Question

i need help with this probelm of geometric sum.. i cannot find the sum. I have tried everything. Please help. Medal and fan

Answer 1

So what is our common ratio here?

Answer 2

.5 since its dividing instead of multiplying right?

Answer 3

Not quite...that is 1 thing to remember, we always go by what is being MULTIPLIED to the previous term

So what times 125 = 25? it would be 1/5 or .20

To check again 25 times 1/5 is indeed 5

So our common ratio is 1/5 or (0.20) right?

Answer 4

oh sorry yes i divided wrong

Answer 5

Not a problem

So, what is the formula for summing an infinite series?

Answer 6

Sn= a(r^n-1)/(r-1)

Answer 7

Okay we can work from that...so let me switch 1 thing around and then yes that is the formula for summing up a series with 'n' number of sums

\[\large \sum_{0}^{\infty} a(\frac{1 - r^n}{1 - r})\]

So...'a' would be our first term in the series...and 'r' would be the common ratio

but what would be 'n'?

Answer 8

we have to find the n by using this formula .... Tn=ar^n-1

Answer 9

Well not here we dont...thats the beautiful thing about summing to infinity...'n' would be infinity

so lets focus on the

\[\large r^n\] up there

We know 'r' is 0.20
And now we know that n is infinity

\[\large 0.20^\infty\]

What does that equal? when you continuously multiply 0.20 by itself over and over again, what do we approach?

Answer 10

the same number?? this is where im stuck

Answer 11

Oh okay

Well we can do it out and see it

\[\large 0.20 * 0.20 = 0.04\]
Then again, that time 0.2
\[\large 0.04 * 0.2 = 0.008\]
and again times 0.2
\[\large 0.008 * 0.2 = 0.0016\]
1 more time because I think it is getting clearer
\[\large 0.0016 * 0.2 = 0.00032\]

So lets see, every time we multiply 0.2 by itself, we get a SMALLER and SMALLER number, so it looks like we are getting closer and closer to 0 right?

Answer 12

yes

Answer 13

So then if that all makes sense to you, it makes sense that

\[\large r^n \rightarrow 0.2^\infty \rightarrow 0\]

right?

Answer 14

right..so \[r^n=0\]

Answer 15

Exactly, which brings us to a nicer looking equation since that is gone now

\[\large \sum_{0}^{\infty} = a(\frac{1}{1 - r})\]

SO

plugging everything in we have

\[\large \sum_{0}^{\infty} 125(\frac{1}{1 - 0.20})\]

Answer 16

so we multiply \[125\times1-.20\]

Answer 17

Not quite, remember that we can factor out the constant

\[\large 125 \sum_{0}^{\infty} \frac{1}{1-0.20}\]

And now we can see that since there is no 'N' to deal with, that summation doesnt mean anything here so
...

\[\large 125(\frac{1}{1 - 0.20})\]

Answer 18

ok so i multiply that equation now?

Answer 19

1-.20

Answer 20

125/.8 ?

Answer 21

Indeed! 125/0.8 is correct...you were throwing me off when you said multiply it

But yes, what is 125/0.8 ?

Answer 22

156.25

Answer 23

Perfect! And that is our answer

Answer 24

Thanks you so much! now i know how to do infinity series

Answer 25

No problem! :)