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

$\dfrac19=3r^3\implies r^3=\dfrac1{27}\implies r=\dfrac13$

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

$a_n=\left(\dfrac13\right)^{n-4}\cdot3\implies\boxed{a_n=\dfrac1{3^{n-5}}}$

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

$S_8-rS_8=(1-r)S_8=a_1(1-r^8)\implies S_8=a_1\dfrac{1-r^8}{1-r}$

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

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