Question

How do you find the 7th term of the geometric sequence with the given terms `a_2 = 12, a_5 = -768`?

How do you find the 7th term of the geometric sequence with the given terms `a_2 = 12, a_5 = -768`?

Answer 1

First find the common ratio `r`...

Explanation:

`r=[(-768)/12]^(1/(5-2))=-4`

`a_7=a_5xx-4xx-4=-768xx16=-12288`

hope that helped

Answer 2

`a_7= -12288`

Explanation:

Given :
`a_2=12`
`a_5=-768`

Known:
Let a constant be k
Let the geometric ratio be r

`a_2 -> kr^2 = 12..............................(1)` but this in not quite correct!

`a_5` is negative so `kr^5!=+768`

However: if we had `a_n=(-1)^nkr^n` then

`a_5-> (-1)^5kr^5=-768` would be correct

Let:
`a_2=(-1)^2kr^2=12..............................(1_a)`
`a_5=(-1)^5kr^5=-768........................(2)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("To find r")`

`color(brown)(Consider: color(white)(...)(Equation (2))/( Equation (1_a))`

`((-1)^5kr^5)/(=(-1)^2kr^2)=(-768)/(12)`

`=>- r^3 =- 64`

Multiply both sides by (-1)

`r^3 = 64`

`color(green)(r= 4 ....................................................(4))`

So `color(green)(a_n=(-1)^nk4^n...................................................(5))`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("To find k")`

`color(brown)("Substitute (4) into " (1_a)" giving")`

`a_2=(-1)^2k(4)^2=12`

`color(green)(k=12/16 = 3/4)........................................(6)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("To find " a_7)`

Substitute (4) and (6) into (5) giving

`a_n=(-1)^nk4^n -> color(blue)(a_7 = (-1)^7(3/4)(4)^7)`

`a_7 = (-1)(3/4)(16384)`

`color(red)(a_7= -12288)`