Question

What is the 12th term of a geometric sequence where `a_1 = 8` and `a_6 = -8,192`?

Answer 1

Negative 33554432

Explanation:

Let K be a constant.
Let the geometric ratio be r.
Let `a_n` be the resulting `n^(th) color(white)(x)`term of the sequence

Then at

`n=1` we have `a_n= kr^n = (+8)`
`n=6` we have `a_n=kr^n=(-8192)`

The fact that `a_6` is negative implies a switching between positive and negative sequence. This is achieved by the incorporation of `(-1)^(n+1)`. It has to be `n+1` or an appropriate variation to force `a_n` to be positive and negative in the appropriate place.

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To find r

Known:
`a_1=(-1)^(1+1)(k)(r^(1)) = (+8) " "-> " "kr=8` ........(1)
`a_6 = (-1)^(6+1)(k)(r^(6)) = -8192" -> " " -kr^(6) = -8192`..(2)

To isolate r use `(2) divide (1)` giving:

`(-kr^6)/(kr) = (-8192)/8`

Thus `r^5 =1025` Note the reversal from negative to positive.

so `r = root(5)(1025) =4`

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To find k by substitution into (1)
`(-1)^2(k)(4)=8`
k=2

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Thus:
`a_12 = (-1)^(12+1) (2)(4)^12= - 33554432`

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