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

1 Answer
Jan 11, 2017

Answer:

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