# 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?

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 $15 t h$ 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 $3 \times 4 = 12$ The third term has 2 factors of four and so on.

Can you see a pattern here? The ${n}^{t h}$ 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 $a {r}^{n - 1} .$ This means that as long as a doesn't contain r in itself, the nth term will have (n-1) factors of r.