arithmetic progression

jvignacio · #1 $Old$ November 10th, 2008, 01:38 AM

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

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

how do we find the number of terms ? cheers

HallsofIvy · #2 $Old$ November 10th, 2008, 03:46 AM

Quote:

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!

jvignacio · #3 $Old$ November 10th, 2008, 03:55 AM

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 $View Post$

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!

mr fantastic · #4 $Old$ November 10th, 2008, 04:01 AM

Quote:

Originally Posted by jvignacio $View Post$

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.

jvignacio · #5 $Old$ November 10th, 2008, 04:13 AM

Quote:

Originally Posted by mr fantastic $View Post$

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?

mr fantastic · #6 $Old$ November 10th, 2008, 04:15 AM

Quote:

Originally Posted by jvignacio $View Post$

so the total number of terms within 200 & 5000 is 30 and the total sum = 71250?

No.

jvignacio · #7 $Old$ November 10th, 2008, 04:25 AM

Quote:

Originally Posted by mr fantastic $View Post$

No.

okay =[

HallsofIvy · #8 $Old$ November 10th, 2008, 04:42 AM

Quote:

Originally Posted by jvignacio $View Post$

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.

mr fantastic · #9 $Old$ November 10th, 2008, 04:48 AM

Quote:

Originally Posted by HallsofIvy $View Post$

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

jvignacio · #10 $Old$ November 10th, 2008, 04:57 AM

Quote:

Originally Posted by mr fantastic $View Post$

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. =[