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?

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Answer 1 · 2017-02-25T14:21:10

$a_n=14(-2)^((n-1))$

We are given $a_1=14$ and as $a_(k+1)=(-2)a_k$ , we have

$a_2=(-2)a_1=(-2)xx14=-28$

$a_3=(-2)a_2=(-2)xx(-28)=56$

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

$a_5=(-2)xx(-112)=224$

It is apparent that it is geometric sequence with first term $a_1=14$ and common ratio as $-2$ . As such $n^(th)$ term $a_n$ is given by

$a_n=a_1xx(-2)^((n-1))=14(-2)^((n-1))$

How do you write the first five terms of the sequence defined recursively a_1=14, a_(k+1)=(-2)a_ka_1=14, a_(k+1)=(-2)a_k, then how do you write the nth term of the sequence as a function of n?