Question

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

Answer 1

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 2

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.