Every term is d more than the one before, so the bars grow (or shrink) in a straight line
Controls
Series type
Readings
- nth term, Tₙ
- 17.0
- First + last term, a + l
- 19.0
- Sum of the first n terms, Sₙ
- 57.0
How to use this simulation
- Start in AP mode and drag the first term a and the common difference d: the bars grow or shrink at a steady, straight-line rate.
- Tick “Show Gauss's pairing trick”: every bar gets covered by a mirrored bar from the other end, and the tops all land at the same height, forming a rectangle.
- Watch the partial-sum curve on the right in AP mode: it keeps curving upward like a parabola and never levels off.
- Switch to GP mode and slide the common ratio r toward 0.5: the curve on the right visibly flattens towards a fixed line, S∞.
- Push r above 1 or below −1: the bars explode in size and the sum curve never flattens, and the “Behaviour of the series” readout switches from converges to diverges.
Saving pocket money and a rumour that spreads
Suppose you save part of your lunch money every week. In week one you save 2 units, and every week after that you save 3 more than the week before: 5 in week two, 8 in week three, and so on. Those numbers, 2, 5, 8, 11, 14, 17, form an arithmetic sequence, because subtracting any term from the next one always gives the same number, 3. How much have you saved after six weeks is the first calculation on this page.
Now flip the story around. A juicy piece of gossip is told to 2 people on day one. The next day each of them tells one more person, and the day after that those new listeners each tell one more — so the number of new listeners each day equals the number from the day before, doubling every time: 2, 4, 8, 16, 32. That is a geometric sequence, because dividing any term by the one before it always gives the same number, 2. Within a handful of days the audience can reach the hundreds of thousands, even though it started small.
Both stories ask the same two questions: how much grows at each step, and how much is the running total after several steps? The first question is answered by the "nth term" formula, the second by the "sum" formula. The simulation above shows the bar chart and the running-sum curve moving together in both modes.
Starting from zero: what a sequence and a series really are
A sequence is a list of numbers built by a fixed rule; each number in the list is called a term. The first term is written a, and the nth term is written Tₙ. Adding up the terms of a sequence gives a series, and the sum of the first n terms is written Sₙ.
In an arithmetic progression (AP), the difference between any term and the one before it is always the same; that constant difference is called the common difference, d. In symbols, Tₙ − Tₙ₋₁ = d for every n. In the savings example above, a = 2 and d = 3.
In a geometric progression (GP), the ratio between any term and the one before it is always the same; that constant ratio is called the common ratio, r. In symbols, Tₙ ÷ Tₙ₋₁ = r for every n. In the gossip example, a = 2 and r = 2.
Both kinds of sequence are built the same way: each term comes from the one before it by one fixed rule, only the rule is "add" for an AP and "multiply" for a GP. That single difference explains every other difference between them, from straight-line versus explosive growth to a sum that never stops growing versus one that settles down.
Key terms
Get the vocabulary straight before the formulas; definition questions come straight from this table.
| Term | Symbol | What it means |
|---|---|---|
| Term | Tₙ or aₙ | One number in the sequence, identified by its position n |
| First term | a or T₁ | The very first number in the sequence |
| Common difference | d | The constant amount added to go from one AP term to the next |
| Common ratio | r | The constant factor multiplied to go from one GP term to the next |
| Number of terms | n | How many terms are being counted |
| Last term | l or Tₙ | The nth term, where the counting stops |
| Partial sum | Sₙ | The sum of the first n terms |
| Sum to infinity | S∞ | The value the sum approaches if the number of terms is allowed to grow forever |
Young Gauss and a ten-second sum
There is a well-known story about the German mathematician Carl Friedrich Gauss (1777–1855). As a schoolboy, his teacher wanted the class occupied and asked them to add up every whole number from 1 to 100. While the other children started adding one number at a time, Gauss reportedly wrote down the answer within seconds.
His trick was simple but beautiful: he paired the numbers from opposite ends, the first with the last, the second with the second-last, and so on — 1 + 100 = 101, 2 + 99 = 101, and every pair gives the same sum, 101. There are 50 such pairs, so the total is 50 × 101 = 5,050.
The same trick works for any arithmetic sequence, because in an AP the sum of the first and last term always equals the sum of the second and second-last term, and so on for every pair. Tick "Show Gauss's pairing trick" in the simulation above and you will see exactly this: every bar gets covered by its mirror-image partner and lands at the same height.
Arithmetic series formulas: the nth term and the sum
The formulas follow straight from the definition. How many times d has been added to the first term decides the nth term.
Tₙ = a + (n − 1)dAdd d exactly (n − 1) times to the first term
Sₙ = n/2 × [2a + (n − 1)d]Number of terms times the average of the first and last
Sₙ = n/2 × (a + l)Equivalent, once l = the last term = Tₙ is known
Why Tₙ = a + (n − 1)d
Going from the first term to the second adds d once (T₂ = a + d). Going to the third adds d again (T₃ = a + 2d). Reaching the nth term means d has been added exactly (n − 1) times, because there are (n − 1) steps between the first term and the nth.
Why Sₙ = n/2 × (a + l): Gauss's rectangle
Write the series once forwards and once backwards, and add the two. The i-th term of the forward list and the i-th term of the backward list always add up to (a + l), because whatever one has gained by moving forward, the other has lost by moving backward the same amount. There are n such positions, so 2Sₙ = n(a + l), which gives Sₙ = n(a + l)/2.
In the example (a = 2, d = 3, n = 6): the last term is l = Tₙ = 17. The pair sum a + l = 19, so Sₙ = 6 × 19 ÷ 2 = 57.
Geometric series formulas: the nth term, the finite sum and the sum to infinity
Multiplying instead of adding at every step changes the shape of the formula.
Tₙ = a rⁿ⁻¹Multiply the first term by r exactly (n − 1) times
Sₙ = a(rⁿ − 1) / (r − 1)When r ≠ 1; when r = 1, Sₙ = na
S∞ = a / (1 − r)Defined only when |r| < 1
Why Tₙ = a rⁿ⁻¹
Going from the first term to the second multiplies by r once (T₂ = ar). Going to the third multiplies by r again (T₃ = ar²). Reaching the nth term means r has been multiplied in exactly (n − 1) times, by the same reasoning that gave the AP formula.
Why Sₙ = a(rⁿ − 1)/(r − 1): multiply and subtract
Write Sₙ = a + ar + ar² + … + arⁿ⁻¹. Multiply both sides by r: rSₙ = ar + ar² + … + arⁿ. Now subtract Sₙ from rSₙ: every middle term cancels, leaving rSₙ − Sₙ = arⁿ − a. The left side is (r − 1)Sₙ, so Sₙ = a(rⁿ − 1)/(r − 1).
In the example (a = 2, r = 0.5, n = 6): Sₙ = 2(0.5⁶ − 1) ÷ (0.5 − 1) = 3.9375. Notice how close that already is to 4 — a hint of the sum-to-infinity coming next.
Taking n to infinity: S∞ = a/(1 − r)
When |r| < 1, the term rⁿ shrinks towards zero as n grows, because a number smaller than 1 gets smaller every time it multiplies itself. Setting rⁿ to zero in Sₙ = a(rⁿ − 1)/(r − 1) leaves S∞ = a(0 − 1)/(r − 1) = −a/(r − 1) = a/(1 − r).
With r = 0.5 here, |r| < 1, so S∞ = 2/(1 − 0.5) = 4. The sum up to n = 6 was already 3.9375, right beside 4; taking more terms pushes it even closer to 4, but it never goes past it.
When |r| ≥ 1, rⁿ never shrinks — it stays constant (r = 1) or grows (|r| > 1) — so Sₙ keeps growing without bound as n grows, never settling near any value. That is called a divergent series.
Arithmetic vs geometric: the differences at a glance
The two behave so differently that laying them side by side makes the contrast obvious.
| Feature | Arithmetic (AP) | Geometric (GP) |
|---|---|---|
| Getting the next term | Add d to the previous term | Multiply the previous term by r |
| Type of growth | Linear (grows or shrinks at a steady rate) | Exponential (grows or shrinks very fast) |
| nth term | Tₙ = a + (n−1)d | Tₙ = a rⁿ⁻¹ |
| Sum Sₙ | n/2 × (a+l) | a(rⁿ−1)/(r−1) |
| Sum as n → infinity | Always grows without bound | Settles at a/(1−r) if |r|<1, else grows without bound |
| Real-world example | Equal loan instalments, seats added per row | Compound interest, virus spread, radioactive decay |
Try these in the simulation
Predict the outcome before you run the simulation, then check.
- In AP mode, set d = 0: every term is equal, the bar chart looks flat, and Sₙ simplifies to na.
- In AP mode, make d negative (say −3): the terms shrink and eventually go negative, and the bars dip below zero.
- Turn on Gauss's trick with n odd, then with n even: with n odd the middle bar pairs with itself, and its height is still exactly half of a+l.
- In GP mode, set r = −0.5: the terms alternate positive and negative, yet the sum still settles at a limit because |r| < 1.
- In GP mode, increase n slowly: with r = 2 the bars grow so fast that the scale keeps changing, while with r = 0.5 the bars shrink so fast they are hard to see after a few terms.
Solved problems
Every solution states what is known, then the formula, then the substitution — write it the same way in an exam.
Problem 1: the simulation's default AP (a = 2, d = 3, n = 6)
Tₙ = a + (n−1)d = 2 + (6−1) × 3 = 2 + 5 × 3 = 17. Sₙ = n/2 × (a+l) = 6/2 × (2+17) = 3 × 19 = 57. Open the simulation and both readings match.
Problem 2: Gauss's sum from 1 to 100
Here a = 1, d = 1, n = 100. Sₙ = n/2 × (a+l) = 100/2 × (1+100) = 50 × 101 = 5,050. That is Gauss's ten-second answer.
Problem 3: a fresh AP, 5, 8, 11, …, its 10th term and sum
Here a = 5, d = 3, n = 10. T₁₀ = 5 + (10−1) × 3 = 5 + 9 × 3 = 32. S₁₀ = 10/2 × (5+32) = 5 × 37 = 185.
Problem 4: how many terms give a sum of 210?
In the series 3, 7, 11, … with a = 3, d = 4, setting Sₙ = n/2 × [2a+(n−1)d] equal to 210 and solving the resulting quadratic in n gives n = 10. Check: T₁₀ = 3 + 9×4 = 39, and S₁₀ = 10/2 × (3+39) = 5 × 42 = 210, which matches.
Problem 5: the simulation's default GP (a = 2, r = 0.5, n = 6)
Tₙ = a rⁿ⁻¹ = 2 × 0.5⁵ = 0.0625. Sₙ = a(rⁿ−1)/(r−1) = 3.9375. And S∞ = a/(1−r) = 2/0.5 = 4. Again, the readings panel matches exactly.
Problem 6: the chessboard and doubling grains
In the old story, the first square of a chessboard holds 1 grain and every next square holds double the one before. That is a GP with a = 1, r = 2. The sum of the first 10 squares is S₁₀ = a(rⁿ−1)/(r−1) = 1 × (2¹⁰−1)/(2−1) = 1,023, and the tenth square alone holds T₁₀ = a rⁿ⁻¹ = 2⁹ = 512 grains. Carried out to all 64 squares, the total outgrows the world's entire wheat harvest — that is the power of exponential growth.
Problem 7: a recurring decimal is an infinite GP
The number 0.333… splits into 0.3 + 0.03 + 0.003 + …, a GP with a = 0.3, r = 0.1. Since |r| < 1, S∞ = a/(1−r) = 0.3/0.9 = 0.333…, which is exactly 1/3. The sum-to-infinity formula is what proves 0.333… is a precise value, not an approximation.
Quick reference: AP vs GP at the simulation's default values
This table makes it obvious how AP's Sₙ grows in a straight line while GP's Sₙ settles near 4 as n grows.
| n | AP Tₙ (a=2, d=3) | AP Sₙ | GP Tₙ (a=2, r=0.5) | GP Sₙ |
|---|---|---|---|---|
| 1 | 2 | 2 | 2.0000 | 2.0000 |
| 2 | 5 | 7 | 1.0000 | 3.0000 |
| 3 | 8 | 15 | 0.5000 | 3.5000 |
| 4 | 11 | 26 | 0.2500 | 3.7500 |
| 5 | 14 | 40 | 0.1250 | 3.8750 |
| 6 | 17 | 57 | 0.0625 | 3.9375 |
Mistakes almost everyone makes
Avoiding these keeps exam marks from slipping away on sequences and series questions.
- Adding d exactly n times instead of (n − 1) times — counting from the first term leaves only (n − 1) steps.
- Dividing by (r − 1) when r = 1. That makes the denominator zero; use Sₙ = na directly instead.
- Plugging into S∞ when |r| ≥ 1. The sum-to-infinity formula is only defined for |r| < 1; otherwise the series diverges.
- Confusing the common difference with the common ratio: d is found by subtracting, r is found by dividing.
- Treating "sequence" and "series" as the same word. A sequence is the list of terms; a series is their sum.
- Mixing up Tₙ and Sₙ — Tₙ is one specific term, Sₙ is the running total of the first n terms; they answer different questions.
Sequences and series in real life
These formulas show up well outside the exam hall.
- Equal loan or rent instalments: if the payment changes by the same amount every month, the payments form an arithmetic sequence.
- Compound interest: a bank balance that earns the same interest rate every year grows as a geometric sequence.
- The spread of a virus or a piece of information: if each person infects (or tells) an average of r others, the total affected grows geometrically.
- Radioactive decay or a drug leaving the bloodstream: losing a fixed fraction in each equal time interval is a geometric sequence with r between 0 and 1.
- Seating in a stadium: if each row has a fixed number of extra seats compared with the row before, the total seat count is an arithmetic sum.
- A bouncing ball: if each bounce reaches a fixed fraction of the previous height, the total distance travelled is found from an infinite geometric series.
Exam tips
Sequences and series appear on almost every school and standardized test syllabus. Two habits earn marks reliably: state which formula applies before substituting numbers, and always check whether |r| < 1 before using the sum-to-infinity formula.
A worked exam-style question
Question: The first sequence has first term 2 and common difference 3. The second sequence has first term 2 and common ratio 0.5. (a) Define common ratio. (b) Find the 6th term of the first sequence. (c) Find the sum of the first 6 terms of the second sequence. (d) Is it true that the second sequence's sum to infinity is a finite number? Justify your answer.
(c): Using Sₙ = a(rⁿ−1)/(r−1) gives S₆ = 3.9375. (d): since r = 0.5 and |r| < 1, the series converges, so S∞ = a/(1−r) = 4 is indeed a finite number — the statement is true.
Revision: one-screen summary
The night before an exam, this list and the comparison table above are enough to refresh everything.
- AP: next term = previous term + d. GP: next term = previous term × r.
- Tₙ(AP) = a + (n−1)d. Tₙ(GP) = a rⁿ⁻¹.
- Sₙ(AP) = n/2 × (a+l) — the formula behind Gauss's rectangle.
- Sₙ(GP) = a(rⁿ−1)/(r−1) when r ≠ 1; when r = 1, Sₙ = na.
- S∞(GP) = a/(1−r), only when |r| < 1; otherwise the series diverges.
- AP grows or shrinks at a steady rate; GP grows or shrinks exponentially — the one idea behind every other difference.
Frequently asked questions
What's the difference between an arithmetic and a geometric sequence?
In an arithmetic sequence (AP), the difference between consecutive terms is constant (the common difference, d); in a geometric sequence (GP), the ratio between consecutive terms is constant (the common ratio, r). An AP grows or shrinks at a steady rate; a GP grows or shrinks exponentially.
What is the sum of an arithmetic series formula?
The sum of the first n terms is Sₙ = n/2 × [2a + (n−1)d], also written n/2 × (a+l) where l is the last term. With a=2, d=3, n=6, this gives Sₙ=57.
What is the formula for the sum of a geometric series?
The sum of the first n terms is Sₙ = a(rⁿ−1)/(r−1), valid when r ≠ 1. When r = 1, every term equals a, so Sₙ = na instead.
What is the sum-to-infinity formula and when does it apply?
The sum of an infinite geometric series is S∞ = a/(1−r), and it is only defined when |r| < 1. When |r| ≥ 1 the terms do not shrink toward zero, so the sum never settles near any value — the series diverges.
How do I find the common difference or common ratio?
Subtract any term from the one after it to get the common difference, d = Tₙ − Tₙ₋₁. Divide any term by the one before it to get the common ratio, r = Tₙ ÷ Tₙ₋₁.
What is Gauss's pairing trick?
Pair the first term with the last, the second with the second-last, and so on: every pair sums to the same value, a+l. With n/2 such pairs, the total sum is n(a+l)/2. As a schoolboy, Gauss used exactly this trick to add 1 through 100 in seconds, getting 5,050.
How do I know if a series converges or diverges?
An arithmetic series (with d ≠ 0) always diverges, because its terms keep growing or shrinking without bound. A geometric series converges only when the common ratio has |r| < 1; otherwise it diverges.
Is 0.333… really exactly equal to 1/3?
Yes. Writing 0.333… as 0.3 + 0.03 + 0.003 + … gives an infinite geometric series with a = 0.3 and r = 0.1. Its sum to infinity is S∞ = a/(1−r) = 0.3/0.9 = 1/3 exactly, not an approximation.
Why does the chessboard grains story give such a huge number?
Doubling the grains on every square is a geometric series with r = 2, which grows exponentially. The first 10 squares alone already sum past a thousand grains, and the full 64 squares add up to a number larger than the entire world wheat harvest — that is how much faster geometric growth is than arithmetic growth.
What's the difference between a sequence and a series?
A sequence is an ordered list of numbers following a rule, such as 2, 5, 8, 11. Adding those terms together, 2+5+8+11, gives a series (or sum).
Keep studying this topic
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