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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.

1.)a17 ≈ 123,802.31

2.)a17 ≈ 30,707.05

3.a17 ≈ 19,684.01

4.)a17 ≈ 216,654.05

Verified Answer :

The geometric sequence formula is expressed as an =  a1 r^(n-1) where n is an integer. In this case, upon substitution, 150.06 = 16 r^(4). extracting r, r is equal to 1.75. Hence the 17th term from the formula is equal to 123802.32.

Answer:

The correct option is 1.

Step-by-step explanation:

It is given that the first term of a geometric sequence is 16 the fifth term of the sequence is 150.06.

\(a_1=16\)

\(a_5=150.06\)

The nth term of a geometric sequence is

\(a_n=a_1r^{n-1}\)                  .... (1)

The fifth term of the sequence is

\(a_5=a_1r^{5-1}\)

Substitute \(a_1=16\) and \(a_5=150.06\).

\(150.06=16r^{4}\)

Divide both sides by 16.

\(9.37875=r^{4}\)

\((9.37875)^{\frac{1}{4}}=r\)

\(r\approx 1.75\)

Substitute n=17, \(a_1=16\) and \(r=1.75\) to find the 17th term.

\(a_{17}=16(1.75)^{17-1}\)

\(a_{17}=123802.31384\)

\(a_{17}\approx 123802.31\)

Therefore the correct option is 1.

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