Arithmetic-geometric progression — finding sum: Visual walkthrough

Before we begin, one promise: no symbol appears until it has a picture. Let's meet the players.

Step 1 — What are we even adding up?

WHAT. We have a list of numbers to add. Each number is a count times a shrinking (or growing) piece. Look at the figure: the count $1, 2, 3, \dots$ climbs like a staircase (that's the arithmetic part), and the tile it sits on scales by the same factor $r$ each step (that's the geometric part).

Figure — Arithmetic-geometric progression — finding sum

WHY start here. You cannot cancel or shift a thing you cannot picture. So we lay the whole sum out as physical bars whose heights are the terms $T_{k}$ .

PICTURE. Each blue bar is one term:

T_{k} = count: 1, 1 + d, 1 + 2 d, \dots (a + (k - 1) d) \cdot scale: 1, r, r^{2}, \dots r^{k - 1}

$a$ ::: the very first count (the height of the staircase's first step).
$d$ ::: how much the count jumps each step ( $== d ==$ is the tilt).
$r$ ::: the multiplier on the tile each step ( $== r ==$ is the shrink/grow factor).

The whole thing we want is

S_{n} = a + (a + d) r + (a + 2 d) r^{2} + \dots + (a + (n - 1) d) r^{n - 1} . (1)

Step 2 — Why multiplying by $r$ is the smart move

WHAT. We build a second copy of the sum, but every bar multiplied by $r$ . Multiplying by $r$ nudges each tile one power higher, which visually slides the whole staircase one step to the right.

WHY this tool and not another? We want the messy " $a + k d$ " counts to disappear. If we simply added or squared, they'd stay tangled. But multiplying by $r$ makes term $k$ of the copy sit directly beneath term $k + 1$ of the original — because both then carry the same power of $r$ . Aligned bars can be subtracted, and subtraction is where cancellation lives. This is the single idea the whole method rests on.

PICTURE. The orange row is the shifted copy:

r S_{n} = a r + (a + d) r^{2} + \dots + (a + (n - 2) d) r^{n - 1} + (a + (n - 1) d) r^{n} . (2)

Every orange bar carries one extra $r$ compared to the blue bar above it.
The orange staircase pokes out one slot to the right — that overhang ( $r^{n}$ ) is a leftover we'll deal with.
The blue staircase pokes out one slot to the left — the lonely $a$ — another leftover.

Step 3 — Subtract, and watch the bulk vanish

WHAT. Compute $S_{n} - r S_{n}$ by subtracting each orange bar from the blue bar sitting above it, power of $r$ by power of $r$ .

WHY. In the overlap region, the blue count is $a + k d$ and the orange count directly below is $a + (k - 1) d$ . Their difference is

(a + k d) - (a + (k - 1) d) = d .

The big, ugly, growing part $a + k d$ cancels — only the constant gap $d$ survives at every overlapping power. That is the payoff of Step 2's alignment.

PICTURE. The tall blue/orange bars collapse into short green bars, all the same height $d$ (times a power of $r$ ):

S_{n} - r S_{n} = lonely left end a + the green survivors [d r + d r^{2} + \dots + d r^{n - 1}] - lonely right overhang (a + (n - 1) d) r^{n} . (3)

The $a$ ::: blue's left end had no orange bar under it, so it survives untouched.
Each $d r^{k}$ ::: an overlap where the count-difference is exactly $d$ .
The final $(a + (n - 1) d) r^{n}$ ::: orange's right overhang had no blue above it, so it survives with a minus sign.

Step 4 — The green survivors are a plain GP

WHAT. Focus only on the green bracket. It is $d r, d r^{2}, \dots, d r^{n - 1}$ : a pure geometric progression — a fixed thing $d r$ multiplied by $r$ again and again.

WHY this is the whole point. A pure GP is the one series we already know how to sum in a single line. By reducing the hard AGP to this friendly GP, we've turned an unknown into a known. (This is why the parent calls the GP "the engine.")

PICTURE. Green bars shrinking (if $∣ r ∣ < 1$ ) by the same ratio $r$ each step. Sum of a GP with first term $g = d r$ , ratio $r$ , and $n - 1$ terms:

d r + d r^{2} + \dots + d r^{n - 1} = \frac{d r ( 1 - r ^{n - 1} )}{1 - r} .

$d r$ ::: the first green bar (the GP's starting height).
$r^{n - 1}$ ::: appears because there are $n - 1$ terms; the standard GP formula $g \frac{1 - r ^{m}}{1 - r}$ with $m = n - 1$ .
The single $(1 - r)$ in the denominator ::: comes only from summing this GP. Remember it — a second $(1 - r)$ is about to arrive.

Step 5 — Solve for $S_{n}$ (the first factor of $(1 - r)$ )

WHAT. Substitute the green GP-sum back into (3) and notice the left side is $S_{n} - r S_{n} = (1 - r) S_{n}$ . Divide by $(1 - r)$ to isolate $S_{n}$ .

WHY. We factored $S_{n}$ out of the left side; dividing by $(1 - r)$ is the only way to peel it off. This division hits the GP-sum's own $(1 - r)$ , producing the famous squared denominator on the arithmetic piece.

PICTURE. Two divisions stacked on the tilt term — that's the $(1 - r)^{2}$ .

(1 - r) S_{n} = a + \frac{d r ( 1 - r ^{n - 1} )}{1 - r} - (a + (n - 1) d) r^{n} .

Divide throughout by $(1 - r)$ :

S_{n} = GP heart \frac{a}{1 - r} + arithmetic tilt \frac{d r ( 1 - r ^{n - 1} )}{( 1 - r ) ^{2}} - finite tail \frac{( a + ( n - 1 ) d ) r ^{n}}{1 - r}

$\frac{a}{1 - r}$ ::: the lonely left-end $a$ , divided once.
$\frac{d r ( 1 - r ^{n - 1} )}{( 1 - r ) ^{2}}$ ::: the green GP; two $(1 - r)$ 's — one from summing greens, one from this division. This is why the parent's mnemonic says "tilt over the square."
$\frac{( a + ( n - 1 ) d ) r ^{n}}{1 - r}$ ::: the lonely right overhang, divided once, with its minus sign.

Step 6 — Degenerate case: what if $r = 1$ ?

WHAT. Every formula above secretly divided by $1 - r$ . If $r = 1$ , that is division by zero — the picture breaks. So handle $r = 1$ separately.

WHY. With $r = 1$ the "geometric" tiles never shrink or grow ( $1, 1, 1, \dots$ ), so the term is just the AP count $a + (k - 1) d$ . There is nothing to shift-and-cancel; instead sum it as a plain Arithmetic Progression.

PICTURE. All tiles equal height $1$ ; the bars are a pure arithmetic staircase:

S_{n}_{r = 1} = k = 1 \sum n (a + (k - 1) d) = \frac{n}{2} (2 a + (n - 1) d) .

The $\frac{n}{2}$ ::: pair-up-the-ends trick of an AP.
$2 a + (n - 1) d$ ::: first term $+$ last term.

The multiply-and-shift picture simply does not apply — no ratio, no slide.

Step 7 — Push $n \to \infty$ when $∣ r ∣ < 1$ (the tail dies)

WHAT. Take the finite $S_{n}$ and let the staircase grow forever, but only when $∣ r ∣ < 1$ . Watch the finite-tail term shrink to nothing.

WHY only $∣ r ∣ < 1$ ? The tail carries $r^{n}$ and a count $(a + (n - 1) d)$ . The count grows linearly; $r^{n}$ decays exponentially. Exponential decay crushes linear growth (see Limits: exponential vs polynomial growth), so $(a + (n - 1) d) r^{n} \to 0$ . If instead $∣ r ∣ \geq 1$ , the tiles never shrink, bars keep piling up, and no finite total exists — the " $S_{\infty}$ " formula is meaningless.

PICTURE. The red tail bar shrinking toward the axis as $n$ climbs; also $r^{n - 1} \to 0$ inside the tilt term. What remains:

S_{\infty} = \frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}} (∣ r ∣ < 1)

$\frac{a}{1 - r}$ ::: the surviving GP heart (the $r^{n - 1}$ inside vanished).
$\frac{d r}{( 1 - r ) ^{2}}$ ::: the tilt, now with its finite-only factor $(1 - r^{n - 1}) \to 1$ .
The whole third term ::: gone, because red $\to 0$ .

The one-picture summary

This final figure compresses all seven moving parts: the blue original, the orange $r$ -shifted copy, the two lonely ends, the green GP survivors, and the arrow that turns the whole thing into the boxed $S_{n}$ — plus the $∣ r ∣ < 1$ gate that lets $n \to \infty$ .

Recall Feynman retelling — the whole walkthrough in plain words

I want to add a staircase where the step number grows but each step is on a shrinking (or growing) tile. Adding it directly is a mess, so I build a second staircase and slide it one step to the right (that slide is multiplying by $r$ ). Now matching steps sit on top of each other, differing only by the tiny constant $d$ . I subtract the two staircases: all the big ugly counts cancel, leaving a neat little pile of equal-height bars — that pile is a plain GP, which I can sum in one breath. Two bits stick out at the ends and refuse to cancel: the very first $a$ on the left and one overhanging tile on the right. Peeling the whole answer apart needs me to divide by $(1 - r)$ — and because the GP already carried one $(1 - r)$ , the tilt part ends up divided twice, giving the squared denominator. If the tiles are exactly the same size ( $r = 1$ ), there's nothing to slide, so I just add the numbers like an ordinary AP. And if the tiles keep shrinking ( $∣ r ∣ < 1$ ), I can let the staircase run forever: the far-off overhang shrinks to zero (exponential beats linear), and only two clean fractions survive: $\frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}}$ .

Connections

Arithmetic-geometric progression — finding sum — the parent result these pictures derive.
Geometric Progression — sum — the green-survivors engine of Step 4.
Arithmetic Progression — the $r = 1$ degenerate case of Step 6.
Sum to infinity of GP — what Step 7 becomes when $d = 0$ .
Power series and generating functions — Example's $\frac{1}{( 1 - x ) ^{2}} = \sum n x^{n - 1}$ .
Limits: exponential vs polynomial growth — why the red tail dies in Step 7.