Let $a_1, a_2, ..., a_n $ be a sequence such that the arithmetic mean of the $n$ terms is $n$. Consider $n = 2009$. Determine the sum of the $2009$ terms of the sequence.
Problem
Source: Paraguayan National Olympiad 2009, Level 3, Problem 4
Tags: algebra proposed, algebra
vinayak-kumar
01.09.2014 07:41
$\frac{a_1+a_2+...+a_{2009}}{2009}=2009$ and so $a_1+a_2+...+a_{2009}=2009^2=\boxed{4036081}$
achen29
16.03.2018 16:54
Seriously wondering if there Is some sort of hidden trick as this looks pretty TOO easy!
adhikariprajitraj
16.03.2018 17:00
LOL I think they have pretty easy Olympiad! Never seen such easy Olympiad questions before.
achen29
16.03.2018 17:16
Hmm.. Moving there and making it to their IMO team lol
AnArtist
16.03.2018 17:24
Leicich wrote: Let $a_1, a_2, ..., a_n $ be a sequence such that the arithmetic mean of the $n$ terms is $n$. Consider $n = 2009$. Determine the sum of the $2009$ terms of the sequence.
$a_{2009} = 4017$.
adhikariprajitraj
16.03.2018 17:37
AnArtist wrote: Leicich wrote: Let $a_1, a_2, ..., a_n $ be a sequence such that the arithmetic mean of the $n$ terms is $n$. Consider $n = 2009$. Determine the sum of the $2009$ terms of the sequence.
$a_{2009} = 4017$.
Yeah, since $a_1=1$ and $a_{2019}=2*2009-1=4017$
# Prove that there are infinites no. of primes in this arithmetic sequence. Also, prove that none of the term in the sequence is even integer.