Question

PLEASE help me with this algebra II question! #5 please!

Answer 1

`a_n` is a geometric sequence, which means consecutive terms occur in a fixed ratio. In other words,

`a_n=ra_(n-1)`

for some fixed number
`r`.

Using this rule, we have

`a_5=ra_4`

`a_6=ra_5=r(ra_4)=r^2a_4`

`a_7=ra_6=r(r^2a_4)=r^3a_4`

So, given that
`a_4=3` and
`a_7=\frac19`, we have

`\frac19=3r^3\implies r^3=\frac1{27}\implies r=\frac13`

We can write
`a_n` in terms of
`a_4`:

`a_n=ra_(n-1)=r^2a_(n-2)=r^3a_(n-3)=\cdots=r^(n-4)a_4`

(notice how the subscript and exponent add up to
`n`) so the sequence is given by the explicit rule

`a_n=\left(\frac13\right)^(n-4)\cdot3\implies\boxed{a_n=\frac1{3^(n-5)}}`

Incidentally, we can pull out the first term from this sequence by plugging in
`n=1` to find
`a_1=81`.

Next, if
`S_n` denotes the
`n`th partial sum of the sequence, then

`S_8=a_1+a_2+\cdots+a_8`

For geometric sequences, we can replace
`a_2` through
`a_8` with terms containing
`a_1`:

`S_8=a_1+ra_1+\cdots+r^7a_1=a_1(1+r+\cdots+r^7)`

Multiply both sides by
`r`:

`rS_8=a_1(r+r^2+\cdots+r^8)`

Subtract
`rS_8` from
`S_8`; a bunch of terms cancel and we're left with

`S_8-rS_8=(1-r)S_8=a_1(1-r^8)\implies S_8=a_1(1-r^8)/(1-r)`

For the sequence at hand, plug in
`a_1=81` and
`r=\frac13`. Then

`S_8=81(1-\left(\frac13\right)^8)/(1-\frac13)\implies\boxed{S_8=(3280)/(27)}`