M + (M+1) + (M+2) +....... + (M+N-1) = 100: Find the value of M + N

[Nkn]

Hello,

I'm new here and have no idea about threads /forums.
Working on this math problem for couple of hours now, but not able to solve it yet. Wondering if someone can help?
M and N are both even positive whole numbers that satisfy
M + (M+1) + (M+2) +....... + (M+N-1) = 100
Find the value of M + N

 

[Deleted member 4993]

Hello,

I'm new here and have no idea about threads /forums.
Working on this math problem for couple of hours now, but not able to solve it yet. Wondering if someone can help?
M and N are both even positive whole numbers that satisfy
M + (M+1) + (M+2) +....... + (M+N-1) = 100
Find the value of M + N

How many terms are there in the arithmetic series given to you?

What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33

 

[Steven G]

Hello,

I'm new here and have no idea about threads /forums.
Working on this math problem for couple of hours now, but not able to solve it yet. Wondering if someone can help?
M and N are both even positive whole numbers that satisfy
M + (M+1) + (M+2) +....... + (M+N-1) = 100
Find the value of M + N

I do not believe there exist M and N both positive that will work. You have tried now for at least two hours so please show us your work.

 

[HallsofIvy]

A nice property of "arithmetic series", such as this, is that the average of all terms is the same as the average of the first and last terms! Here, the first term is M and the last term is M+ N- 1 so the average value is (M+ M+N- 1)/2= (2M+ N- 1)/2= M+ (N-1)/2. The sum is equal to that average times the number of terms. So, as Jomo asked, "How many terms are there in the arithmetic series given to you?" Set the sum equal to 100 to get an equation in M and N. (Since you have essentially one equation in two unknowns, there may be an infinite set of answers.)

 

[Steven G]

A nice property of "arithmetic series", such as this, is that the average of all terms is the same as the average of the first and last terms! Here, the first term is M and the last term is M+ N- 1 so the average value is (M+ M+N- 1)/2= (2M+ N- 1)/2= M+ (N-1)/2. The sum is equal to that average times the number of terms. So, as Jomo asked, "How many terms are there in the arithmetic series given to you?" Set the sum equal to 100 to get an equation in M and N. (Since you have essentially one equation in two unknowns, there may be an infinite set of answers.)

The sum of 1 + 2 + ... + N-1 = N(N-1)/2 < 100. Then N(N-1)< 200. Hence N<= 14. So N can only be 2, 4, 6, ... , or 14. Try finding an even number M that works in MN + N(N-1)/2 = 100 using the allowable values for N.

 

[Nkn]

Thank you! I'm impressed with the replies because I thought it'll never be seen/ replied back, but it actually works. I'm still trying to figure out how this site/ forum works.
Yes, I already read the rules, but didn't included my work as I thought it's meaningless. But, if it's that important here then I'm including my scrap work (sorry about the mess).
The 'terms' are not given. I know the answer (it's given), but still not able to solve it. Trying to recall the basics, and learn about arithmetic series.

How many terms are there in the arithmetic series given to you?

What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33

 

[Nkn]

I do not believe there exist M and N both positive that will work. You have tried now for at least two hours so please show us your work.

Thanks for the response! I'm still trying to figure out how this forum /site works.
My work is a scrap (sorry about the mess), but I'm attaching it if it's that important. I'm trying to recall the basics and trying to learn how to solve arithmetic series.
I think you may be right that both M and N can't be positive. Looks like some error in the question (although it's not possible). Because the answer is given- it's an odd number. How can the sum of two even numbers be an odd number?

 

[Nkn]

A nice property of "arithmetic series", such as this, is that the average of all terms is the same as the average of the first and last terms! Here, the first term is M and the last term is M+ N- 1 so the average value is (M+ M+N- 1)/2= (2M+ N- 1)/2= M+ (N-1)/2. The sum is equal to that average times the number of terms. So, as Jomo asked, "How many terms are there in the arithmetic series given to you?" Set the sum equal to 100 to get an equation in M and N. (Since you have essentially one equation in two unknowns, there may be an infinite set of answers.)

Thank you so much for your help! I'm trying to learn from your explanation. The 'terms' are not given. The question is exactly as I have posted.

 

[Nkn]

The sum of 1 + 2 + ... + N-1 = N(N-1)/2 < 100. Then N(N-1)< 200. Hence N<= 14. So N can only be 2, 4, 6, ... , or 14. Try finding an even number M that works in MN + N(N-1)/2 = 100 using the allowable values for N.

Thanks, I'll try