What is the geometric sequence for #3, 6, 12, 24, ...#?

1 Answer
Feb 10, 2017

Answer:

The geometric sequence is #a_n=ar^(n-1)=3*2^(n-1)#.

Explanation:

In a geometric sequence, the terms are separated by a common ratio #r#. So, for example, the 4th term #a_4# will be #rxx a_3#, the 3rd term #a_3=r xx a_2#, and so on. From this we can get a general formula for the #n^"th"# term in terms of #r# and the first term #a_1#:

#a_n= r xx a_(n-1)"                  "=r^1a_(n-1)#
#color(white)(a_n)=r xx(r xx a_(n-2))"        "=r^2a_(n-2)#
#color(white)(a_n)=r xx r xx (r xx a_(n-3))" "=r^3a_(n-3)#
#color(white)(a_n)=...#
#a_n=r^(n-1)xxa_(n-(n"-"1))"     "=r^(n-1)a_1#

This is often written with the initial value #a_1# (often just called #a#) in front, like this:

#a_n=ar^(n-1)#

For the sequence #3, 6, 12, 24#, we are given the first term #a=3#. Now all we need is the common ratio #r#. This can be found by computing the ratio of any two successive terms.

(That is, since #a_2=r xx a_1# for any geometric sequence, we can find #r# by solving #r=a_2/a_1#, or #r=a_3/a_2#, or in general #r=a_k/a_(k-1)#.)

Using #a_2=6# and #a_1=3#, we get

#r=a_2/a_1=6/3=2#.

Thus, the common ratio is 2, the first term is 3, and so the formula for this geometric sequence is

#a_n=ar^(n-1)#
#color(white)(a_n)=3*2^(n-1)#.