How many terms are in the following sequence?14348907, ..., 9, 3, 117151618

Accepted Solution

Answer:There are 16 terms in the sequence.Step-by-step explanation:The given sequence is 14348907, ..., 9, 3, 1The 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]