+0

-1
1673
5
+2096

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
+220
+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!

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

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

Guest Mar 17, 2020
#3
+118613
0

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
+2096
+1

I am with my new account now

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
+118613
0

Welcome aboard CTG