Associated Topics || Dr. Math Home || Search Dr. Math

Geometric Series and Sequences

```
Date: 8/9/96 at 20:45:44
From: Anonymous
Subject: Geometric Series and Sequences

1. What term of the geometric sequence 3,6,12... is equal to 768?

Result so far:

Tn = ar^n-1
768 = 3 x 2^n-1
2^n-1 = 256

I'm stuck from here (not even sure if I'm on the right track!).

2. For what range of values for y will the series y+y^2+y^3+....y^n
have a limiting sum?  Find the limiting sum if y=3/4.

3. If the limiting sum of the geometric series 1+x+x^2+x^3+... is 20,
find x.

4. Given the geometric series 2,-6,18,... find the smallest value of n
such that |Tn|>3000.

5. Starting with 5, how many consecutive multiples of 5 must be added
to make their sum exceed 1000?

6. Express 0.27(repeated) in rational form.  (I'm not sure where to
start on this one).

Sorry to have so many questions, but I'm having real trouble with
this.  I study by correspondence here in Australia and my tutor is on
leave this weekend so I'm really stumped.  I'd be very grateful for
any assistance - even a starting point.

Melanie
```

```
Date: 8/11/96 at 9:11:34
From: Doctor Anthony
Subject: Re: Geometric Series and Sequences

Question 1:
You are on the right track.  Since 256 = 2^8  we have 2^(n-1) = 2^8

This means that n-1 = 8  or n = 9.

If you could not see this by inspection, you could use logs to find
the value of (n-1).

2^(n-1) = 256               Take ln of both sides:
We have (n-1)ln(2) = ln(256)
n-1 = ln(256)/ln(2)
= 5.545/0.69315
= 8

and so n = 9 as before.

Question 2:
The geometric series converges if -1 <or= y < 1

The sum to infinity is given by  a/(1-r)  = (3/4)/(1 - 3/4)
= (3/4)/(1/4)
= 3

Question 3:
S(inf.) = 1/(1-x) = 20

1 = 20 - 20x

20x = 19    and so  x = 19/20

Question 4:
Here the common ratio is -3 and T(n) = ar^(n-1) > 3000
2*(-3)^(n-1) > 3000

In fact we require  3^(n-1) > 1500
(n-1)ln(3) > ln(1500)
n-1 > ln(1500)/ln(3)
n-1 > 6.6567
n > 7.6567

n = 8

Question 5:
The sequence would be 5 + 5^2 + 5^3 + etc, so this is a GP with common
ratio 5 and first term 5.

S(n) = a(r^n -1)/(r-1)  = 5(5^n - 1)/4  > 1000
5^n -1 > 4000/5
5^n > 801
n*ln(5) > ln(801)
n > ln(801)/ln(5)
n > 4.154

So n = 5

Question 6:
The number is 0.272727.... etc.

You can do this in two ways.  Probably the easier is to let

x = 0.272727....
100x = 27.272727....

Subtracting  99x = 27    and so x = 27/99 = 3/11

Alternatively, we can consider this as a GP with terms:

27/100 + 27/10000 + etc to infinity with a = 27/100 and r = 1/100

Sum to infinity = a/(1-r)  =  (27/100)/(1-1/100)

= (27/100)/(99/100)

= 27/99

= 3/11     as before.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Sequences, Series

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search