Question

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.

Answer 1

Common ratio = 1.75

17th term = 123,802.31

Explanations:

Given the following parameters:

`\begin{gathered} a_1=16 \\ a_5=150.06 \end{gathered}`

Since the sequence is geometric, the nth term of the sequence is given as;

`a_n_{}=a_{}r^(n-1)`

a is the first term

r is the common ratio

n is the number of terms

If the first term a1 = 16, then;

`\begin{gathered} a_1=ar^(1-1)_{} \\ 16=ar^0 \\ a=16 \end{gathered}`

Similarly, if the fifth term a5 = 150.06, then;

`\begin{gathered} a_5=ar^(5-1) \\ a_5=ar^4 \\ 150.06=16r^4 \\ r^4=(150.06)/(16) \\ r^4=9.37875 \\ r=1.74999271132 \\ r\approx1.75 \end{gathered}`

Hence the common ratio to the nearest hundredth is 1.75

Next is to get the 17th term as shown;

`\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}`

Hence the 17th term of the sequence to the nearest hundredth is 123,802.31