The prize of a lottery is $100 in the first week, $200 in the second week, $400 in the third week,

eddy2017 said:
let me see if, after an strenuous workout suggested by jonah, and some teasing, ( not specifically brainteasing [but same effect;)] rendered by my better half, I manage to get this:
1 week =100
2 week=200
3 week= 400
4 week = 800
5 week=1600
6 week=3200
7 week=6400
8 week=12800
9 week=25000
10 week= 51200
That is the prize money accumulated. If it is not, then neither the workout nor the teasing did me any good. lol
Thanks for always being there for me.
Now, if this is correct and I think it is, i would like to give the sum of a geometric progression a try. let me see if I can.
Click to expand...
There is a typo in the amount for the 9th week.
Also, I assumed that the "accumulated" means the sum of the amounts for each week.
 
blamocur said:
There is a typo in the amount for the 9th week.
Also, I assumed that the "accumulated" means the sum of the amounts for each week.
Click to expand...
Yes, got it.
9 week = 25600
10 week= 51200
Thanks a lot.
Yes, accumulated means that. The sum of the amount for each week which is a series or progression of amounts given, as I studied in sources found on the net when you mentioned the sum of GP. Very interesting and yet uncharted territory for me until yesterday when you mentioned it.
 
Formula for sum of geometric progression.
S = a + a(r) + a(r)^2 + . . . + a(r)^n−1
where a = first term
r =ratio
n = the number of terms
In the problem at hand, I am having trouble identifying the ratio? ( maybe a power of 2?).
The ratio being doubled the previous amount, But I do not how to express it. I have studied geometric sequence but the ratio is always the same there( in the tutorial that i watched) but here I have 100 the first week and 200 the second week and then doubles up every week. That throws me off.
 
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I think that ratio is common to all the terms in both an arithmetic and a geometric sequence, the difference being that in arithmetic sequence we are dealing with addition and subtraction while in a geometric sequence we are dealing with multiplication and division between terms and that is our case here. But then again, I am failing to recognize the ratio here.
By the way,
Happy thanksgiving to all teachers and tutors on this site!. Big thanks for all of you.
 
eddy2017 said:
Formula for sum of geometric progression.
S = a + a(r) + a(r)^2 + . . . + a(r)^n−1
where a = first term
r =ratio
n = the number of terms
In the problem at hand, I am having trouble identifying the ratio? ( maybe a power of 2?).
The ratio being doubled the previous amount, But I do not how to express it. I have studied geometric sequence but the ratio is always the same there( in the tutorial that i watched) but here I have 100 the first week and 200 the second week and then doubles up every week. That throws me off.
Click to expand...
Look up the "direct" formula for the sum.
 
eddy2017 said:
I think that ratio is common to all the terms in both an arithmetic and a geometric sequence, the difference being that in arithmetic sequence we are dealing with addition and subtraction while in a geometric sequence we are dealing with multiplication and division between terms and that is our case here. But then again, I am failing to recognize the ratio here.
By the way,
Happy thanksgiving to all teachers and tutors on this site!. Big thanks for all of you.
Click to expand...
I believe they call it "common difference" instead of "ratio" in arithmetic progressions.
As for geometric progression, ratio means the ratio of the [imath]n+1[/imath]th term to the [imath]n[/imath]th term. In your case it would be the ratio of, for examply, 6th week's amount to that of the 5th week.

Happy Thanksgiving everybody, at the very least everybody in the US!
 
blamocur said:
I believe they call it "common difference" instead of "ratio" in arithmetic progressions.
As for geometric progression, ratio means the ratio of the [imath]n+1[/imath]th term to the [imath]n[/imath]th term. In your case it would be the ratio of, for examply, 6th week's amount to that of the 5th week.

Happy Thanksgiving everybody, at the very least everybody in the US!
Click to expand...
Thanks!
 
I just watched and studied a video where a teacher proved how we get to this formula.
I think this is the 'direct' formula' you asked me to look up, Mr lev.
Sn =a(1-r^n) / 1-r
 
blamocur said:
I believe they call it "common difference" instead of "ratio" in arithmetic progressions.
As for geometric progression, ratio means the ratio of the [imath]n+1[/imath]th term to the [imath]n[/imath]th term. In your case it would be the ratio of, for examply, 6th week's amount to that of the 5th week.

Happy Thanksgiving everybody, at the very least everybody in the US!
Click to expand...
Yes, in arithmetic sequence and in geometric sequence they call it 'common difference', but
an arithmetic sequence has a common difference between terms and a geometric sequence has a common ratio between terms.
 
eddy2017 said:
I'm still ahving trouble identifying what the ratio is.
Click to expand...
In the formulas, the common ratio is r.

Note that r is also the base of the powers.

What is the base of the powers in your first post?

That is your common ratio. Remember, not all ratios look like fractions.

:)
 
The base of powers in my initial post is 2, then, 2 is the common ratio.
Now, my question: is 2 the common ratio because that is the common difference between the terms?. I think the answer is yes.
 
eddy2017 said:
Yes, got it.
9 week = 25600
10 week= 51200
Thanks a lot.
Yes, accumulated means that. The sum of the amount for each week which is a series or progression of amounts given, as I studied in sources found on the net when you mentioned the sum of GP. Very interesting and yet uncharted territory for me until yesterday when you mentioned it.
Click to expand...
Not so fast

The accumulated sum is :

S = 100 * (2^10 - 1) =100 * (1024 -1) = 102300

Look up sum of first 'n' terms of a GP series with initial value 'a' and common ratio 'r':

Sn = a * \(\displaystyle \frac{r^n - 1}{r - 1} \)

In your case, n = 10, a=100, & r = 2

Check the answer 102300 using the equation [1]
 
eddy2017 said:
I just watched and studied a video where a teacher proved how we get to this formula.
I think this is the 'direct' formula' you asked me to look up, Mr lev.
Sn =a(1-r^n) / 1-r
Click to expand...
You need grouping symbol in the denominator

Sn =a(1- r^n) / (1-r) .......................... when r < 1

Sn =a(r^n - 1) / (r - 1 ) .......................... when r > 1
 
Subhotosh Khan said:
Not so fast

The accumulated sum is :

S = 100 * (2^10 - 1) =100 * (1024 -1) = 102300

Look up sum of first 'n' terms of a GP series with initial value 'a' and common ratio 'r':

Sn = a * \(\displaystyle \frac{r^n - 1}{r - 1} \)

In your case, n = 10, a=100, & r = 2

Check the answer 102300 using the equation [1]
Click to expand...
I saw your reply Mr Khan. I'm at dinner now so I will check the result with the equation first thing tomorrow morning.
Thanks.
Second equation because r happens to be greater than 1.
 
eddy2017 said:
Now, my question: is 2 the common ratio because that is the common difference between the terms?. I think the answer is yes.
Click to expand...
No, Eddy. You're confusing the geometric sequence with an arithmetic sequence. The terms in an arithmetic sequence differ by the same amount (that's why it's called a common difference).

With geometric sequences, we call r the common ratio because (starting with the second term) the ratio of each term to the prior term is always the same: r. Here are four examples:

800/400 = 2
400/200 = 2
200/100 = 2
an / an-1 = 2

I forgot. Are you working with math textbooks now?

:)
 
Subhotosh Khan said:
You need grouping symbol in the denominator

Sn =a(1- r^n) / (1-r) .......................... when r < 1

Sn =a(r^n - 1) / (r - 1 ) .......................... when r > 1
Click to expand...
Greetings Mr Khan, I did check it out with the formula and it checks out. The result is the same.
Sn= a(1-r^2) / 1-r
Sn = 100 ( 1-2^n) / (1-2)
Sn =102300
 
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