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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.

Help Me With Number 4 Please Identify The 17th Term Of A Geometric Sequence Where A1 16 And A5 15006 Round The Common Ratio And 17th Term To The Nearest Hundred class=

Sagot :

Answer:

Common ratio = 1.75

17th term = 123,802.31

Explanations:

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

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