# How do you write the first five terms of the sequence defined recursively a_1=14, a_(k+1)=(-2)a_k, then how do you write the nth term of the sequence as a function of n?

Feb 25, 2017

${a}_{n} = 14 {\left(- 2\right)}^{\left(n - 1\right)}$

#### Explanation:

We are given ${a}_{1} = 14$ and as ${a}_{k + 1} = \left(- 2\right) {a}_{k}$, we have

${a}_{2} = \left(- 2\right) {a}_{1} = \left(- 2\right) \times 14 = - 28$

${a}_{3} = \left(- 2\right) {a}_{2} = \left(- 2\right) \times \left(- 28\right) = 56$

a_4=(-2)xxa_3=(-2)xx(56=-112 and

${a}_{5} = \left(- 2\right) \times \left(- 112\right) = 224$

It is apparent that it is geometric sequence with first term ${a}_{1} = 14$ and common ratio as $- 2$. As such ${n}^{t h}$ term ${a}_{n}$ is given by

${a}_{n} = {a}_{1} \times {\left(- 2\right)}^{\left(n - 1\right)} = 14 {\left(- 2\right)}^{\left(n - 1\right)}$