Identify the 19th term of a geometric sequence where [tex]$a_1 = 14$[/tex] and [tex]$a_9 = 358.80$[/tex]. Round the common ratio and 19th term to the nearest hundredth.

A. [tex]a_{19} = 9,195.53[/tex]
B. [tex]a_{19} \approx 31,035.73[/tex]
C. [tex]a_{19} = 13,185.66[/tex]
D. [tex]a_{19} \approx 20,690.49[/tex]



Answer :

To identify the 19th term of a geometric sequence where [tex]\( a_1 = 14 \)[/tex] and [tex]\( a_9 = 358.80 \)[/tex], we need to follow these steps:

1. Determine the common ratio [tex]\( r \)[/tex]:

We know that for a geometric sequence, the [tex]\( n \)[/tex]-th term [tex]\( a_n \)[/tex] is given by:
[tex]\[ a_n = a_1 \cdot r^{(n-1)} \][/tex]

Given [tex]\( a_9 = 358.80 \)[/tex]:
[tex]\[ a_9 = a_1 \cdot r^{(9-1)} \implies 358.80 = 14 \cdot r^8 \][/tex]

Solve for [tex]\( r \)[/tex]:
[tex]\[ r^8 = \frac{358.80}{14} \implies r^8 = 25.6285714 \][/tex]
[tex]\[ r = \sqrt[8]{25.6285714} \implies r \approx 1.5 \][/tex]

2. Calculate the 19th term [tex]\( a_{19} \)[/tex]:

Now, with [tex]\( r \approx 1.5 \)[/tex], we determine [tex]\( a_{19} \)[/tex]:
[tex]\[ a_{19} = a_1 \cdot r^{(19-1)} \implies a_{19} = 14 \cdot (1.5)^{18} \][/tex]
[tex]\[ a_{19} \approx 20689.88 \][/tex]

Hence, after rounding to the nearest hundredth, the 19th term of the geometric sequence, [tex]\( a_{19} \)[/tex], is [tex]\( \approx 20,689.88 \)[/tex].

The correct answer is:

[tex]\[ a_{19} \approx 20,689.88 \][/tex]