Arithmetic progression (AP) — nth term, sum of n terms — derivations: Worked examples

Before any symbol appears in a new form, we rebuild what it means. Every $a$ is the first term (where you stand), every $d$ is the common difference (the fixed jump), every $n$ is a count of terms (so it must be a whole positive number), and $S_{n}$ is the running total of the first $n$ terms.

The scenario matrix

Every AP behaviour is decided by just two switches: the sign of the start $a$ (do you begin below, at, or above zero?) and the sign of the step $d$ (do the terms climb, stay flat, or fall?). The figure below turns that idea into a picture: the horizontal axis is the sign of $d$ , the vertical axis is the sign of $a$ , and each cell is a distinct AP "shape". The red cell is the one most often forgotten — a progression that both starts negative and keeps falling.

Figure — Arithmetic progression (AP) — nth term, sum of n terms — derivations

The examples below are tagged with the cell they hit, and together they fill the whole grid.

Cell	What makes it different	Watch out for
A. $d > 0$ (increasing)	terms climb	ordinary case
B. $d < 0$ (decreasing, starts positive)	terms fall, can cross zero	sign of $d$ , terms crossing zero
C. $d = 0$ (degenerate / constant)	every term equal	formulas still work, sum $= na$
D. $a < 0, d > 0$ (starts negative, climbs)	start below zero, climb up	don't drop the minus sign
I. $a < 0, d < 0$ (starts negative, keeps falling)	always negative, sinks further	every term negative; sum grows more negative
E. find $n$ (unknown count)	quadratic in $n$	reject non-integer / negative roots (Quadratic Equations)
F. $S_{n}$ given as a formula	recover $a_{n}$	use $a_{n} = S_{n} - S_{n - 1}$ , linearity test (Linear Functions)
G. word problem (real world)	translate story → $a, d, n$	identify what is asked
H. limiting / exam twist	e.g. sum turns negative, or "which term is zero?"	boundary behaviour

Case A — increasing AP ( $d > 0$ )

Look at the figure below. The first few terms are drawn as a staircase: you start at height $a = 3$ (the black dot) and every tread rises by the same $d = 4$ (the red rise-arrows). The height of the $n$ th step is $a_{n}$ , which is why the coefficient of $d$ is $(n - 1)$ — the very first step cost zero climbs. The staircase makes the "linear in $n$ " nature of $a_{n}$ visible as a straight ramp.

Case B — decreasing AP, crossing into negative territory ( $d < 0$ )

Worked example

B · For $17, 13, 9, \dots$ find the first negative term, and the sum up to that term. Forecast: falling by 4 each step from 17 — it should cross zero somewhere around the 5th or 6th term.

$a = 17$ , $d = 13 - 17 = - 4$ . Why negative? Terms shrink, so the "jump" is downward; $d$ carries a minus sign (Cell B).
Solve $a_{n} < 0$ : $17 + (n - 1) (- 4) < 0 \Rightarrow 17 - 4 n + 4 < 0 \Rightarrow 21 < 4 n \Rightarrow n > 5.25$ . Why an inequality? We want the first $n$ making the term drop below zero.
Smallest integer $n$ is $6$ . $a_{6} = 17 + 5 (- 4) = 17 - 20 = - 3$ . Why check integer? $n$ counts terms, so we round up from $5.25$ to $6$ .
$S_{6} = \frac{6}{2} (a + a_{6}) = 3 (17 + (- 3)) = 3 \cdot 14 = 42$ . Why still valid with a negative term? The formulas never assumed positivity — they only need constant $d$ .

Verify: list them $17, 13, 9, 5, 1, - 3$ — the 5th term $1 > 0$ , the 6th $- 3 < 0$ ✓ first negative is term 6. Sum $17 + 13 + 9 + 5 + 1 - 3 = 42$ ✓.

Look at the figure below. The black bars are the still-positive terms; the single red bar is the first term to dip below the zero line (term 6). The dashed line is $a_{n} = 0$ — the moment the AP crosses from positive to negative. This is exactly the "crossing" that Cell B warns about, and locating it is the whole task.

Case I — starts negative and keeps falling ( $a < 0, d < 0$ )

Worked example

I · A bank account is overdrawn at $-\$ 40 $an d e a c h w ee k$ $15 $m or e i ss p e n t :$ -40, -55, -70, \dots$ (dollars). What is the balance in week 5 (the 5th reading), and the total of the first 5 readings? Forecast: it starts negative and sinks further — every term should be negative, and the running sum should get more negative, never turning around.

$a = - 40$ , $d = - 55 - (- 40) = - 15$ (Cell I — negative start, negative step). Why is every term negative? You begin below zero and always subtract; there is no way back up, so the AP never crosses zero.
$a_{5} = a + (5 - 1) d = - 40 + 4 (- 15) = - 40 - 60 = - 100$ . Why $4$ steps? Reaching reading 5 is $n - 1 = 4$ jumps, each of size $- 15$ .
$S_{5} = \frac{5}{2} (a + a_{5}) = \frac{5}{2} (- 40 + (- 100)) = \frac{5}{2} (- 140) = - 350$ . Why negative sum? Adding only negative terms drives the total downward — the opposite of Case A.

Verify: list $- 40, - 55, - 70, - 85, - 100$ — all negative ✓, 5th is $- 100$ ✓. Sum $- 40 - 55 - 70 - 85 - 100 = - 350$ ✓. Compare with the average form: average $= \frac{- 40 + ( - 100 )}{2} = - 70$ , times $5 = - 350$ ✓.

Look at the figure below. Every bar hangs below the zero line — there is no red "crossing" bar here because the sequence never rises to meet it. The single red bar marks the 5th reading, the deepest so far. This is the mirror image of Case B: instead of terms diving through zero once, they start below it and only sink.

Case D — AP that starts negative and climbs ( $a < 0$ , $d > 0$ )

Worked example

D · A diver holds at depth $- 30$ m (below sea level) at the first reading, then rises $4$ m during each following minute: readings $- 30, - 26, - 22, \dots$ (metres, surface $= 0$ ). Reading number $n$ is taken at the start of minute $n$ , so reading 1 = 0 minutes elapsed, reading 7 = 6 minutes elapsed. What is the depth at reading 7, and at which reading does the diver first reach or break the surface? Forecast: starting deep negative and adding 4 each step — it should climb toward zero and cross it after several readings.

$a = - 30$ , $d = - 26 - (- 30) = 4$ (Cell D — negative start, positive step). Why keep the minus? The first term is genuinely below zero; dropping the sign would place the diver above water at the start.
Reading 7 (= 6 minutes elapsed): $a_{7} = a + (7 - 1) d = - 30 + 6 \cdot 4 = - 30 + 24 = - 6$ (still $6$ m under water). Why $6$ steps not 7? The reading index is $n = 7$ , but the number of jumps since the first reading is $n - 1 = 6$ — matching the 6 minutes of elapsed rising. Distinguishing "reading index" from "intervals elapsed" is the trap of Cell D.
Surface reached when $a_{n} \geq 0$ : $- 30 + (n - 1) 4 \geq 0 \Rightarrow (n - 1) \geq 7.5 \Rightarrow n \geq 8.5$ , so reading $9$ . Why round up? $n$ indexes readings; the first reading at or above the surface is reading 9.

Verify: list by reading index $1 : - 30, 2 : - 26, 3 : - 22, 4 : - 18, 5 : - 14, 6 : - 10, 7 : - 6, 8 : - 2, 9 : + 2$ — reading 7 is $- 6$ ✓, reading 8 is $- 2$ (still under), reading 9 is $+ 2 \geq 0$ ✓ first at/above surface. The signs never got dropped, which is the whole point of Cell D.

Look at the figure below. It is a number line of depth (metres). Each reading is a dot placed at its value; the arrows between consecutive dots are the $+ 4$ jumps. Reading 1 sits deepest at $- 30$ ; the dots march rightward and the red dot is reading 9, the first to land on the non-negative side of the $0$ mark. The picture separates index (which dot) from elapsed intervals (how many arrows) — exactly the confusion the wording flags.

Case C — the flat AP ( $d = 0$ )

Look at the figure below. All eight bars have the same height $37$ — a perfectly flat top (the red dashed line). The sum is then just one rectangle: width $n = 8$ times height $37$ , an "area" reading of $S_{n} = na$ . The flatness is why the general sum formula collapses.

Case E — find the number of terms ( $n$ unknown)

Worked example

E · How many terms of $5, 8, 11, \dots$ are needed to reach a sum of $440$ ? Forecast: each term is around 5–50, so we might need somewhere near 15 terms. Expect a quadratic in $n$ .

$a = 5$ , $d = 3$ , and we want $S_{n} = 440$ .
$440 = \frac{n}{2} [2 (5) + (n - 1) 3] = \frac{n}{2} (3 n + 7)$ . Why this form? We know $a$ and $d$ but not the last term, so use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ .
Multiply by 2: $880 = 3 n^{2} + 7 n \Rightarrow 3 n^{2} + 7 n - 880 = 0$ . Why a quadratic? $n$ appears squared — this is the Quadratic Equations link.
$n = \frac{- 7 \pm 49 + 4 \cdot 3 \cdot 880}{6} = \frac{- 7 \pm 10609}{6} = \frac{- 7 \pm 103}{6}$ . Why the $\pm$ ? Both roots must be examined.
$n = \frac{96}{6} = 16$ or $n = \frac{- 110}{6}$ (reject — negative). Why reject? $n$ counts terms; it must be a positive integer (Cell E's whole point).

Verify: $S_{16} = \frac{16}{2} [10 + 15 \cdot 3] = 8 (55) = 440$ ✓. Discriminant $10609 = 10 3^{2}$ is a perfect square, guaranteeing an integer — always check this.

Look at the figure below. The black curve is $S_{n} = \frac{n}{2} (3 n + 7)$ drawn as a smooth parabola in $n$ ; the horizontal dashed line is the target $440$ . The red dot is where they meet — at $n = 16$ , a whole number. This is the quadratic being solved: the intersection is the root we keep, and the parabola's other arm (negative $n$ ) is the root we throw away.

Case F — sum given as a formula, recover the AP

Worked example

F · Given $S_{n} = 2 n^{2} + 3 n$ , find $a_{1}$ , the common difference $d$ , and $a_{10}$ . Forecast: a quadratic $S_{n}$ with no constant term almost always hides an AP. The slope-like coefficient should surface as $d$ .

$a_{n} = S_{n} - S_{n - 1}$ . Why? $S_{n}$ counts everything up to term $n$ ; $S_{n - 1}$ up to term $n - 1$ ; the leftover is term $n$ .
$S_{n - 1} = 2 (n - 1)^{2} + 3 (n - 1) = 2 n^{2} - 4 n + 2 + 3 n - 3 = 2 n^{2} - n - 1$ . Why expand? We must subtract like-for-like.
$a_{n} = (2 n^{2} + 3 n) - (2 n^{2} - n - 1) = 4 n + 1$ . Why is this an AP? $a_{n} = 4 n + 1$ is linear in $n$ — a straight line — so the difference is constant (Linear Functions).
$a_{1} = 4 (1) + 1 = 5$ ; $d = a_{n + 1} - a_{n} = 4$ ; $a_{10} = 4 (10) + 1 = 41$ . Why $d = 4$ ? The slope of the line $4 n + 1$ is the common difference.

Verify: direct check $S_{1} = 2 + 3 = 5 = a_{1}$ ✓. $S_{2} = 8 + 6 = 14$ , so $a_{2} = 14 - 5 = 9 = 4 (2) + 1$ ✓, and $d = 9 - 5 = 4$ ✓.

Look at the figure below. The terms $a_{n}$ are plotted as dots against $n$ and they fall exactly on a straight red line of slope $4$ . That straightness is the visual signature of an AP: a linear $a_{n}$ forces a constant gap $d$ between neighbours, which is the whole content of the "linear ⇔ AP" test.

Case G — real-world word problem

Worked example

G · A theatre has 20 seats in the front row, and each row behind has 2 more seats than the row in front. With 15 rows, how many seats total? Which row first exceeds 40 seats? Forecast: rows grow by 2 — the last row should have around 48 seats, and the total a few hundred.

Translate: $a = 20$ (front row), $d = 2$ (extra per row), $n = 15$ rows. Why an AP? "Same increase every row" is a constant difference.
Last row $a_{15} = 20 + 14 \cdot 2 = 20 + 28 = 48$ seats. Why $14$ ? Reaching row 15 is $n - 1 = 14$ steps from the front.
Total $S_{15} = \frac{15}{2} (20 + 48) = \frac{15}{2} \cdot 68 = 15 \cdot 34 = 510$ seats. Why the average form? We know first (20) and last (48).
First row exceeding 40: solve $20 + (n - 1) 2 > 40 \Rightarrow (n - 1) > 10 \Rightarrow n > 11$ , so row $12$ . Why round up? $n = 11$ gives exactly $40$ (not "exceeds"); the first strictly-above is row 12.

Verify: row 12 $= 20 + 11 \cdot 2 = 42 > 40$ ✓, row 11 $= 40$ (not exceeding) ✓. Total: average $34 \times 15 = 510$ ✓.

Look at the figure below. Each row is a black bar whose height is its seat count, widening steadily from $20$ to $48$ . The stack of bars is the total $S_{15}$ ; the red bar is row 12, the first to poke above the dashed $40$ -seat line. Seeing the sum as a pile of bars makes "terms × average" concrete: level them off and every bar reaches the average height $34$ .

Case H — the exam twist (limiting behaviour)

Worked example

H · For $20, 17, 14, \dots$ , for which $n$ is the sum $S_{n}$ maximum, and what is that maximum? Then show the sum eventually turns negative. Forecast: a decreasing AP starting positive — the sum keeps growing only while terms are still positive, then starts shrinking. The peak is where terms cross zero.

$a = 20$ , $d = - 3$ (Cell B behaviour driving Cell H).
Terms stay positive while $a_{n} = 20 + (n - 1) (- 3) \geq 0 \Rightarrow 23 \geq 3 n \Rightarrow n \leq 7.67$ . Why does this locate the peak? Adding a positive term grows $S_{n}$ ; adding a negative term shrinks it. The maximum sits at the last positive term.
Last positive term: $n = 7$ . $a_{7} = 20 + 6 (- 3) = 2 > 0$ ; $a_{8} = 20 + 7 (- 3) = - 1 < 0$ . Why 7 not 8? Term 8 is negative — including it would reduce the total.
$S_{7} = \frac{7}{2} [2 (20) + 6 (- 3)] = \frac{7}{2} (40 - 18) = \frac{7}{2} \cdot 22 = 77$ (the maximum).
When does the sum go negative? Write $S_{n} = \frac{n}{2} (40 - 3 (n - 1)) = \frac{n}{2} (43 - 3 n)$ . We want $S_{n} < 0$ . Why can we ignore the $\frac{n}{2}$ factor? For any real count of terms $n \geq 1$ we have $\frac{n}{2} > 0$ — it is a strictly positive multiplier. Multiplying by a positive number never flips an inequality, so the sign of $S_{n}$ is decided entirely by the other bracket $(43 - 3 n)$ . Thus $S_{n} < 0 ⟺ 43 - 3 n < 0 ⟺ n > \frac{43}{3} = 14.33$ , so from $n = 15$ onward.

Verify: $S_{7} = 20 + 17 + 14 + 11 + 8 + 5 + 2 = 77$ ✓ and $S_{8} = 77 + (- 1) = 76 < 77$ ✓ (peak confirmed). $S_{14} = \frac{14}{2} (40 - 39) = 7 > 0$ , $S_{15} = \frac{15}{2} (40 - 42) = - 15 < 0$ ✓ (first negative sum at $n = 15$ ).

The figure below plots the running sum $S_{n}$ against $n$ . The black curve rises, tops out, then falls. The red dot marks the peak at $n = 7$ (the last positive term); to its right the curve slides back down and eventually crosses the dashed zero line at $n = 15$ . This picture is the argument: sums grow while terms are positive and shrink once terms turn negative.

Recall

Recall Which formula for which cell?

Unknown last term, know $a, d, n$ ? ::: Use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ . Know first and last term? ::: Use $S_{n} = \frac{n}{2} (a + a_{n})$ — fastest. Given $S_{n}$ as a formula, want a term? ::: $a_{n} = S_{n} - S_{n - 1}$ . Asked "how many terms"? ::: Set $S_{n}$ equal, solve the quadratic, keep only positive integer roots. Decreasing AP, want maximum sum? ::: Stop at the last non-negative term. Starts negative and $d < 0$ ? ::: Every term is negative and the sum only sinks further. Sum turns negative when? ::: Solve $S_{n} < 0$ ; since $\frac{n}{2} > 0$ the sign comes from the bracket alone.

Connections

Parent — AP nth term & sum — the engines used everywhere above.
Quadratic Equations — Cell E "how many terms" reduces to a quadratic in $n$ .
Linear Functions — Cell F: $a_{n}$ linear in $n$ ⇔ AP.
Sigma Notation — $S_{n} = \sum_{k = 1}^{n} a_{k}$ compactly.
Gauss Summation Trick — the pairing behind $S_{n}$ .
Arithmetic Mean — the "average term" reappears as first-plus-last over two.
Geometric Progression (GP) — contrast: multiply-by-ratio, so you'd divide to find the ratio.

First negative term of

17, 13, 9, \dots

Term 6, which equals

- 3

Diver at

- 30

m rising

4

m/min: depth at reading 7 (6 minutes elapsed)?

- 6

m (still under water).

AP starting

- 40

with

d = - 15

: balance at reading 5?

- 100

(and the sum of 5 readings is

- 350

For a decreasing AP starting positive, where is

S_{n}

maximum?

At the last non-negative term.

Given

S_{n} = 2 n^{2} + 3 n

, what is

d

d = 4

(slope of the linear

a_{n} = 4 n + 1

How many terms of

5, 8, 11, \dots

sum to 440?

16

terms.

Sum of 8 constant readings of 37?

296

, since

S_{n} = na

when

d = 0

Why does the sign of

S_{n} = \frac{n}{2} (43 - 3 n)

depend only on

(43 - 3 n)

Because

\frac{n}{2} > 0

for

n \geq 1

, and multiplying by a positive number can't flip the sign.

Worked examples — Arithmetic progression (AP) — nth term, sum of n terms — derivations

The scenario matrix

Case A — increasing AP ( $d > 0$ )

Case B — decreasing AP, crossing into negative territory ( $d < 0$ )

Case I — starts negative and keeps falling ( $a < 0, d < 0$ )

Case D — AP that starts negative and climbs ( $a < 0$ , $d > 0$ )

Case C — the flat AP ( $d = 0$ )

Case E — find the number of terms ( $n$ unknown)

Case F — sum given as a formula, recover the AP

Case G — real-world word problem

Case H — the exam twist (limiting behaviour)

Recall

Connections

Concept Map