That's it

Number Theory Level 3

Suppose $a_1,a_2,\ldots,a_{200}$ are integers arranged on a circle such that $\displaystyle a_n= \frac{a_{n-1} + a_{n+1}}2$ and $\displaystyle \sum_{n=1}^{100} a_{2n} = 1234$ . Find the value of $\displaystyle \sum_{n=1}^{200} a_n$ .

The answer is 2468.

1 solution

Nguyen Thanh Long
Aug 15, 2015

It is very clear to get the expresses: $a_2=\frac{a_1+a_3}{2}$ $a_4=\frac{a_3+a_5}{2}$ $...$ $a_{200}=\frac{a_{199}+a_1}{2}$ $\rightarrow (a_2+…a_{200})=a_1+a_3+…+a_{199}$ $SUM=1234 \times 2 = \boxed{2468}$

Or just notice that $a_1=a_2=\cdots=a_{200}$ .

mathh mathh - 5 years, 10 months ago

Shortest method.I think nothing could be better than this observation.Thanks.

Vishal Yadav - 5 years, 7 months ago

$Upvoted!!!!$

Akshat Sharda - 5 years, 10 months ago

In this case, a solution is impossible. Setting a(1)=x and a(2)=y gives a(n) = (n-1)x - (n-2)y by induction. This also gives the sum from 1 to 100 of a(2n) as 10000x - 9900y. x and y are integers, as a(1) and a(2) are stated to be integers in the question. This means that the sum divides through by 100 to give the integer 100x - 99y. 1234 does not divide by 100, so either the sum cannot equal 1234, or the restriction of a(n) to integers in the question should be removed.

Edwin Fennell - 5 years, 9 months ago

That's it

The answer is 2468.

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