Given the two terms in a geometric sequence how do you find the recursive formula... a1=-4 and a4=-500?

2 Answers
Apr 27, 2016

Answer:

Recursive formula is #a_1=-4# and #a_(n+1)=5a_n#

Explanation:

In a geometric sequence, if #a# is the first term and #r# is the ratio between a term and its preceding term, then #n^(th)# term is given by #axxr^(n-1)#.

Here #a=a_1=-4# and as fourth term is #-500#

#-4xxr^(4-1)=-500# or #4r^3=500# or #r^3=125# and hence

#r=root(3)(125)=root(3)(5xx5xx5)=5#

Hence recursive formula is #a_1=-4# and #a_(n+1)=5a_n#

May 1, 2016

Answer:

Recursive formulation:

#{ (a_1 = -4), (a_(n+1) = 5a_n color(white)(000) (n = 1,2,3,...)) :}#

Explanation:

We are given:

#{ (a_1 = -4), (a_4 = -500) :}#

The general formula for the #n#th term of a geometric series is:

#a_n = a r^(n-1)#

where #a# is the initial term and #r# is the common ratio.

A recursive formula can be given as:

#{ (a_1 = a), (a_(n+1) = ra_n color(white)(000) (n = 1,2,3,...)) :}#

In our example:

#5^3 = 125 = (-500)/(-4) = r_4/r_1 = (color(red)(cancel(color(black)(a))) r^(4-1))/(color(red)(cancel(color(black)(a))) r^(1-1)) = r^3#

So the only possible Real value for #r# is #root(3)(5^3) = 5#.

#color(white)()#
Footnote

There are two other possibilities for a geometric sequence with #a_1 = -4# and #a_4 = -500#, which are sequences of Complex numbers.

This is because #5^3# has two other cube roots, namely #5omega# and #5omega^2#, where #omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#. Either of these will also work as a suitable common ratio.