Proof that the Partial Sums of an Arithmetic Sequence Also Form an Arithmetic Sequence
Theorem
An arithmetic sequence $a_n = a + (n-1)d$, its partial sum $\displaystyle S_n = \sum_{k=1}^{n} a_k $, and a certain natural number $m$ for $A_n = S_{mn} - S_{m(n-1)} $ form an arithmetic sequence.
Explanation
It’s really tough if you don’t know.
For example, consider the sequence formed by summing every three natural numbers: $(1 + 2+ 3)= 6 $, $(4+5+6)=15$, $(7+8+9)=24 \cdots$ form an arithmetic sequence with the first term 6 and common difference 9.
This property is also possessed by geometric sequences. The principle is actually simple, so read carefully once and memorize the facts for next time.
Proof
$$ A_n = S_{mn} - S_{m(n-1)} = \left\{ a + (mn-1)d \right\} + \left\{ a + (mn-2)d \right\} + \cdots + \left\{ a + (mn-m)d \right\} $$ grouping $a$ and $d$ respectively and arranging the formula yields $$ \begin{align*} A_n =& ma + \left\{ m^{2} n - { {m(m+1)} \over 2} \right\} d \\ =& ma + \left\{ m^{2} n - m^2 + m^2 - { {m(m+1)} \over 2} \right\} d \\ =& ma + ( m^{2} n - m^2 ) d + \left\{ m^2 - { {m(m+1)} \over 2} \right\} d \\ =& ma + ( m^{2} n - m^2 ) d + { {m(m-1)} \over 2} d \\ =& {m \over 2} \left\{ 2a + (m-1)d \right\} + (n-1) m^{2} d \end{align*} $$ Therefore, $A_n$ is an arithmetic sequence with the first term $\displaystyle {m \over 2} \left\{ 2a + (m-1)d \right\}$ and common difference $m^{2} d$. It’s not necessary to know exactly what the first term and common difference are.
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