Cannot understand certain aspects when proving the sum of an arithmetic sequence

Question

To start off with, I already know what the equation for the sum of an arithmetic sequence is, this being $S_n = {n\over 2}[2a + (n - 1)d]$, however I cannot understand certain aspects when proving this equation.

For example when:

$S_n = a + (a + d) + (a + 2d) + ... + [a + (n - 2)d] + [a + (n - 1)d]$

why is it that the value $n$ is taken away by decreases each time? Surely the value of $d$ increases over time, so that:

$S_n = a + (a + d) + (a + 2d) + (a + 3d) ...$ and so on.

Another problem is found in this stage of the proof:

$2S_n = [2a + (n - 1)d] + [2a + (n - 1)d] + ... + [2a + (n - 1)d]$

I don't understand why $S_n$ and $a$ are multiplied by $2$, however I think I do understand why $(n - 1)$ and $d$ are not.

The rest of the proof I can understand, it's just grasping the above that I cannot seem to do.

Answer 1

Try substituting $n = 5$, say, in $$ S_n = a + (a + d) + (a + 2d) + \dots + (a + (n - 2)d) + (a + (n - 1)d) $$ and see what happens. Note that $n - 1 > n - 2 > n - 3 > \dots > 3 > 2 > 1 > 0$.

$$ \newcommand{bs}{\!\!\!}\newcommand{cp}{\bs&\bs +\bs&} 2 S_n = \begin{matrix} a\cp (a + d)\cp\dots\cp (a + (n - 2)d)\cp (a + (n - 1)d)\ +\\ (a + (n - 1)d)\cp (a + (n - 2)d)\cp \dots\cp (a + d) \cp a. \\ \end{matrix} $$ Now sum first over columns.

Answer 2

Of course $n{-}1$ is one greater than $n{-}2$, so going from $(a + (n - 2)d)$ to $(a + (n - 1)d)$ represents an increase of $d$.

The next value in the sequence would be $(a+nd)$. So going backwards from that, you will see smaller values, which you can produce by subtracting more from $n$.

Guessing slightly at the proof, it appears that you have produced $2S_n$ by adding two copies, one of which is reversed, to produce a set of constant terms. Adding the reversed sequence means that every summed term gets $2$ copies of the value $a$, one from each contributing term, and a varying number of increments from teh two contributors which always sums to $(n-1)d$.

Answer 3

I guess you are looking for a proof of the formula of the sum of the first $n$ terms of an AP.

So, $S_n= a + (a+d) + \dots +{a+(n-1)d}$

Also, $S_n = {a+(n-1)d} + {a+(n-2)d} + \dots + (a+d) + a$ (after reversing the order)

Now, if you add the corresponding terms of the two equations, you get $$2S_n = n[2a + (n-1)d]$$

$$\implies S_n = \frac {n}{2} [2a + (n-1)d]$$