Arithmetic Progression

A sequence $a_1, a_2, a_3, \ldots$ is called an Arithmetic Progression (A.P.) if the difference between every pair of consecutive terms is the same constant, called the common difference, usually denoted $d$:

If the first term is $a$ (also written $a_1$), the A.P. is written out as

so that each term is obtained from the previous one by adding $d$. If $d>0$ the A.P. is increasing; if $d<0$ it is decreasing; if $d=0$ every term equals $a$.

The $n$th (general) term. Reading off the pattern above, the first term needs $0$ copies of $d$ added to $a$, the second term needs $1$ copy, the third needs $2$ copies, and in general the $n$th term needs $(n-1)$ copies:

This can be proved formally by induction on $n$: it holds for $n=1$ (giving $a_1=a$); and if it holds for some $n$, then $a_{n+1} = a_n + d = [a+(n-1)d]+d = a+nd$, which is exactly the formula with $n$ replaced by $n+1$ — so it holds for every $n$.

Sum of the first $n$ terms — derivation by "reverse and add". Let

Write the same sum with its terms listed in the reverse order:

Now add these two expressions for $S_n$ term by term. In each column, one term contributes $a+kd$ from the first line and $a+(n-1-k)d$ from the second, and these add to exactly $2a+(n-1)d$ — the same value in every one of the $n$ columns, because the $d$-parts always add up to $(n-1)d$ regardless of $k$. So

This is the celebrated trick attributed to the young Gauss, who is said to have summed $1+2+\cdots+100$ instantly this way.

An equivalent form using the last term. If $l = a_n = a+(n-1)d$ is the last (i.e. $n$th) term of the sum, the formula above can be written as

i.e. $n$ times the average of the first and last terms — since $2a+(n-1)d = a + [a+(n-1)d] = a+l$. This form is often quicker to use when the last term is already known, and both forms must always agree.

Recovering a term from consecutive sums. As noted in Section 1, $a_n = S_n - S_{n-1}$; for an A.P. this gives an independent check on the sum formula, since substituting $S_n=\frac n2[2a+(n-1)d]$ and $S_{n-1}=\frac{n-1}{2}[2a+(n-2)d]$ and simplifying does indeed recover $a_n=a+(n-1)d$. …