Arithmetic Progressions — the constant-difference engine

A sequence $a_1, a_2, a_3, \ldots$ lists terms by position; $a_n$ (also written $T_n$) is the nth term or general term. A series is the sum $a_1 + a_2 + \cdots + a_n$, written compactly as $S_n = \sum_{k=1}^{n} a_k$. Two bridges connect them:

• $S_n = a_1 + a_2 + \cdots + a_n$ (add up the first $n$ terms).
• $a_n = S_n - S_{n-1}$ for $n \ge 2$, and $a_1 = S_1$ (each term is the jump in the running total).

An arithmetic progression has first term $a$ and common difference $d = a_{k+1} - a_k$ (constant). Then:

• nth term: $a_n = a + (n-1)d$.
• Sum of $n$ terms: $S_n = \dfrac{n}{2}\,[\,2a + (n-1)d\,] = \dfrac{n}{2}(a + l)$, where $l = a_n$ is the last term.

The $\tfrac{n}{2}(a+l)$ form is fastest whenever you can see the first and last terms.

Given $S_n$ as a function of $n$, the nth term is $a_n = S_n - S_{n-1}$. A quadratic $S_n = An^2 + Bn$ (no constant term) always describes an AP with common difference $2A$ and first term $A + B$. A non-zero constant term means the very first term breaks the pattern (use $a_1 = S_1$ separately).

The arithmetic mean of $a$ and $b$ is $\dfrac{a + b}{2}$; inserting it between them makes three terms in AP. Two symmetry facts solve most AP word-problems:

• Equidistant terms sum to a constant: $a_k + a_{n+1-k} = a_1 + a_n$ for every $k$.
• Symmetric choice: pick 3 unknown terms as $a-d,\ a,\ a+d$ and 4 as $a-3d,\ a-d,\ a+d,\ a+3d$ — then the sum gives $a$ immediately.

Numbers $a, b, c$ are in AP $\iff 2b = a + c$. Operations that keep an AP an AP (they map a constant difference to a constant difference):

• adding/subtracting a constant $k$: $a+k,\ b+k,\ c+k$ (or $k-a,\ k-b,\ k-c$);
• multiplying/dividing by a non-zero constant $k$: $ka,\ kb,\ kc$ and $\tfrac{a}{k},\ \tfrac{b}{k},\ \tfrac{c}{k}$.

Squaring the terms or taking reciprocals generally breaks the AP. Extremising a product of AP terms: the product of two terms $(a + pd)(a + qd)$, or of a term and its "partner" summing to a constant, is a quadratic in the unknown — its minimum or maximum is at the vertex $x = -\tfrac{B}{2A}$, value $-\tfrac{D}{4A}$, with no calculus needed. Two numbers with a FIXED SUM have the largest product when equal; with a FIXED PRODUCT, the smallest sum when equal (AM–GM).

For one AP, $\dfrac{S_p}{S_q} = \dfrac{p[\,2a + (p-1)d\,]}{q[\,2a + (q-1)d\,]}$; a given value forces a relation between $a$ and $d$. For two APs with $\dfrac{S_n}{S_n'} = \dfrac{f(n)}{g(n)}$, the ratio of nth terms is

Memorable consequences (each provable in two lines):

• If $S_m = n$ and $S_n = m$ (with $m \ne n$), then $S_{m+n} = -(m+n)$.
• If $p\,a_p = q\,a_q$ (with $p \ne q$), then $a_{p+q} = 0$.
• If $S_p = S_q$ (with $p \ne q$), then $S_{p+q} = 0$.

All three come from the same idea: a linear/quadratic in the index, pinned by two conditions.

Rule-based counting sum: identify the qualifying numbers as an AP (first term, common difference, last term), count them, then apply $S_n = \tfrac{n}{2}(a+l)$. Alternating sum: group consecutive terms into pairs of equal value, then add the leftover. First negative term: solve $a + (n-1)d < 0$ for the smallest integer $n$.

If AP$_1$ has common difference $d_1$ and AP$_2$ has $d_2$, their common terms form an AP with common difference $\operatorname{lcm}(d_1, d_2)$, starting at the first shared value. To count them, take the new AP up to the smaller of the two last terms and apply $n = \big\lfloor \tfrac{L - a}{\operatorname{lcm}(d_1,d_2)} \big\rfloor + 1$, where $a$ is the first common term and $L$ is the largest value not exceeding either list's last term.