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What is the greatest positive integer that must divide the sum of the first ten terms of any arithmetic sequence whose terms are positive integers?

 Mar 17, 2020
 #1
avatar+219 
+1

Hello Guest.

 

I believe that the answer to this question is 5.

 

This is because in any arithmetic sequence, there are 5 pairs of equal sums. So if the sum is any natural number(positive integer), then the 5 pairs' sum is also divisible by 5. 

 

Hope this makes sense.

 

Cheers!laugh

 Mar 17, 2020
edited by Hypotenuisance  Mar 17, 2020
edited by Hypotenuisance  Mar 17, 2020
 #2
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+1

Thank you! Can you explain a little more? I am not very familiar with Arithmetic Sequences. Thanks!

Guest Mar 17, 2020
 #3
avatar+110829 
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The terms of an AP are 

\(T_n=a+(n-1)d\)

a is the first term and d is what keeps getting added to give the next term

 

for example

if a = -3 and d=5 the sequence would be     -3, 2, 7, 12 etc     see it starts at -3 and 5 keeps getting added.

 

With yours a is a positive integer and to make it easy d can be too

so the first 10  the sequence is

a, a+d, a+2d, .......a+9d

Add all those terms up and you get   10a+ 1d+2d+3d+4d+5d+6d+7d+8d+9d = 10a+ 45

What can be factorized out of that?

 Mar 18, 2020
 #4
avatar+2314 
+3

I am with my new account nowwink 

 

Thank you Melody! I think I got it! I greatly appreciate all of you who helped me with this problem!

CalTheGreat  Mar 19, 2020
 #5
avatar+110829 
0

Welcome aboard CTG    laugh

 

I am glad that I could help you.

 

 I hope you enjoy it here and learn a lot while you are at it :)

Melody  Mar 20, 2020
edited by Melody  Mar 20, 2020

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