Help with finding the number of terms in a geometric sum

Question

I'm currently stuck on a problem involving a geometric sum, and I was hoping to get some assistance with it. Here's the problem:

A geometric sum is equal to 215. The first term is 5, and the last term is 320. I need to find the number of terms in the sum.

Here's what I have so far:

\[215 = 5 \cdot (r^n - 1) / (r - 1)\] \[320 = 5 \cdot r^{n-1}\]

I'm having trouble finding the value of the common ratio (r) and the number of terms (n) in the sequence. I'm not sure where to begin or how to proceed.

One user suggested using the fact that the first term is 5 and the last term is 320 to find the ratio (r) in terms of the number of terms (n) and vice versa. However, I'm not sure how to do that.

I tried multiplying r on both sides of the second equation (320 = 5 * r^(n-1)), but I'm not sure how that helps me find the ratio (r) or the number of terms (n).

I would greatly appreciate any guidance or insights on how to approach this problem and find the number of terms in the geometric sum. If you could provide a step-by-step explanation or walk me through the solution, that would be incredibly helpful.

Answer 1

Since $r^{n-1}=\frac{320}5=64$,$$43=\frac{215}5=\frac{r^n-1}{r-1}=\frac{64r-1}{r-1},$$and therefore $r=-2$. And, since $(-2)^{n-1}=64$, $n=7$.

Answer 2

$320=5\times r^{n-1}$

$64=r^{n-1}$ and $64r=r^n$

$215=\frac{5(r^n - 1)}{r-1}$

$215=\frac{5(64r - 1)}{r-1}$

$43=\frac{64r - 1}{r-1}$

$r=-2$

then

$320=5\times (-2)^{n-1}$

$64=(-2)^6=(-2)^{n-1}$

$n=7$