Question

What is the 9th term of the geometric sequence where a1 = -5 and a6 = -5,120?

Answer 1

You can first find the common ratio r by seeing the number of spaces apart the 2 sums are

Answer 2

\[a_{n2}-a_{n1} = 6-1 = 5 \]

Answer 3

So now we know
\[a_6 = a_1r^5\]

Answer 4

Solve for r and then use the first term a1 to solve for the nth term (9th) using your geometric formula

Answer 5

Let the GP be
a,ar,ar^2........ar^8
given that
a1=a=-5
a6=(-5)r^5=-5120
r^5=1024=2^10
r=4
Therefore required term i.e a9 = a r^8 = > (-5) (4)^8

Answer 6

\[a_6 = a_1r^5 \]
\[-5,120 = (-5)r^5\]
Dividing both sides by (-5)\[1024 = r^5 \]
Taking the 5th root of both sides\[\sqrt[5]{1024} = r = 4\]

Answer 7

\[a_{nth}= a_1(r^{n-1})\]
\[a_9= (-5)(4^{9-1})\]