Question

Find an equation for the nth term of a geometric sequence where the second and fifth terms are -2 and 16, respectively.

Answer 1

we need to know the first term and the common ratio

Answer 2

let a be the first term and r be the common ratio then we have
ar^4= -2
and ar^7=16
thus dividing the two eq.
we hve
r^3= 16/(-2)= -8
hence r =-2
now
a(-2)^4= -2
or 16 a= -2
or a= -2/16
a=-1/8
so
a_n= (-1/8)*(-2)^(n-1)

Answer 3

If a is the first term, and r is the ratio, it looks to me like you are using the fact that the second term a_2 = -2. Wouldn't your identity, then, be -2 = a(r)^2-1 which is -2 = a(r)^1? I'm just confused as to where you get the exponent on the r in each one.

Answer 4

The nth term of a geometric sequence is \(a_n = ar^{n-1}\)
\(
a_2 = ar \\
a_5 = ar^4 \\
a_5/a_2 = r^3 = 16/(-2) = -8 \\
r = -2 \\
a_2 = ar = -2 \\
a = -2/r = -2/-2 = 1 \\
a_n = (-2)^{n-1}
\)

Answer 5

Gotcha, mosaic. Thank you for that explanation...whoa

Answer 6

You are welcome.

Answer 7

Thank you!

Answer 8

np.