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/