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  #1  
Old 11-10-2008, 02:38 AM
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Default arithmetic progression

\frac {numberOfTerms}{2} \cdot (firstTerm+lastTerm)


\frac {numberOfTerms}{2} \cdot (200+5000)

how do we find the number of terms ? cheers
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Old 11-10-2008, 04:46 AM
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Originally Posted by jvignacio View Post
\frac {numberOfTerms}{2} \cdot (firstTerm+lastTerm)


\frac {numberOfTerms}{2} \cdot (200+5000)

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!
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Old 11-10-2008, 04:55 AM
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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

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Originally Posted by HallsofIvy View Post
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!
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Old 11-10-2008, 05:01 AM
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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 t_n = t_1 + (n-1)d.

Therefore 5000 = 200 + (n - 1) 150.

Solve for n.
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Old 11-10-2008, 05:13 AM
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You're expcted to know that t_n = t_1 + (n-1)d.

Therefore 5000 = 200 + (n - 1) 150.

Solve for n.

so the total number of terms within 200 & 5000 is 30 and the total sum = 71250?
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Old 11-10-2008, 05:15 AM
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so the total number of terms within 200 & 5000 is 30 and the total sum = 71250?
No.
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Old 11-10-2008, 05:25 AM
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No.
okay =[
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Old 11-10-2008, 05:42 AM
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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.
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Old 11-10-2008, 05:48 AM
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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 5000 = 200 + (n - 1) 150 .....
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Old 11-10-2008, 05:57 AM
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Which is the answer I was actually hoping the OP would get for themselves by correctly solving the simple equation 5000 = 200 + (n - 1) 150 .....
yes sorry its 33. I had written 4500 for 4950. =[
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