I posted a question last night but all that happened was I became even more confused.

The homework question is:

Find the sum of each geometric series.
a) S10 in the series $2500 +$2562.50 + \$2626.56 + ...

now i would normally use the formula

tn= ar^n-1 except I do not know what tn equals. To find the ^n I need to know what tn is. Now can some one please help me.

2. Hello, VDestinV!

Find the sum of the geometric series.

a) $S_{10}$ in the series: . $\2500 + \2562.50 + \2626.56 + \hdots$
First, find the common ratio $r.$

We have: . $\frac{2562.50}{2500} \:=\:1.025$

. . . .and: . $\frac{2626.56}{2562.50} \:=\:1.024999024$

The common ratio appears to be: . $r \:=\:1.025$

Then: . $S_{10} \;=\;2500\,\frac{1.1025^{10}-1}{1.025-1} \;=\;128,008.4544 \:\approx\:\128,008.45$

3. Originally Posted by Soroban
Hello, VDestinV!

First, find the common ratio $r.$

We have: . $\frac{2562.50}{2500} \:=\:1.025$

. . . .and: . $\frac{2626.56}{2562.50} \:=\:1.024999024$

The common ratio appears to be: . $r \:=\:1.025$

Then: . $S_{10} \;=\;2500\,\frac{1.1025^{10}-1}{1.025-1} \;=\;128,008.4544 \:\approx\:\128,008.45$
wow thanks... but uhh what about the n?

4. 10 is n, as I explained in the other thread.

5. Originally Posted by Soroban
Hello, VDestinV!

First, find the common ratio $r.$

We have: . $\frac{2562.50}{2500} \:=\:1.025$

. . . .and: . $\frac{2626.56}{2562.50} \:=\:1.024999024$

The common ratio appears to be: . $r \:=\:1.025$

Then: . $S_{10} \;=\;2500\,\frac{1.1025^{10}-1}{1.025-1} \;=\;128,008.4544 \:\approx\:\128,008.45$
I was looking over your calculations...

shouldn't . $S_{10} \;=\;2500\,\frac{1.1025^{10}-1}{1.025-1} \;=\;128,008.4544 \:\approx\:\128,008.45$

actually be . $S_{10} \;=\;2500\,\frac{1.025^{10}-1}{1.025-1} \;=\;28,008.4544 \:\approx\:\28,008.45$

6. Yes, that's correct.

7. okay but i tried to do that with this one but it didnt work

s12 in the series 60, 12, 2.4, ...

8. Originally Posted by VDestinV
okay but i tried to do that with this one but it didnt work

s12 in the series 60, 12, 2.4, ...
$a = 60$

$r = \frac{12}{60} = 0.2$

$n = 12$

$S_{12} = \frac{60(1 - 0.2^{12})}{1 - 0.2}$

$S_{12} = 75$