Question

what is the 19th term of a geometric sequence where a1=8 and a9=360. Round the common ratio and 19th term to the nearest hundredth.

REALLY need help on this one please please.

Answer 1

Answer:

Step-by-step explanation:

$a_{1} = 8\\a_{9} = 360 = a_{1} \cdot r^9 = 8 \cdot r^9\\r = \sqrt[9]{\frac{360}{9} }  \cong 1.53$

By the formula of a geometric series:

$S_{n} = a \cdot \frac{1-r^n}{1-r} => S_{19} = 8 \cdot \frac{1-1.53^{19}}{1-1.53}  = 6091.54$

Answer 2

Answer:

r = 1.6

a19 = 37800

Step-by-step explanation:

First term $a_1 = 8$ and $a_9 = 360$ of a geometric sequence and we are to find the 19th term along with the common ratio for this sequence.

We know that the general term of a Geometric Sequence is given by:

$a_n= a \cdot r^{n-1}$

where $a_n$ is the nth term, $a$ is the first term, $n$ is the number of terms and $r$ is the common ratio.

Substituting the given values in the above formula to get:

$360 = (8)*(r)^(9-1)$

$360 = 8*(r^8)$

$r^8 = \frac{360}{8}$

$r^8 = 45$

r = 1.6

Finding the 19th term:

$a_n= a \cdot r^{n-1}$

$a_{19}= 8 \cdot 1.6^{19-1}$

a19 = 37800