How do you write an nth term rule for $a_2=-20$ and $a_4=-5$ ?

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Answer 1 · 2017-01-11T13:42:23

$a_n=-40/2^(n-1)$ or $a_n=40(-1/2)^(n-1)$

Although it is not specifically mentioned as the question is framed under "Geometric Series", it is assumed to be so.

In a geometric series if firs term is $a_1$ and common ratio is $r$

the $n^(th)$ term $a_n=a_1xxr^(n-1)$

As $a_2=-20$ and $a_4=-5$ , we have

$a_1xxr=-20$ .................(1) $-$ and $a_1=-20/r$

$a_1xxr^3=-5$ .................(2)

Dividing (2) by (1), we get $r^2=(-5)/(-20)=1/4$

Hence $r=+-1/2$

If $r=1/2$ , $a_1=-20/(1/2)=-40$ and $a_n=-40xx(1/2)^(n-1)=-40/2^(n-1)$

and if $r=-1/2$ , $a_1=-20/(-1/2)=40$ and $a_n=40(-1/2)^(n-1)=40(-1/2)^(n-1)$

How do you write an nth term rule for a_2=-20a_2=-20 and a_4=-5a_4=-5?