First, WLOG $b_i \in [0,1)$. Now note that: \[|X_n - n(a_1 + ... + a_k)| \le 2k\] Thus we get: \[{ | \frac{X_{ak} - X_k}{(a-1)k}} - (a_1 + a_2 + ... + a_k) | \le \frac{4}{(a-1)}\] Note that ${\frac{X_{ak} - X_k}{(a-1)k}}$ is an integer. Thus by setting $a \to \infty$ we get $a_1 + ... + a_k$ is an integer as well, as desired.

$Y_n = \sum_{i=1}^{k} a_i n + b_i$ is also arithmetic progression, so $Y_n - X_n$, which is the sum of decimal part of $a_i n + b_i$ is also in arithmetic progression. $Y_n - X_n$ is bounded by $0, k$ since the 'decimal part' is between $0, 1$. Bounded arithmetic progression is constant so $Y_n - X_n$ is constant. The common difference of $Y_n$ which is $\sum a_i$ is the same with that of $X_n$, but $X_n$ is an integer sequence so $\sum a_i$ is an integer

Define $\textstyle \sum a_i=A,\sum b_i=B$ and suppose $X_i=X_1+(i-1)d$. Since $X_1$ and $X_2$ are integers, so $X_2-X_1=d$ is also an integer. $\textstyle \sum a_in+b_i+1>\sum \lfloor a_in+b_i\rfloor>\sum a_in+b_i-1$ $\implies nA+B+k>X_n>nA+B-k\quad (1)$ Putting $n=1$ we deduce that $A+B+k>X_1>A+B-k\implies k-A-B>-X_1>-A-B-k\quad (2)$ Adding (1) & (2), we get $(n-1)A+2k>X_n-X_1>(n-1)A-2k\implies 2k>(n-1)(d-A)>-2k$ So if $d-A$ is positive (resp. negative) taking $n\to \infty$ would cross the upper (resp. lower) bound. So $d-A=0\implies d=A$ Since $d$ is integer so $A$ is also integer.

As always, let $\displaystyle\sum_{i=1}^{k}a_i=A,\displaystyle\sum_{i=1}^{k}b_i=B$, and let $X_n = \sum_{i=1}^k [a_in + b_i] = Sn+T$ (as it is an arithmetic series). So, we have $\sum (a_in+b_i) \geq \sum [a_in + b_i] \geq \sum (a_in + b_i -1) $ $\Rightarrow An +B \geq Sn+T \geq An+B - k$ $\Rightarrow A+ \frac B n \geq S +\frac Tn \geq A + \frac Bn - \frac kn$ Taking $n \rightarrow \infty$ implies $S = A$, and as $S$ is an integer (common difference from an integer arithmetic sequence), so is $A$.

Take any $i$ . We have : $$\frac{ [a_in+b_i]-b_i+1}{n} \geq a_i \geq \frac{ [a_in+b_i]-b_i}{n}$$.Summing over from $i=1 \text{ to } k$ we have , $$\frac{X_n - B_k+k}{n} \geq A_k \geq \frac{X_n - B_k}{n}$$. Here , $B_k= \sum_{i=1 }^k b_i$ and $A_k= \sum_{i=1 }^k a_i$ . Now from the fact that the $X_i 's$ are in A.P with integral common difference $d$ . , it boils down to $$ \frac{t}{n} + \frac{n-1}{n} \cdot d \geq A_k \geq \frac{t-k}{n} + \frac{n-1}{n} \cdot d $$. Letting $n \rightarrow \infty$ yields $A_k=d \in \mathbb{Z}$ as desired . $ \text{Q.E.D.} \blacksquare$.

Does this work? It seems far too easy for a #3. Note that $X_n=(a_1+a_2+\hdots+a_k)n+O(1)$. So by taking $n\to\infty$, the common difference must be $a_1+\hdots+a_k$. But each term is an integer so the commom difference must be an integer.

dame dame

@above, can you explain a bit more about the $O(1)$ thing you used? I do not see how you can treat the $b_i$ as if they do not exist for large values of $n$? What am I missing?

Define another sequence \[ Y_n = \sum_{i=1}^k a_in + b_i \quad (n=1,2,...). \]Since both $X_n, Y_n$ are arithmetic progressions even $Z_n = Y_n - X_n$ is an arithmetic progression. However it is bounded from both sides; (values lie in $[0, k)$) and thus it is constant. So $Y_2 - Y_1 = X_2 - X_1 \in \mathbb{Z}$. But $Y_2 - Y_1 = \sum_{i=1}^k a_i$ and thus we are done.

Not sure if this is correct, if you notice anything wrong pls lmk. Sps $X_n$ form an arithmetic progression, i.e, there is some real $a$ such that $X_n=a(n-1)+X_1, \forall n \in \mathbb{N}$, which can be rewritten as: $R_n=\sum_{i=1}^k\left(\frac{[a_in+b_i]-[a_i+b_i]}{n-1}\right)=a$. But: $a_i(n-1)-1<[a_in+b_i]-[a_i+b_i]<a_i(n-1)+1$, summing up these inequalities we get: $\sum_{i=1}^ka_i-\frac{k}{n-1} < R_n <\sum_{i=1}^ka_i+\frac{k}{n-1}$. Taking $n\to\infty$, we get: $a=\sum_{i=1}^ka_i$, but then $X_2-X_1=a=\sum_{i=1}^ka_i$ which is surely an integer. Done

Nice question. Same solution as (10)

Main observation

does this actually just work Treat $k$, all $a_i$, and all $b_i$ as constant and $n$ as variable. Let $S$ be the sum of all $a_i$. Since the addition of $b_i$ and the flooring can have at most a $O(1)$ effect, $$X_n=Sn+O(1).$$However, the arithmetic progression condition implies $X_n=mn+O(1)$ for some integer $m$, so therefore $S$ is an integer.

Notice we have $[a_in + b_i] \in a_in + b_i + (-1, 0]$. Thus $X_n \in (\sum a_i)n + (\sum b_i) + (-k, 0]$. Suppose $\sum a_i$ is not an integer, than $X_n$ has an integer slope $d$, then take $n = 1, \frac{10000k}{|d - \sum a_i|}$ to finish since the two lines cannot remain within $k$ of each other.