SOLUTION: If Sarah was saving for a computer that cost$1200 the first month she saved$20 and doubled the amount each month after that how many months will it take her to save the money?

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Question 1055333: If Sarah was saving for a computer that cost$1200 the first month she saved$20 and doubled the amount each month after that how many months will it take her to save the money?
Found 3 solutions by Boreal, stanbon, MathTherapy:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
Two ways to do this.
First is brute force
$20+$40+$80+$160+$320+$640=$1260. She will make it in the 5th month.
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Algebraically
sum of 20*2^(n-1)>=$1200, for n=1 to some number k
factor out a 20 and divide both sides by 20
sum of 2^(n-1), for n=1 to k=60
It is a geometric series where a1=2 and the ratio is 2.
Sum is a1{{1-r^k)/(1-r)}=2(1-r^k)/-1 is greater than = 60
2(1-r^k) is less than or equal to -60
(1-2^k) is <=-30
-2^k <=-31
2^k>=31; k=5

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
If Sarah was saving for a computer that cost $1200 the first month she saved$20 and doubled the amount each month after that how many months will it take her to save the money?
-------
1st mth:: 20
2nd mth:: 40
3rd mth:: 40
etc
----
20 + 40x = 1200
40x = 1180
x = 29.5
Ans:: 30 mths when rounded up
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Cheers,
Stan H.
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Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
If Sarah was saving for a computer that cost$1200 the first month she saved$20 and doubled the amount each month after that how many months will it take her to save the money?
Month 1: $20 saved
Month 2: 2(20), or $40 saved
Month 3: 2(40), or $80 saved

The above represents a GP with the following sequence: a%5Bn%5D+=+20%282%29%5E%28n+-+1%29, with:
n = number (month) of the sequence

We now use the sum of a GP, or S%5Bn%5D+=+%28a%5B1%5D+-+a%5Bn%5D+%2A+r%29%2F%281+-+r%29   
%221%2C200%22+=+%2820+-+a%5Bn%5D+%2A+2%29%2F%281+-+2%29 ------- Substituting 
%221%2C200%22+=+%2820+-+2a%5Bn%5D%29%2F%28-+1%29%29
20+-+2a%5Bn%5D+=+-+%221%2C200%22 -------- Cross-multiplying
-+2a%5Bn%5D+=+-+%221%2C200%22+-+20
-+2a%5Bn%5D+=+-+%221%2C220%22
a%5Bn%5D+=+%22-+1%2C220%22%2F%28-+2%29, or $610

As seen above, the term, or month that she will have saved 1,200 or more is the one in which she has saved $610, or more

Now we see that: a%5Bn%5D+=+20%282%29%5E%28n+-+1%29 becomes:
610+=+20%282%29%5E%28n+-+1%29
610%2F20+=+2%5E%28n+-+1%29 ------ Dividing by 20
61%2F2+=+2%5E%28n+-+1%29
log+%282%2C+%2861%2F2%29%29+=+n+-+1 ------- Converting to LOGARITHMIC form
n, or month in which she will have saved $610 = log+%282%2C+%2861%2F2%29%29+%2B+1, or highlight_green%28matrix%281%2C6%2C+5.930737%2C+or%2C+close%2C+to%2C+month%2C+6%29%29. 
This means that in month 5, she would NOT HAVE realized her goal of $1,200, but in month 6, she would’ve saved more than the $1,200 (see table below).


OR


The EASIEST METHOD:
Month     Amount Saved     Total Saved
  1           $20              $20
  2      2(20)  =  $40         $60
  3      2(40)  =  $80        $140
  4      2(80)  = $160        $300
  5      2(160) = $320        $620
  6      2(320) = $640      $1,260