Geometric series: how to find $r$ and $n$ given the last term, the first term and the sum of terms $S_n=76$, $a=36$, last term is $16$ given that this is a geometric series how to find $r$ and number of terms  I have tried to solve simultaneously but failed
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 A: HINT : 
If $n$ is the number of terms, we have $$16=36\cdot r^{n-1}$$ and $$76=36\cdot \frac{r^n-1}{r-1}$$
With $r^n=\frac{16}{36}\cdot r$, which you can insert into the second equation, you can easily calculate $r$.
A: Hints, assuming $\;a=a_1\;$ is the first element of the series:
$$76=S_n=36\frac{r^n-1}{r-1}\;,\;\;a_n=36r^{n-1}=16\implies r^{n-1}=\frac49\implies$$
$$76=36\frac{\frac49r-1}{r-1}\implies...$$
A: Suppose geometric sequence has general term $a_n = a_0r^n$ and it has $k$ terms. We are given $a_0 = 36$ (first term) so general term becomes $a_n = 36r^n$. Now, we have to use:
$$S_k = 36 \cdot \frac{r^{k}-1}{r-1}$$
for sum of the terms and
$$a_k = 36r^{k-1}$$
to find the last term. Notice that $a_k = 16$ and $S_k = 76$ are already given so we have $r^{k-1} = \frac{4}{9}$ and $\frac{r^{k}-1}{r-1} = \frac{19}{9}$. I'm leaving rest to you.
A: The first term is $a$
Then last term is $l = ar^n$
And Sum is $\sum_{k=0}^n ar^k$
I should leave it at that but I'm kind of gabby.  So I trust you stopped reading and will work the rest on your own....
$ = a\sum_{k=1}^n(r^k)= a\frac{(r^{n+1} -1)}{r-1}$
... right.  You aren't reading this.  Because you worked it out on your own, right?......
$= \frac {ar^n*r - a}{r-1} = \frac {lr - a}{r-1}$.
.... Okay, I really need to stop.  The rest you've done on your own... right?
So $(r-1)S = lr -a$
So $r = \frac {S-a}{S-l}$
.......
So $r = \frac {76-36}{76-16} = \frac 23$.
And the series is $36, 24, 16$
