Question

What is the 6th term of the geometric sequence where a1 = 1,024 and a4 = −16? (6 points)A 1B −0.25C −1D 0.25

Answer 1

given :

`\begin{gathered} a_1=1024\text{ and} \\ a_4=-16 \end{gathered}`

In a GP,

`a_n=ar^(n-1)`

in this problem

`\begin{gathered} n=1\Rightarrow a_1=ar^0 \\ n=4\Rightarrow a_4=ar^3 \end{gathered}`

we divide both equations, we get

`\begin{gathered} (ar^3)/(a)=-(16)/(1024) \\ \Rightarrow r^3=-(1)/(64) \\ \Rightarrow r=-(1)/(4) \end{gathered}`

here a=1024 and r =-1/4 and let n=6

`\begin{gathered} a_6=1024(-(1)/(4))^5 \\ a_6=1024(-(1)/(1024)) \\ a_6=-1 \end{gathered}`

hence the solution is c) -1