Arithmetic Sequences and Series

An arithmetic sequence is the simplest kind of pattern: you add the same number each time. The common difference $d$ is what defines it, and also what makes it so easy to work with algebraically.

The defining property:

• A sequence $T_1, T_2, T_3, \ldots$ is an AP if and only if $d = T_2 - T_1 = T_3 - T_2 = T_4 - T_3 = \cdots$ (all consecutive differences equal).
• In symbols: $$d = T_n - T_{n-1} \quad \text{for all } n \geq 2$$
• Test: compute all consecutive differences. If they are all equal, it's an AP.

Examples:

• 3, 7, 11, 15, 19, …, differences are all 4. AP with $d = 4$, $a = 3$.
• 20, 17, 14, 11, 8, …, differences are all $-3$. AP with $d = -3$, $a = 20$.
• 1.5, 2.0, 2.5, 3.0, …, differences are all 0.5. AP with $d = 0.5$, $a = 1.5$.

Not APs:

• 2, 4, 8, 16, …, ratios are constant (not differences). This is a geometric sequence.
• 1, 4, 9, 16, …, differences are 3, 5, 7 (increasing). This is a perfect-square sequence.

AP definition: $d = T_n - T_{n-1}$ constant for all $n$.; Test: compute all consecutive differences, must all be equal.

Pause, copy the AP definition: common difference $d = T_n - T_{n-1}$ is constant for all $n$; test by computing all consecutive differences, they must all be equal, into your book.

We just saw that an AP has a constant common difference $d = T_n - T_{n-1}$. That raises a question: instead of stepping through term by term, can we jump directly to any position in the sequence? This card answers it → the $n$th term formula $T_n = a + (n-1)d$, which also reveals that an AP is a linear function of $n$ with gradient $d$.

Instead of listing terms one by one, the nth term formula jumps directly to any position. It also reveals that an AP is secretly a linear function in disguise.

Derivation: Starting from $T_1 = a$ and adding $d$ each time:

• $T_1 = a$
• $T_2 = a + d$
• $T_3 = a + 2d$
• $T_4 = a + 3d$
• $T_n = a + (n-1)d$, adding $d$ exactly $(n-1)$ times to reach the $n$th term.

Worked example 1, finding a term and a position:

Worked example 2, finding $a$ and $d$ from two terms:

$T_n = a + (n-1)d$ where $a = T_1$ (first term) and $d$ = common difference.; Rewritten: $T_n = dn + (a-d)$, linear in $n$ (straight-line graph, gradient $= d$).

Pause, copy the $n$th term formula $T_n = a + (n-1)d$ and its linear form $T_n = dn + (a-d)$, showing an AP has gradient $d$ when graphed against $n$, into your book.

We just saw that $T_n = a + (n-1)d$ finds any individual term instantly. That raises a question: to find the total of the first $n$ terms, do we have to add them all up one by one, or is there a shortcut? This card answers it → Gauss's pairing trick gives $S_n = \dfrac{n}{2}(a+l)$ (use when you know the last term $l$) and $S_n = \dfrac{n}{2}(2a+(n-1)d)$ (use when you know $d$).

Adding up $n$ terms of an AP looks tedious, but Gauss's pairing trick reduces it to a single multiplication. The result gives two equivalent formulas, each useful in different situations.

Gauss's derivation:

• Write $S_n = a + (a+d) + (a+2d) + \cdots + l$ where $l = T_n = a + (n-1)d$.
• Write $S_n$ again in reverse: $S_n = l + (l-d) + (l-2d) + \cdots + a$.
• Add term by term: each of the $n$ pairs sums to $(a + l)$.
• So $2S_n = n(a + l)$, giving: $$S_n = \dfrac{n}{2}(a + l)$$
• Substitute $l = a + (n-1)d$: $$S_n = \dfrac{n}{2}(2a + (n-1)d)$$

Example: Sum of the first 20 terms of 3, 7, 11, …

Two equivalent AP sum formulas: $S_n = \frac{n}{2}(a+l)$ and $S_n = \frac{n}{2}(2a+(n-1)d)$.; Gauss's trick: write $S_n$ forwards and backwards, add pairs, each pair sums to $a+l$.

Pause, copy both AP sum formulas: $S_n = \dfrac{n}{2}(a+l)$ (when last term $l$ is known) and $S_n = \dfrac{n}{2}(2a+(n-1)d)$ (when $d$ is known) into your book.

We just saw the two AP sum formulas $S_n = \dfrac{n}{2}(a+l)$ and $S_n = \dfrac{n}{2}(2a+(n-1)d)$. That raises a question: in HSC problems where $n$ is unknown, or where you must find $T_n$ from a sum formula, which technique applies and what kind of equation do you end up solving? This card answers it → finding $n$ from $S_n = k$ produces a quadratic; finding $T_n$ from $S_n$ uses $T_n = S_n - S_{n-1}$ for $n \geq 2$.

Most AP exam problems involve finding unknowns from partial information. The key is recognising which formula applies and what equation structure it produces.

Strategy, the 3-unknown rule: The variables in AP problems are $a$, $d$, $n$, $T_n$ and $S_n$. Each formula gives one equation in these unknowns. Two pieces of information give two equations, solve simultaneously.

Type A, find $n$ such that $S_n$ equals a given value (produces a quadratic in $n$):

Type B, find $T_n$ from a given $S_n$ formula:

Type C, how many terms until the sum exceeds a threshold:

Finding $n$ from $S_n = k$: substitute into AP sum formula, rearrange to quadratic, solve, take positive integer root.; Finding $T_n$ from $S_n$ formula: $T_n = S_n - S_{n-1}$ for $n \geq 2$; check $T_1 = S_1$ separately.

Pause, copy the three AP problem types: finding $n$ from $S_n = k$ (rearrange to quadratic, take positive integer root); finding $T_n$ from a sum formula ($T_n = S_n - S_{n-1}$ for $n \geq 2$); threshold problems ($S_n > k$), into your book.

We just saw the algebraic AP problem types, finding $n$ from $S_n = k$ and extracting $T_n$ from a sum formula. That raises a question: when a real-world situation involves a fixed-dollar salary rise, a fixed-unit fuel consumption decrease, or any constant periodic change, how do you identify it as an AP and set up the calculation? This card answers it → identify $a$ (initial value), $d$ (fixed change per period), and $n$; use $T_n$ for a specific period, $S_n$ for the total.

Real situations where a quantity changes by the same fixed amount each period are AP models. Recognising this structure lets you skip building a table, just identify $a$, $d$ and $n$, then apply the formulas.

Common AP models:

• Salary increasing by a fixed dollar amount each year (not a percentage).
• Rent increasing by a fixed dollar amount each month.
• Temperature dropping by a fixed number of degrees each hour.
• Savings increasing by a fixed extra amount each period (the hook question!).

Setup strategy: Identify the initial value ($a = T_1$), the fixed change per period ($d$), and what you need: a specific term ($T_n$) or a total ($S_n$).

AP applications: fixed-amount change per period. Identify $a$ (initial value), $d$ (fixed change), $n$ (number of periods).; Use $T_n$ for a specific period's value; use $S_n$ for total accumulated over $n$ periods.

Pause, copy the AP application framework: identify $a$ (initial value), $d$ (fixed change per period), $n$; use $T_n = a+(n-1)d$ for a specific period's value; use $S_n$ for total accumulated, into your book.