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

Question

Natali

2 months ago

9

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

t hundredth.

Mathematics

2 answers:

tester [12.3K] · Answer 1 · 2026-03-26T17:23:20+03:00

tester [12.3K]2 months ago

3 0

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

r signifies the ratio between any two successive terms

We are provided
$a_{1}=14$ and
$a_{9}=358.8$
substitute 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 the eighth root of both sides (raising (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
$a_{19}= 20690.49$

Zina [12.3K] · Answer 2 · 2026-03-26T17:24:41+03:00

An = ar^(n-1)

where, an refers to the nth term, n signifies the number of terms, r denotes the common ratio 
a represents the 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....common ratio 

to find the 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...result. 

therefore, the 19th term is 20 690.49