# Thread: geometric series...need major help.

1. ## geometric series...need major help.

Find the sum of the following infinite
geometric series. Use a formula and

(100)+(-40)+(16)+(-32)+etc...forever
5
Hint: r = (-2)
5

thats -32/5.... and on the hint is -2/5

2. Do you mean its - $32/5$?

Since the hint says to try $r =-2/5$, we can plug that in to our values and see what happens
$100(-2/5) = -40$
$
-40(-2/5) = 16
$

So we see that each term is changing by a factor of $-2/5$. Now we know the equation for a geometric series is $a/(1-r),$where a is the first term and r is the ratio. Plug in the first term in the series for a, and plug in the ratio r, and then solve the equation to find the sum of the series.

3. $\sum_{k=0}^{n-1} ar^k=a\frac{1-r^n}{1-r}$
When $\lim_{n\rightarrow \infty} a\frac{1-r^n}{1-r} = \frac{a}{1-r}$ since r < 1

4. so how would i solve that, I dont even know what the E thing is...

5. Hello, Steph07!

Find the sum of the following infinite geometric series:
. . $100 -40 + 16 - \tfrac{32}{5} + \hdots$

You're expected to know the formula . . .

. . $S \;=\;\frac{a}{1-r}$ . . . where $a$ is the first term and $r$ is the common ratio.

We have: . $a = 100,\;r = \text{-}\tfrac{2}{5}$

Plug them into the formula . . .

6. $
\sum
$

means the summation, but I guess you don't really need to know that, just plug into the formula will get you your answer.

7. so is the answer 100.4 ?

8. or is it 71.4 ?

9. I think it should be 71.4.

$
100/(1-(-2/5)) = 100/(1+2/5) = (100/(7/5)) = 500/7
$

10. oOo... I see now... thank u