Answer:
There are 16 terms in the sequence.
Step-by-step explanation:
The given sequence is
14348907, ..., 9, 3, 1
The first term of the sequence is
[tex]a_1=14348907[/tex]
The last term of the sequence is [tex]l=1[/tex]
The common ratio is [tex]r=\frac{1}{3}[/tex]
The nth term of this sequence is;
[tex]a_n=a_1(r)^{n-1}[/tex]
We plug in the common ratio and the first term to get;
[tex]a_n=14348907(\frac{1}{3})^{n-1}[/tex]
The find the number of terms in the sequence , we plug in the last term of the sequence.
This implies that;
[tex]1=14348907(\frac{1}{3})^{n-1}[/tex]
[tex]\frac{1}{14348907}=(\frac{1}{3})^{n-1}[/tex]
[tex]\Rightarrow 3^{-15}=3^{-(n-1)}[/tex]
Since the bases are the same, we equate the exponents;
[tex]-15=-(n-1)[/tex]
[tex]15=n-1[/tex]
[tex]15+1=n[/tex]
[tex]n=16[/tex]
