The 7-Year-Old Who Added 1 to 100 In 30 Seconds
In 1786, a German schoolteacher named Büttner asked his class to add the integers from 1 to 100. He expected to have a quiet hour. Then his 9-year-old student Carl Friedrich Gauss handed in the answer — 5050 — in roughly 30 seconds. Gauss had spotted the symmetry: pair the first with the last (1 + 100 = 101), the second with the second-last (2 + 99 = 101), and so on. There are 50 pairs, each summing to 101. Total: $50 \times 101 = 5050$. The trick generalises to the formula $S_n = \tfrac{n}{2}(a_1 + a_n)$ — the sum of any arithmetic progression. Every formula in this article is, at heart, a pairing or a multiplication trick that turns a long addition into a short one.
A progression in maths is a sequence in which every term after the first is obtained from the previous term by a fixed rule. The three classical progressions — arithmetic, geometric, and harmonic — are the cases where that rule is add a constant, multiply by a constant, or take a reciprocal-of-arithmetic.
Arithmetic Progression (AP) — The Constant Difference
An arithmetic progression is a sequence where each term differs from the previous one by a fixed value $d$, called the common difference.
General form: $a_1, a_1 + d, a_1 + 2d, a_1 + 3d, \ldots$
$n$-th term formula:
$$a_n = a_1 + (n - 1)d$$
Sum of first $n$ terms:
$$S_n = \frac{n}{2}(2a_1 + (n - 1)d) = \frac{n}{2}(a_1 + a_n)$$
The two sum formulas are equivalent — use whichever fits the given information. If you know $a_1$, $d$, and $n$, use the first. If you know $a_1$ and $a_n$, use the second (it's literally Gauss's pairing trick generalised).
Examples: $5, 9, 13, 17, \ldots$ ($a_1 = 5$, $d = 4$). $20, 15, 10, 5, 0, -5, \ldots$ ($a_1 = 20$, $d = -5$).
Geometric Progression (GP) — The Constant Ratio
A geometric progression is a sequence where each term equals the previous term multiplied by a fixed value $r$, called the common ratio.
General form: $a_1, a_1 r, a_1 r^2, a_1 r^3, \ldots$
$n$-th term formula:
$$a_n = a_1 \cdot r^{n - 1}$$
Sum of first $n$ terms (for $r \neq 1$):
$$S_n = a_1 \cdot \frac{r^n - 1}{r - 1} = a_1 \cdot \frac{1 - r^n}{1 - r}$$
Sum of an infinite GP (only when $|r| < 1$):
$$S_\infty = \frac{a_1}{1 - r}$$
Examples: $3, 6, 12, 24, \ldots$ ($a_1 = 3$, $r = 2$). $81, 27, 9, 3, 1, \tfrac{1}{3}, \ldots$ ($a_1 = 81$, $r = \tfrac{1}{3}$). The second one has an infinite sum: $S_\infty = \tfrac{81}{1 - 1/3} = \tfrac{81}{2/3} = 121.5$.
Harmonic Progression (HP) — The Reciprocal Of An AP
A harmonic progression is a sequence whose reciprocals form an arithmetic progression. There is no direct closed-form sum for an HP — to work with one, convert to its AP of reciprocals first.
General form: $\tfrac{1}{a_1}, \tfrac{1}{a_1 + d}, \tfrac{1}{a_1 + 2d}, \tfrac{1}{a_1 + 3d}, \ldots$
$n$-th term formula: Take the reciprocal of the $n$-th term of the corresponding AP:
$$a_n^{\text{HP}} = \frac{1}{a_1^{\text{AP}} + (n - 1)d}$$
Example: $1, \tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4}, \tfrac{1}{5}, \ldots$ — its reciprocals are the AP $1, 2, 3, 4, 5, \ldots$. The 100th HP term is $\tfrac{1}{100}$.
The HP shows up in music (string-length ratios for harmonic intervals), in optics (lens equations), and in machine learning (the learning-rate decay schedule $1/t$).
Quick — Standard — Stretch: three worked examples
Quick — find the 20th term of the AP $5, 9, 13, 17, \ldots$
$a_1 = 5$, $d = 4$, $n = 20$.
$a_{20} = 5 + (20 - 1)(4) = 5 + 76 = 81$.
Final answer: $a_{20} = 81$.
Standard (Wrong-Path-First) — find the 10th term of the harmonic progression $1, \tfrac{1}{4}, \tfrac{1}{7}, \tfrac{1}{10}, \ldots$
Wrong path. First instinct — treat the HP as if it had its own "common difference." Differences: $1 - \tfrac{1}{4} = \tfrac{3}{4}$, then $\tfrac{1}{4} - \tfrac{1}{7} = \tfrac{3}{28}$. Different. So no AP structure. Give up.
Hold on. An HP isn't an AP — but its reciprocals are an AP. Take the reciprocals: $1, 4, 7, 10, \ldots$. That is an AP with $a_1 = 1$ and $d = 3$. The wrong path tried to find the common difference in the wrong sequence.
Correct method. Step 1: take the reciprocal of every term — get the AP $1, 4, 7, 10, \ldots$. Step 2: find the 10th term of this AP: $a_{10}^{\text{AP}} = 1 + (10 - 1)(3) = 1 + 27 = 28$. Step 3: take the reciprocal back: $a_{10}^{\text{HP}} = \tfrac{1}{28}$.
Final answer: $a_{10} = \tfrac{1}{28}$.
This is the most common HP slip in Grade 11 cohorts at our McKinney TX center — roughly four of every ten students try to find a common difference in the HP itself rather than in its AP-of-reciprocals. The two-line check — "is this an HP? If so, reciprocate first" — fixes the error.
Stretch — find the sum of the infinite GP $1, \tfrac{2}{3}, \tfrac{4}{9}, \tfrac{8}{27}, \ldots$
$a_1 = 1$, $r = \tfrac{2}{3}$. Since $|r| < 1$, the sum converges.
$$S_\infty = \frac{1}{1 - 2/3} = \frac{1}{1/3} = 3$$
Final answer: $S_\infty = 3$. Add the first few terms to verify: $1 + 0.667 + 0.444 + 0.296 + 0.197 + \ldots$ — already past 2.6, climbing toward 3. ✓
Which Progression Do I Have? A 30-Second Diagnostic
Given a sequence, identify the type by checking three things in order:
Compute consecutive differences. Constant? It's an AP.
Compute consecutive ratios. Constant? It's a GP.
Take the reciprocals and re-test (1) and (2). Reciprocals form an AP? It's an HP.
If none of the three checks fires, it's something else — quadratic, exponential-but-not-pure-geometric, recursive (Fibonacci-style), or simply not a classical progression at all.
Why Progressions Matter — From Finance To Music
Progressions sit underneath several practical models.
Compound interest. A balance growing at rate $r$ each year is a geometric progression: $P, P(1+r), P(1+r)^2, \ldots$. After $n$ years, $A = P(1+r)^n$ — the $n$-th term of a GP.
Loan amortisation. Each monthly payment splits between principal and interest in proportions that change geometrically. The sum-of-a-finite-GP formula is what computes your monthly EMI.
Music intervals. The fundamental frequency $f$ and its harmonics $2f, 3f, 4f, \ldots$ form an AP — but the string lengths that produce them form an HP. Pythagoras's monochord experiment is the earliest documented use of harmonic progression in music theory (around 530 BCE).
Population doubling. Bacterial growth in a Petri dish: each generation doubles. A GP with $r = 2$. Within hours, the GP outpaces any AP — which is why exponential growth feels so different from linear growth, and why early-pandemic case counts surprised everyone who was thinking in AP terms.
Half-life and decay. Radioactive decay is a GP with $0 < r < 1$. Carbon dating uses the GP formula to compute the age of organic material from the ratio of remaining $^{14}$C.
The destination of this concept: every "growth or decay over time" question is, in disguise, a progression — and the right formula is the one that pairs (AP) or multiplies (GP) instead of grinding through term by term.
Where Students Lose Marks On Progressions
Mistake 1: Trying to find a common difference in a geometric progression (or a common ratio in an arithmetic progression)
Where it slips in: First encounter with a sequence whose type isn't named.
Don't do this: Subtract consecutive terms of $3, 6, 12, 24, \ldots$ — get $3, 6, 12$ — and conclude "differences not constant, no progression."
The correct way: Run both checks in order. If the difference isn't constant, check the ratio: $\tfrac{6}{3} = 2, \tfrac{12}{6} = 2, \tfrac{24}{12} = 2$ — constant ratio 2, so it's a GP. The rusher who only checks one of the two diagnostics misses the GP case.
Mistake 2: Using the $n$-th term formula but counting positions wrong
Where it slips in: When asked for the 10th term, the student plugs in $n = 10$ but uses the formula with $n$ instead of $(n - 1)$.
Don't do this: Write $a_{10} = a_1 + (10)(d)$.
The correct way: $a_{10} = a_1 + (10 - 1)d = a_1 + 9d$. The off-by-one in progressions is the same off-by-one in explicit-sequence formulas — count steps from $a_1$, not positions. The memorizer who recalls "plug in $n$" without the $(n-1)$ refinement loses a mark on every problem.
Mistake 3: Applying the infinite-GP sum formula when $|r| \geq 1$
Where it slips in: GP problems with ratios like $r = 2$ or $r = 1.5$.
Don't do this: Plug $a_1 = 1$ and $r = 2$ into $S_\infty = \tfrac{a_1}{1 - r} = \tfrac{1}{-1} = -1$. The sum can't be negative when every term is positive — the formula doesn't apply.
The correct way: Check the convergence condition first: $|r| < 1$ is required. If $|r| \geq 1$, the infinite sum diverges (grows without bound for $r > 1$, oscillates for $r \leq -1$). The second-guesser who notices the negative answer for a sum of positives is right to recheck — but they need the criterion to know where to look.
The real-world version of Mistake 3 — applying a formula outside its convergence domain — has driven an entire branch of mathematical analysis. When 18th-century mathematicians worked with divergent series as if they converged, they got contradictions (e.g., $1 - 1 + 1 - 1 + \ldots = \tfrac{1}{2}$ by formal manipulation). It took Cauchy's careful 1821 work on convergence to fix the foundation.
Aryabhata and Gauss — A Short History
Aryabhata (476–550, India). In the Aryabhatiya (499 CE), Aryabhata gave compact rules for the sum of an arithmetic progression and the sum of squares — the latter the foundation for what we now call $\sum_{k=1}^{n} k^2 = \tfrac{n(n+1)(2n+1)}{6}$. Indian astronomers used his AP sum to predict planetary positions over long calendar cycles.
Carl Friedrich Gauss (1777–1855, Germany). The schoolboy pairing trick — adding 1 to 100 in 30 seconds — produced the formula $S_n = \tfrac{n}{2}(a_1 + a_n)$ that every AP sum problem still uses. The pairing logic generalises Aryabhata's earlier prose rule into the symbolic formula students learn today.
Why it matters. Aryabhata gave the sum rules a precise algorithmic form 1,200 years before Gauss; Gauss gave them the symbolic compactness that turned them into the formulas in every textbook. Both contributions sit behind the AP sum formula you'll use today.
Bottom Line
A progression in maths is a sequence whose terms follow a fixed rule from the previous term.
The three classical types are arithmetic (common difference), geometric (common ratio), and harmonic (reciprocals form an AP).
AP $n$-th term: $a_n = a_1 + (n-1)d$. GP $n$-th term: $a_n = a_1 r^{n-1}$. HP: solve in the AP of reciprocals.
The infinite GP sum $S_\infty = \tfrac{a_1}{1 - r}$ works only when $|r| < 1$ — otherwise the sum diverges.
Compound interest, music, decay, and growth all reduce to one of these three progressions.
Sharpen your progression skills — three practice problems
If you get stuck on Problem 3, return to the Standard worked example.
Find the sum of the first 50 terms of the AP $4, 9, 14, 19, \ldots$.
The third term of a GP is 12 and the sixth is 96. Find the first term, the common ratio, and the sum to infinity (if it exists).
The first three terms of an HP are $\tfrac{1}{2}, \tfrac{1}{6}, \tfrac{1}{10}$. Find the 12th term.
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