3, 12, 48 are the first three terms of the geometric sequence. What is the number of factors of 4 that is in the 15th term?

2 Answers
May 1, 2018

#14#

Explanation:

The first term, #3#, does not have #4# as a factor. The second term, #12#, has #4# as one factor (it is #3# multiplied by #4#). The third term, #48#, has #4# as its factor twice (it is #12# multiplied by #4#). Therefore, the geometric sequence must be created by multiplying the preceding term by #4#. Since each term has one less factor of #4# than its term number, the #15th# term must have #14# #4#s.

May 1, 2018

The fifteenth term's factorization will contain 14 fours.

Explanation:

The given sequence is geometric, with the common ratio being 4 and the first term being 3.

Note that the first term has 0 factors of four. The second term has one factor of four, as it is #3xx4 = 12# The third term has 2 factors of four and so on.

Can you see a pattern here? The #n^(th)# term has (n-1) factors of four. Thus the 15th term will have 14 factors of four.

There is also another reason for this. The nth term of a G.P is #ar^(n-1).# This means that as long as a doesn't contain r in itself, the nth term will have (n-1) factors of r.