Question

Identify the 19th term of a geometric sequence where a1 = 14 and a9 = 358.80. Round the common ratio and 19th term to the nearest hundredth.

Answer 1

an = ar^(n-1) 

where, an = nth term, n=number of terms, r=common ratio 
a = first term 

given: a =14, a9= 358.80, n = 9, r=? 

an = ar^(n-1) 
358.80 = (14)*(r)^(9-1) 
358.80 = 14*(r^8) 
r^8 = 358.80/14 
r^8 = 25.63 
r = 1.5 .. . .commonratio 

solve for 19th term... 
a19=? , n = 19, r = 1.5, a=14 
an = ar^(n-1) 
a19 = (14)*(1.5)^(18) 
a19=14*(1477.89) 
a19 = 20 690.49 . . .ans. 

hence, 19th term is 20 690.49

Answer 2

geometric sequence
$a_{n}= a_{1}r^{n-1}$

r is the ratio between any 2 succesive terms

we are given
$a_{1}=14$ and
$a_{9}=358.8$
sub and solve for r, since we know $a_{1}=14$
$358.8=a_{9}= 14*r^{9-1}$
$358.8= 14*r^{8}$
divide both sides by 14
$358.8/14= r^{8}$
take eight root of both sides (put (358.8/14) to the (1/8) power)
1.5=r

sub
$a_{19}= 14*(1.5^{19-1})$
$a_{19}= 14*(1.5^{18})$
$a_{19}= 20690.486320497$
hudnretht
$a_{19}= 20690.49$