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`