Visual walkthrough — Geometric progression (GP) — nth term, sum of n terms — derivations

---

Step 1 — What a "constant-ratio" chain looks like

WHAT: We line up a chain starting at the first term $a$, giving $a,ar,a{r}^2,\dots$, and mark the jump between each pair.

WHY: Before touching any formula, we must see that every jump is the identical operation "$\times r$". That single repeated act is the whole story — everything later is bookkeeping about how many times we did it.

PICTURE (s01): Four black dots in a row are the terms $a,ar,a{r}^2,a{r}^3$. Between each neighbouring pair a red arrow points rightward, each one labelled "$\times r$" — the same label on every arrow. That visual sameness of the arrows is exactly what makes the chain a GP.

Here $a$ (defined above) is where the chain starts, and each red $\times r$ arrow is one jump to the next term.

---

Step 2 — Counting jumps: why the exponent is $n-1$

WHAT: We reach term number $n$ by starting at $a$ and applying $\times r$ exactly $(n-1)$ times.

WHY this counting: Between $n$ dots there are always $(n-1)$ gaps — like fence posts and the gaps between them. Each gap is one $\times r$. So the power on $r$ must equal the gap count, not the dot count.

PICTURE (s02): The same four dots, now labelled $a_1=a,a_2=ar,a_3=a{r}^2,a_4=a{r}^3$. Under each dot a red note counts the arrows behind it: dot 1 has $0$ arrows, dot 2 has $1$, dot 3 has $2$, dot 4 has $3$. The red arrow-count under each dot is the exponent on $r$ for that term.

$$a_n=\underset{\text{start value}}{\underbrace{a}}\cdot \underset{\begin{array}{c}r\text{ applied}\\ (n-1)\text{ times}\end{array}}{\underbrace{r^{n-1}}}$$

• $a_n$ — the value sitting on dot number $n$.
• $a$ — the starting value, multiplied by $r$ zero times so far ($r^0=1$).
• $r^{n-1}$ — the exponent literally counts the arrows between dot 1 and dot $n$.

Sanity check on the first dot: put $n=1$: $a_1=a{r}^0=a$. Zero arrows, still standing at the start. ✓

---

Step 3 — Setting up the total: the pile we want to add

WHAT: We stack the first $n$ terms into one bar-chart pile and call its total $S_n$.

WHY: Adding $a+ar+a{r}^2+\cdots$ one bar at a time works but is slow, and gives no formula. We want a shortcut that works for a million terms. To find it, first we must see the whole pile at once.

PICTURE (s03): Five black-outlined bars stand side by side, labelled $a,ar,a{r}^2,a{r}^3,a{r}^4$. Each bar is taller than the one before by the factor $r$ (here $r>1$, so they climb upward). A red callout box in the middle names their combined height $S_n$ — the total of all five bars.

$$S_n=\underset{\text{bar }1}{\underbrace{a}}+\underset{\text{bar }2}{\underbrace{ar}}+\underset{\text{bar }3}{\underbrace{a{r}^2}}+\cdots +\underset{\text{bar }n}{\underbrace{a{r}^{n-1}}}\text{\hspace{1em}(1)}$$

Every bar is the previous one times $r$ — the same rule from Step 1, now stacked.

---

Step 4 — The magic move: copy the pile and slide it

WHAT: We multiply equation (1) by $r$, producing a second pile $r{S}_n$.

WHY this tool and not another: We could try adding averages (that's the Arithmetic Progression (AP) trick, which works because AP terms are evenly spaced). But GP terms are not evenly spaced — they explode. What is regular here is the factor $r$ linking neighbours. So the natural tool is: exploit that factor by multiplying through by $r$.

PICTURE (s04): Two sets of bars share the same slots. The black-outlined bars are the original pile $S_n$; the red bars are every term grown by $\times r$, i.e. $r{S}_n$. Line them up: each red bar sits at the height of the next original bar — so they overlap term-for-term, except a brand-new red bar $a{r}^n$ sticks out at the far right end.

$$r{S}_n=\underset{\text{was bar }1}{\underbrace{ar}}+\underset{\text{was bar }2}{\underbrace{a{r}^2}}+\cdots +\underset{\text{was bar }n-1}{\underbrace{a{r}^{n-1}}}+\underset{\text{NEW end bar}}{\underbrace{a{r}^n}}\text{\hspace{1em}(2)}$$

• Every term of (2) is one power higher than the term below it in (1).
• A brand-new term $a{r}^n$ pops out at the far end.
• The original first term $a$ has no partner in (2) — nothing produced it.

---

Step 5 — Subtract, and watch the middle vanish

WHAT: Compute $S_n-r{S}_n$: line (1) minus line (2).

WHY: Every interior term $a{r}^k$ appears in both lines (once in (1), once in (2)), so it cancels. What can't cancel? The lonely $a$ from (1) and the lonely $a{r}^n$ from (2).

PICTURE (s05): Six bars for the powers $a,ar,a{r}^2,a{r}^3,a{r}^4,a{r}^5$. The four interior bars are greyed out and marked "cancels" — they appear in both lines and annihilate. Only the two end bars stay bold red: the first, $a$, and the last, $a{r}^n$. A red box names them the survivors.

$$S_n-r{S}_n=\underset{\text{survivor from (1)}}{\underbrace{a}}-\underset{\text{survivor from (2)}}{\underbrace{a{r}^n}}$$

Now pull $S_n$ out on the left:

$$\underset{\text{factor the left}}{\underbrace{S_n(1-r)}}=\underset{\text{factor the right}}{\underbrace{a(1-r^n)}}$$

• $(1-r)$ — appears because $S_n-r{S}_n=S_n(1-r)$.
• $a(1-r^n)$ — the two lonely ends, with the common $a$ pulled out.

---

Step 6 — Divide out, and the formula appears

WHAT: Divide both sides by $(1-r)$.

WHY carefully: Division is only legal when the divisor isn't zero. Here $1-r=0$ exactly when $r=1$ — so we must set that case aside (Step 7 handles it).

PICTURE (s06): A clean text panel showing the algebra collapse: the line $S_n(1-r)=a-a{r}^n$ on top, then the boxed red result $S_n=a(1-r^n)/(1-r)$ below it, with a caption reminding us this is "the two survivors, over the gap $(1-r)$."

$$\boxed{S_n=\frac{a(1-r^n)}{1-r}=\frac{a(r^n-1)}{r-1}}(r\ne 1)$$

• $a$ — first term, sets the scale of the whole pile.
• $r^n$ — the one place $n$ appears: it's the next term after the pile ($a{r}^n$ without the $a$).
• $1-r$ (or $r-1$) — the leftover from the telescoping; it never contains $n$.

---

Step 7 — The degenerate case $r=1$ (the formula's blind spot)

WHAT: Handle $r=1$ directly, since Step 6 divided by $1-r=0$.

WHY it must be separate: Division by zero is undefined — the boxed formula is silent at $r=1$. But the sum is trivially easy here: $n$ equal bars of height $a$.

PICTURE (s07): Six red bars all exactly the same height $a$, forming a flat rectangle. A label in the middle reads "$n$ equal bars, total $=na$" — since nothing grows, the sum is just $a$ added $n$ times.

$$S_n=\underset{n\text{ copies}}{\underbrace{a+a+\cdots +a}}=\boxed{na}$$

---

Step 8 — The other awkward ratios: $r<0$ and $r=0$

WHAT: Check the formula on two ratios the bar-picture couldn't draw.

WHY: A reader might think the drawings are the proof. They aren't — the algebra of Steps 4–6 is. Here we confirm the formula survives where bars can't.

PICTURE (s08): Two side-by-side panels. Left ($r=-2$): the terms $3,-6,12,-24$ zig-zag across the zero line — positive bars in black point up, negative bars in red point down, showing the $+,-,+,-$ flip. Right ($r=0$): only the first bar, height $a=7$ in red, stands tall; the remaining slots are flat zero.

Case $r<0$ (alternating signs). Take $a=3,r=-2$: the terms are $3,-6,12,-24,\dots$ — they alternate in sign but the derivation never cared. Sum of the first $4$:

$$S_4=\frac{3(1-(-2{)}^4)}{1-(-2)}=\frac{3(1-16)}{3}=\frac{3(-15)}{3}=-15.$$

Direct check: $3-6+12-24=-15$ ✓. The formula is fine — the bars were the only thing that couldn't be drawn.

Case $r=0$ (everything after the first vanishes). With $r=0$ the chain is $a,0,0,0,\dots$ Take $a=7,n=4$:

$$S_4=\frac{7(1-0^4)}{1-0}=\frac{7(1)}{1}=7=a.$$

Direct check: $7+0+0+0=7$ ✓. (The parent note excludes $r=0$ from being a "proper" GP because you can no longer divide by a term to recover $r$ — but the sum formula, taken as pure algebra, still returns the right answer $a$.)

---

Step 9 — What happens as $n$ grows (a peek beyond)

WHAT: Watch the boxed formula's behaviour as $n\to \infty$ (as the number of terms grows without end).

WHY it matters: This is the bridge to the Sum of infinite GP and to Exponential functions — same machinery, infinite pile.

PICTURE (s09): Two curves plotted against $n$. The red curve ($r>1$) shoots upward without limit — the pile explodes. The black curve ($r<1$) climbs but flattens, hugging a dashed grey ceiling line marked $a/(1-r)$ — the pile settles to a finite total even with more and more terms.

• If $|r|<1$: $r^n\to 0$, so $S_n\to \frac{a}{1-r}$ — a finite total even with infinitely many bars.
• If $|r|>1$: $r^n\to \infty$, so $S_n$ grows without bound.
• If $r=1$: $S_n=na$ grows steadily (a straight ramp), not a curve.

---

The one-picture summary

PICTURE (s10): The whole derivation on one chart. The black-outlined bars are the original pile $S_n$; the red bars are its $\times r$ (shifted-by-one) copy $r{S}_n$. An arrow points to the lonely first term $a$ ("no partner") and another to the new end term $a{r}^n$ — the two survivors. A red box in the centre holds the finished formula $S_n=a(1-r^n)/(1-r)$, showing how the two survivors over the gap $(1-r)$ become the total.

This single diagram carries the whole derivation: the original pile $S_n$ (black), its shifted-by-one copy $r{S}_n$ (red), the matching bars cancelling in the middle, and the two lonely survivors $a$ and $a{r}^n$ that become the formula.

Recall Feynman retelling — the whole walk in plain words

Picture a row of blocks. Each block is the one before it, grown (or shrunk) by the same magic factor $r$. To reach the $n$-th block you apply that factor $n-1$ times — one fewer than the block count, because the very first block was free. Remember $n$ is always a counting number ($1,2,3,\dots$) — there is no "block $2.5$". That's why the height of block $n$ is $a$ (the value we started at) times $r$ to the power $n-1$: count the arrows, not the blocks.

Now you want the total height of all the blocks. Adding them one by one is a chore. So here's the trick: make a photocopy of the whole row and slide it one block to the right — which is exactly the same as multiplying every block by $r$. The copy and the original are almost twins: every block in the middle appears in both. Line them up and subtract: the entire middle wipes out, like a telescope collapsing. Only two blocks survive — the very first one of the original ($a$) and the brand-new one at the end of the copy ($a{r}^n$). Their difference, shared over the gap $(1-r)$, is the total. That's the formula $S_n=a(1-r^n)/(1-r)$.

A few warnings. If the magic factor is exactly $1$, no block grows, every block is height $a$, and you just have $n$ equal blocks — total $na$ (the formula would divide by zero there, so we sidestep it). If the factor is negative, the blocks flip sign up and down — you can't draw them as heights, but the formula still spits out the right number. If the factor is $0$, only the first block is left and the total is just $a$. And if the factor is a fraction between $-1$ and $1$, the blocks shrink so fast that even an endless row of blocks has a finite total — the shrinking bars pile up to a ceiling of $a/(1-r)$ and never pass it. That last case is the door to the infinite sum, and you can walk through it on the Sum of infinite GP page.

---

Connections

• Parent — GP nth term & sum — the algebra behind these pictures.
• Arithmetic Progression (AP) — its sum uses reverse-and-add; ours uses shift-and-subtract.
• Sum of infinite GP — Step 9's shrinking case taken to the limit.
• Geometric Mean — the balance point of a 3-term GP.
• Exponential functions — the continuous cousin of $a{r}^{n-1}$.
• Compound interest — real money piling up as a GP.
• Logarithms — the tool to solve $r^{n-1}=k$ for the exponent $n$.

---

Flashcards