How do you write an nth term rule for $a_{2} = - 20$ $a_2=-20$ and $a_{4} = - 5$ $a_4=-5$ ?

Question

How do you write an nth term rule for $a_{2} = - 20$ $a_2=-20$ and $a_{4} = - 5$ $a_4=-5$ ?

1 Answer

Impact of this question

1580 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-11T13:42:23

$a_{n} = - \frac{40}{2^{n - 1}}$ $a_n=-40/2^(n-1)$ or $a_{n} = 40 {(- \frac{1}{2})}^{n - 1}$ $a_n=40(-1/2)^(n-1)$

Explanation:

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 $a_{1}$ $a_1$ and common ratio is $r$ $r$

the $n^{t h}$ $n^(th)$ term $a_{n} = a_{1} \times r^{n - 1}$ $a_n=a_1xxr^(n-1)$

As $a_{2} = - 20$ $a_2=-20$ and $a_{4} = - 5$ $a_4=-5$ , we have

$a_{1} \times r = - 20$ $a_1xxr=-20$ .................(1) $-$ $-$ and $a_{1} = - \frac{20}{r}$ $a_1=-20/r$

$a_{1} \times r^{3} = - 5$ $a_1xxr^3=-5$ .................(2)

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

Hence $r = \pm \frac{1}{2}$ $r=+-1/2$

If $r = \frac{1}{2}$ $r=1/2$ , $a_{1} = - \frac{20}{\frac{1}{2}} = - 40$ $a_1=-20/(1/2)=-40$ and $a_{n} = - 40 \times {(\frac{1}{2})}^{n - 1} = - \frac{40}{2^{n - 1}}$ $a_n=-40xx(1/2)^(n-1)=-40/2^(n-1)$

and if $r = - \frac{1}{2}$ $r=-1/2$ , $a_{1} = - \frac{20}{- \frac{1}{2}} = 40$ $a_1=-20/(-1/2)=40$ and $a_{n} = 40 {(- \frac{1}{2})}^{n - 1} = 40 {(- \frac{1}{2})}^{n - 1}$ $a_n=40(-1/2)^(n-1)=40(-1/2)^(n-1)$

Science

Math

Humanities

... and beyond

How do you write an nth term rule for $a_{2} = - 20$ $a_2=-20$ and $a_{4} = - 5$ $a_4=-5$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you write an nth term rule for a_2=-20a2=−20a_2=-20 and a_4=-5a4=−5a_4=-5?

1 Answer

Explanation:

Related questions

Impact of this question

How do you write an nth term rule for $a_{2} = - 20$ $a_2=-20$ and $a_{4} = - 5$ $a_4=-5$ ?