}} = r$

So, if we call ${r}_{a}$ the ratio of the first series and ${r}_{b}$ the ratio of the second series, and we use the first two terms to compute it, we have

${r}_{a} = {a}_{2} / {a}_{1} = - \frac{4}{4} = - 1 , \setminus q \quad {r}_{b} = {b}_{2} / {b}_{1} = \frac{8}{16} = \frac{1}{2}$

So, we have the general rules:

${a}_{n} = 4 \cdot {\left(- 1\right)}^{n - 1} , \setminus q \quad {b}_{n} = 16 \cdot {\left(\frac{1}{2}\right)}^{n - 1} = {2}^{4} \cdot {2}^{- n + 1} = {2}^{5 - n}$

To compute the eight element, simply plug $n = 8$:

${a}_{8} = 4 \cdot {\left(- 1\right)}^{7} = 4 \cdot \left(- 1\right) = - 4$

${b}_{8} = {2}^{5 - 8} = {2}^{- 3} = \frac{1}{8}$