# How do you write the first five terms of the geometric sequence a_1=5, a_(k+1)=-2a_k and determine the common ratio and write the nth term of the sequence as a function of n?

Sep 29, 2017

first 5 terms

$\left\{5 , - 10 , 20 , - 40 , 80\right\}$

common ratio

$r = - 2$

nth term

${a}_{n} = 5 {\left(- 2\right)}^{n - 1}$

#### Explanation:

${a}_{1} = 5$

${a}_{k + 1} = - 2 {a}_{k}$

${a}_{1} = 5$

${a}_{2} = - 2 {a}_{1} = - 2 \times 5 = - 10$

${a}_{3} = - 2 {a}_{2} = - 2 \times - 10 = 20$

${a}_{4} - - 2 {a}_{3} = - 2 \times 20 = - 40$

${a}_{5} = - 2 {a}_{4} = - 2 \times - 40 = 80$

first 5 terms

$\left\{5 , - 10 , 20 , - 40 , 80\right\}$

common ratio

is ${a}_{k + 1} / {a}_{k} = \frac{- 2 {a}_{k}}{a} _ k$

$r = - 2$

nth term of any Gp

${a}_{n} = {a}_{1} {r}^{n - 1}$

${a}_{n} = 5 {\left(- 2\right)}^{n - 1}$

Oct 8, 2017

${a}_{1} = 5$ and ${a}_{k + 1} = - 2 {a}_{k}$

so, ${a}_{2} = - 2 {a}_{1}$

=> ${a}_{2} = - 10$

Also, ${a}_{3} = - 2 {a}_{2}$

i.e. ${a}_{3} = 20$

so, our terms are $5 , - 10 , 20 , - 40 , \ldots$

Hence, 1st term 5 and common ratio -2

Now, nth term is $a {r}^{n - 1}$

i.e. 5(-2)^(n - 1))

or, $\left(- \frac{5}{2}\right) \left(- {2}^{n}\right)$

With n = 7 we have:

$\left(- \frac{5}{2}\right) \left(- {2}^{7}\right) = 320$

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