Question

Geometric Sequence help?

Answer 1

\[n _{4} = -18\]\[n _{7} = 2/3\]
What is\[n _{6}\]

Answer 2

you have 3 geometric means missing between n4 and n7.

We can write an equation, \(\large\color{black}{ a_7=a_4\times r^3\LARGE\color{white}{ \rm │ }}\) to solve for r.

Answer 3

When you solve and find r, \(\large\color{black}{ a_4\times r^2=a_6\LARGE\color{white}{ \rm │ }}\)

Answer 4

let me know if you have any questions.

Answer 5

I'm still having issues... \[a _{7}=a _{4}*r ^{3}\]\[a _{4}*r ^{2}=a _{6}\]\[r ^{3}=\frac{ 52 }{ 3 }\] so the second equation becomes: \[-18*(\frac{ 52 }{ 3 })^{\frac{ 2 }{ 3 }} = a _{6}\] right?

Answer 6

But that answer makes no sense.

Answer 7

\(\large a_7=a_4\times r^3\)

\( \dfrac{2}{3} = -18 \times r^3 \)

What is \(r^3\) = ?

Answer 8

-52/3

Answer 9

nevermind 55/3

Answer 10

Divide both sides by -18:

\(\dfrac{2}{3} \times \dfrac{1}{-18} =\cancel{ -18}\times \dfrac{1}{\cancel{-18}} \times r^3\)

Answer 11

BTW, 18 * 3 = 54

Answer 12

don't you get \(-\frac{1}{27}\) or am i nuts?

Answer 13

Sorry I'm not thinking today.....

Answer 14

\(-\dfrac{2}{54} = r^3\)

\(-\dfrac{1}{27} = r^3\)

\(r = -\dfrac{1}{3} \)

Answer 15

Now use

\(a_6 = a^4 \times r^2\)

to get

\(a^6\)

Answer 16

Thank you.

Answer 17

You're welcome.