How do you write a rule for the nth term of the geometric term given the two terms #a_3=25, a_6=-25/64#?

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Answer 1 · 2018-02-24T23:20:17

#400(-1/4)^(n-1)#

The nth term of a geometric series is given by:

#ar^(n-1)#

Where #bba# is the first term, #bbr# is the common ratio and #bbn# is the nth term.

We have:

Third term #= 25#

Sixth term #= -25/64#

#:.#

#a_3=ar^(2)=25color(white)(888888.8)[1]#

#a_6=ar^(5)=-25/64color(white)(8888)[2]#

Solving for #bba# and #bbr#

#[1]#

#a=25/r^2#

Substituting in #[2]#

#(25/r^2)r^5=-25/64#

Multiply by #r^2/25#

#r^5=-r^2/64#

#64r^5+r^2=0#

Factor out #bbr#

#r(64r^4+r)=0#

We know #bbr# can't be zero.

#(64r^4+r)=0#

Factor out #bbr# again.

#r(64^3+1)=0#

As before:

#64r^3+1=0#

#r^3=-1/64#

#r=root(3)(-1/64)=-1/4#

Using this in #[1]#

#a(-1/4)^2=25#

#a=25/(-1/4)^2=25/(1/64)=400#

So we have:

#400(-1/4)^(n-1)# for our nth term.