Help me with number 4 please Identify the 17th term of a geometric sequence where a1 = 16 and a5 = 150.06 Round the common ratio and 17th term to the nearest hundredth.
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Common ratio = 1.75
17th term = 123,802.31
Given the following parameters:
[tex]\begin{gathered} a_1=16 \\ a_5=150.06 \end{gathered}[/tex]Since the sequence is geometric, the nth term of the sequence is given as;
[tex]a_n_{}=a_{}r^{n-1}[/tex]a is the first term
r is the common ratio
n is the number of terms
If the first term a1 = 16, then;
[tex]\begin{gathered} a_1=ar^{1-1}_{} \\ 16=ar^0 \\ a=16 \end{gathered}[/tex]Similarly, if the fifth term a5 = 150.06, then;
[tex]\begin{gathered} a_5=ar^{5-1} \\ a_5=ar^4 \\ 150.06=16r^4 \\ r^4=\frac{150.06}{16} \\ r^4=9.37875 \\ r=1.74999271132 \\ r\approx1.75 \end{gathered}[/tex]Hence the common ratio to the nearest hundredth is 1.75
Next is to get the 17th term as shown;
[tex]\begin{gathered} a_{17}=ar^{16} \\ a_{17}=16(1.75)^{16} \\ a_{17}=16(7,737.6446) \\ a_{17}\approx123,802.31 \end{gathered}[/tex]Hence the 17th term of the sequence to the nearest hundredth is 123,802.31