| 
November 10th, 2008, 01:38 AM
| | Super Member | | Join Date: Oct 2007 Location: Santiago
Posts: 359
Country: Thanks: 77
Thanked 3 Times in 3 Posts
| | arithmetic progression
how do we find the number of terms ? cheers | 
November 10th, 2008, 03:46 AM
| | MHF Contributor | | Join Date: Apr 2005
Posts: 2,051
Thanks: 130
Thanked 752 Times in 697 Posts
| | Quote:
Originally Posted by jvignacio
how do we find the number of terms ? cheers | What, exactly did the problem say? What you've written here makes no sense.
In the sequence 200, 5000, 200 is the first term , 5000 is the last term and there are 2 terms. The sum is 5200 or, using your formula, (2/2)(200+ 5000)= 5200.
In the sequence 200, 2600, 5000, 200 is the first term, 5000 is the last term and there are 3 terms. The sum is 7800 or, using your formula, (3/2)(200+ 5000)= (3/2)(5200)= 3(2600)= 7800.
You cannot find either the number of terms nor the sum of an arithmetic series just from the first and last numbers. You have to know what the series is! | 
November 10th, 2008, 03:55 AM
| | Super Member | | Join Date: Oct 2007 Location: Santiago
Posts: 359
Country: Thanks: 77
Thanked 3 Times in 3 Posts
| | the first term is 200, the last term is 5000. The difference between consecutive terms equal to 150.
so basically the sequence is 200, 350, 500, 650 etc etc all the way to 5000!
now i need to find the number of terms within this sequence Quote:
Originally Posted by HallsofIvy What, exactly did the problem say? What you've written here makes no sense.
In the sequence 200, 5000, 200 is the first term , 5000 is the last term and there are 2 terms. The sum is 5200 or, using your formula, (2/2)(200+ 5000)= 5200.
In the sequence 200, 2600, 5000, 200 is the first term, 5000 is the last term and there are 3 terms. The sum is 7800 or, using your formula, (3/2)(200+ 5000)= (3/2)(5200)= 3(2600)= 7800.
You cannot find either the number of terms nor the sum of an arithmetic series just from the first and last numbers. You have to know what the series is! | | 
November 10th, 2008, 04:01 AM
|  | Flow Master | | Join Date: Dec 2007 Location: Zeitgeist
Posts: 10,029
Country: Thanks: 1,999
Thanked 3,949 Times in 3,481 Posts
| | Quote:
Originally Posted by jvignacio the first term is 200, the last term is 5000. The difference between consecutive terms equal to 150. so basically the sequence is 200, 350, 500, 650 etc etc all the way to 5000! now i need to find the number of terms within this sequence | You're expcted to know that  .
Therefore  .
Solve for n.
__________________ There are two things you should never try to prove ...... the impossible and the obvious.
Things everyone should read: | 
November 10th, 2008, 04:13 AM
| | Super Member | | Join Date: Oct 2007 Location: Santiago
Posts: 359
Country: Thanks: 77
Thanked 3 Times in 3 Posts
| | Quote:
Originally Posted by mr fantastic You're expcted to know that  .
Therefore  .
Solve for n. |
so the total number of terms within 200 & 5000 is 30 and the total sum = 71250? | 
November 10th, 2008, 04:15 AM
|  | Flow Master | | Join Date: Dec 2007 Location: Zeitgeist
Posts: 10,029
Country: Thanks: 1,999
Thanked 3,949 Times in 3,481 Posts
| | Quote:
Originally Posted by jvignacio so the total number of terms within 200 & 5000 is 30 and the total sum = 71250? | No.
__________________ There are two things you should never try to prove ...... the impossible and the obvious.
Things everyone should read: | 
November 10th, 2008, 04:25 AM
| | Super Member | | Join Date: Oct 2007 Location: Santiago
Posts: 359
Country: Thanks: 77
Thanked 3 Times in 3 Posts
| | Quote:
Originally Posted by mr fantastic No. | okay =[ | 
November 10th, 2008, 04:42 AM
| | MHF Contributor | | Join Date: Apr 2005
Posts: 2,051
Thanks: 130
Thanked 752 Times in 697 Posts
| | Quote:
Originally Posted by jvignacio so the total number of terms within 200 & 5000 is 30 and the total sum = 71250? | From 200 to 5000 is a distance of 4800 and 4800/150= 32. Since that counts only the steps and not the first number in the sequence the number of terms is 33. | 
November 10th, 2008, 04:48 AM
|  | Flow Master | | Join Date: Dec 2007 Location: Zeitgeist
Posts: 10,029
Country: Thanks: 1,999
Thanked 3,949 Times in 3,481 Posts
| | Quote:
Originally Posted by HallsofIvy From 200 to 5000 is a distance of 4800 and 4800/150= 32. Since that counts only the steps and not the first number in the sequence the number of terms is 33. | Which is the answer I was actually hoping the OP would get for themselves by correctly solving the simple equation  .....
__________________ There are two things you should never try to prove ...... the impossible and the obvious.
Things everyone should read: | 
November 10th, 2008, 04:57 AM
| | Super Member | | Join Date: Oct 2007 Location: Santiago
Posts: 359
Country: Thanks: 77
Thanked 3 Times in 3 Posts
| | Quote:
Originally Posted by mr fantastic Which is the answer I was actually hoping the OP would get for themselves by correctly solving the simple equation  ..... | yes sorry its 33. I had written 4500 for 4950. =[ | | Thread Tools | | | | Display Modes | Linear Mode |
Posting Rules
| You may not post new threads You may not post replies You may not post attachments You may not edit your posts HTML code is Off | | | All times are GMT -7. The time now is 07:06 PM. | | |