# Sequence difference and proving

Prove that if the sequence $a_{n}$ is arithmetic, then for any positive integer k the sequence: $$b_{1} = a_{1}+a_{2} +...+ a_{k}, \\ b_{2}=a_{k+1}+...+ a_{2k}, \\ b_{3}= a_{2k+1}+...+ a_{3k}$$ is arithmetic as well. What is the difference of the sequence $b_{n}$ ?

I figured that a given $b_n$ has a form of $b_n=a_{(n-1)k+1}+a_{(n-1)k+2}+...+a_{nk}$. Hence, $b_{n+1}=a_{nk+1}+a_{nk+2}+...+a_{(n+1)k}$. We know that $a_n$ is arithmetic so we may use the sum formula and we get: $$b_{n+1}=\frac{a_{nk+1}+a_{(n+1)k}}{2}\cdot k \\ b_{n}=\frac{a_{(n-1)k+1}+a_{nk}}{2}\cdot k$$ And further: $$b_{n+1}-b_n=\frac{ka_{nk+1}+ka_{(n+1)k}-ka_{(n-1)k+1}-a_{nk}}2=\frac{1}{2}k(a_{nk+1}-a_{(n-1)k+1}+a_{(n+1)k}-a_{nk})$$ And I believe it's enough to show that $b_n$ is arithmetic as well. $\frac{1}{2}k$ is constant, $a_{nk+1}-a_{(n-1)k+1}$ are two adjacent terms in an arithmetic progression so their difference is certainly constant and $a_{(n+1)k}-a_{(n-1)k}$ are separated by some term in the middle but as it's arithmetic sequence, $(a_{(n+1)k}-a_{nk}=const \land a_{nk}-a_{(n-1)k}=const)\rightarrow a_{(n+1)k}-a_{(n-1)k}=const.$ Is it enough to prove or this should be done some other way? [this paragraph may be a little off now as I found an error - there is no $a_{(n-1)k}$ here but $a_{nk}$ instead which I plugged into the equations above]

And as regards the difference - I thought about using the formula that $a_n=a_1+(n-1)d$ so we have: $$b_{n+1}-b_n=\frac{1}{2}k([a_1+nkd]+[a_1+(nk+k-1)d]-[a_1+(nk-k)d]-[a_1+(nk-1)d])=\frac{k}{2}\cdot 2kd=k^2d$$

So the difference of $b_n$ is $k^2d$. Is it OK?

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You are working too hard to find the difference for the $b_n$. What is $a_{k+s}-a_s$? To get from $a_s$ to $a_{k+s}$ we added $d$ $k$ times. So $a_{k+1}-a_1=kd$, $a_{k+2}-a_2=kd$, and so on up to $a_{2k}-a_k=kd$, total $k^2d$. Same difference for the next bunch, forever. – André Nicolas May 1 '12 at 18:12
OK, thank you :) – Straightfw May 1 '12 at 18:27
$a_{nk+1}-a_{(n-1)k+1}$ is not the difference of adjacent terms of $a$ – Thomas Andrews May 1 '12 at 19:17
You're absolutely right. Thank you for pointing that out! – Straightfw May 1 '12 at 19:19

Let $d$ be the common difference of the original sequence, i.e., $d=a_{n+1}-a_n$ for each $n$. For any $n$ we have $a_{n+k}-a_n=kd$, so for any $n$ we have
\begin{align*} b_{n+1}-b_n&=\sum_{i=1}^ka_{nk+i}-\sum_{i=1}^ka_{(n-1)k+i}\\ &=\sum_{i=1}^k(a_{nk+i}-a_{(n-1)k+i})\\ &=\sum_{i=1}^kkd\\ &=k^2d\;. \end{align*}
This is constant, so the sequence $\langle b_1,b_2,\dots\rangle$ is arithmetic with common difference $dk^2$.