Sequence & Series Basics

Switch sequence type, control parameters, and track how terms accumulate.

Worked example — Sequence & Series Basics

terms	3, 5, 7, 9, 11, 13, 15, 17

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Sequence & Series Basics

A sequence is the list; a series is what you get by adding the list up. This page shows both at once — a_n is the nth item of the list, S_n is the running total of the first n of them — and confusing the two is the most common mistake in this topic.

What each symbol means

d: the common difference of an arithmetic sequence — what gets added each step. 2 here, giving 3, 5, 7, 9, …
r: the common ratio of a geometric sequence — what it gets multiplied by. Switch mode and r = 1.5 gives 3, 4.5, 6.75, …
Sₙ: the sum of the first n terms. For the arithmetic defaults, S₈ = 80 while a₈ is only 17.

Where the formula comes from

Write the arithmetic sum forwards and then backwards, one under the other: a₁ + (a₁+d) + ⋯ + a_n above a_n + (a_n−d) + ⋯ + a₁.
Add the two rows column by column. Every column gives the same total, a₁ + a_n, because each step up on one row is matched by a step down on the other — and there are n columns.
So twice the sum is n(a₁ + a_n), giving S_n = (n/2)(a₁ + a_n). At the defaults, (8/2)(3 + 17) = 80 — the formula in the table above, and the trick a young Gauss is said to have used on the numbers 1 to 100.
Now switch the mode selector to geometric: the terms become 3, 4.5, 6.75, … and both formulas in the table change. This sum needs a different trick, because its terms do not pair up — each is a multiple of the last, not a fixed step from it. Write the sum out, and underneath it write r times the same sum: S_n = a₁ + a₁r + ⋯ + a₁rⁿ⁻¹ above r·S_n = a₁r + a₁r² + ⋯ + a₁rⁿ.
Subtract the second line from the first. Every term in the middle appears in both and cancels, leaving only the ends: S_n − r·S_n = a₁ − a₁rⁿ. Factor both sides and divide by (1 − r): S_n = a₁(1 − rⁿ)/(1 − r) — the expression the table shows in geometric mode, giving 147.773438 at a₁ = 3, r = 1.5, n = 8.

Assumes: A constant difference or a constant ratio, forever. Any real sequence that merely looks arithmetic for a few terms will part company from these formulas immediately.
Breaks when: The arithmetic formula has no failure of this kind: S_n = (n/2)(a₁ + a_n) divides only by 2, and 2 is never zero. The geometric one divides by (1 − r), which is exactly why it catches people. Switch to geometric and set r = 1: the sequence becomes 3, 3, 3, …, and the formula shown changes to S_n = n·a₁, giving 8 × 3 = 24. That is not a patch, it is the correct limit — when nothing grows, the sum is just the term repeated.

Adding forever can still land on a whole number 🖖

It sounds like a contradiction that adding endlessly many positive terms could produce a finite total, but a geometric series with |r| < 1 does exactly that, and the sum is a/(1 − r). The most familiar example is hiding in plain sight: 0.999… is the series 0.9 + 0.09 + 0.009 + …, so a = 0.9 and r = 0.1, giving 0.9/(1 − 0.1) = 1. Not approaching 1, not almost 1 — the sum is 1, and 0.999… is another way of writing the number one. Zeno’s runner covering half the remaining distance forever is the same series with a = r = 1/2, which totals exactly 1: he does finish the race.

Adding versus multiplying 🖖

Both sequence types start from a first term, but they grow by different rules. An arithmetic sequence adds the same amount each step (like saving $50 every month), while a geometric sequence multiplies by the same factor (like a balance earning 5% interest yearly). The nth-term formula hands you any single value without listing them all, and the partial sum adds up the first n. Watch how one starting value diverges wildly depending on which rule you pick.

The only sequence that is both 🖖

Try to build a sequence that is simultaneously arithmetic and geometric. It turns out only one family works: a constant sequence like 7, 7, 7, …, where the difference is d = 0 and the ratio is r = 1. Any genuine change breaks one rule or the other, because steady adding always curves away from steady multiplying. It is a rare case where two seemingly independent growth laws are forced to collapse into a single, motionless value.

Practice

Check yourself

Predict the answer first, then use the controls above to find out. Reveal only after you have committed to a guess — that is what makes it practice.

Press geometric decay, then set the ratio to 0.99, then to 1.01. The eight terms barely differ between the two. Which one has a finite infinite sum — and can you tell from the terms on screen?

Set it up:

Show answer

0.99 converges and 1.01 does not, and nothing in the terms or the eight-term sum gives it away: eighth terms of 18.641307 against 21.442707, sums of 154.510611 against 165.713411. Only the infinite-sum row flips, from S∞ = 2000 to Diverges. Convergence is settled by the ratio alone, and no finite stretch of a sequence can reveal it.
Stay at r = 0.99, where the infinite sum is 2000 and eight terms have reached 154.510611. Guess how far along the way that is.

Set it up:

Show answer

About 7.7% — under a thirteenth. Compare r = 0.7, where the same eight terms reach 62.823466 of an infinite 66.666667, or 94%. Both series converge. One is essentially finished after eight terms and the other has hardly begun, so convergence promises a destination and says nothing whatever about the speed of arrival.
Press geometric growth and push n to 30. The thirtieth term on its own is 255668.078977. Predict the sum of all thirty.

Set it up:

Show answer

767000.236932 — almost exactly three times that single last term, so the twenty-nine terms of history together are worth only twice the newest one. Read the sum backwards to see why: at r = 1.5 each term is two thirds of the one after it, making the total a₃₀·(1 + 2/3 + 4/9 + ⋯) = 3a₃₀, a geometric series in its own right.

Problems solved in full

The 8 terms adding up to 80 without adding 8 numbers 5 steps
The 8 terms 3, 5, 7, 9, 11, 13, 15, 17 add up to 80. Get there without adding 8 numbers, then work out what happens to the sum when the list is twice as long.
1. The term formula counts steps, not terms. Getting from a₁ to a₈ takes 7 steps of 2, not 8, and that off-by-one is the only awkward part of the whole topic.
  
  $a_{n} = a_{1} + (n - 1) d = 3 + 2 (n - 1) = 2 n + 1 ⟹ a_{8} = 17$
2. Pair the list from the outside in. 3 and 17 make 20, and so do 5 and 15, 7 and 13, 9 and 11. They have to: moving one place right adds 2 and moving one place left takes 2 away, so each pair loses exactly what the other gains.
  
  $3 + 17 = 20, 5 + 15 = 20, 7 + 13 = 20, 9 + 11 = 20$
3. 8 terms make 4 pairs, and every pair is worth 20. Nobody added 8 numbers.
  
  $S_{8} = 4 \times 20 = 80$
4. Nothing in that argument used 3 or 2. Any arithmetic list of n terms splits into n/2 pairs each worth a₁ + a_n, and for this list the algebra collapses to n(n + 2).
  
  $S_{n} = \frac{n}{2} (a_{1} + a_{n}) = \frac{n}{2} (2 n + 4) = n (n + 2)$
5. Test the closed form somewhere it matters. Doubling the list to 16 terms gives 16 × 18 = 288, which is 3.6 times the sum of 8 terms rather than twice it.
  
  $S_{16} = 16 \times 18 = 288 = 3.6 \times S_{8}$
Answer

The panel prints a₈ = 17 and S₈ = 80, and the 4 pairs of 20 are the reason. Doubling the length multiplies the sum by 4(n + 1)/(n + 2), which is 3.6 here and creeps toward 4 as the list grows — an arithmetic sum is quadratic in the number of terms, because a longer list is also a list of bigger terms. Push the closed form past anything on the page and it stays sharp: n(n + 2) hits 9999 exactly at n = 99, and first clears 10000 at n = 100. This page will not take more than 30 terms, where the sum is only 960, so almost everything this sequence does happens past the last value the input accepts.

$\frac{S _{2 n}}{S _{n}} = \frac{4 ( n + 1 )}{n + 2} ⟶ 4, 99 \times 101 = 9999$
Where pairing the ends broke on a sequence that returns 86.588344 6 steps
Pairing the ends worked on an arithmetic list. Try it on 20, 14, 9.8, 6.86, 4.802, 3.3614, 2.35298, 1.647086 and it returns 86.588344 — more than this sequence can reach if you let it run forever. Find the real sum, and find where the pairing broke.
1. Check the pairs instead of assuming them. The outer pair is 21.647086 and the next one in is 16.35298. Stepping inward multiplies one side by 0.7 and divides the other by 0.7, and a factor gained is not a factor lost — that cancellation only works for adding and subtracting.
  
  $a_{1} + a_{8} = 20 + 1.647086 = 21.647086, a_{2} + a_{7} = 14 + 2.35298 = 16.35298$
2. What does work is a shift. Write the sum out, then write 0.7 times the sum underneath it, lined up one place to the right. Every term in the second row is already sitting in the first.
  
  $S_{8} = 20 + 14 + \dots + 20 \times 0. 7^{7}, 0.7 S_{8} = 14 + 9.8 + \dots + 20 \times 0. 7^{8}$
3. Subtract the rows and everything in the middle annihilates. Only the two overhanging ends survive: the leading 20 from the top and the trailing 20 × 0.7⁸ from the bottom. Divide by 1 − r and you have a closed form for any first term and any ratio.
  
  $S_{8} - 0.7 S_{8} = 20 - 20 \times 0. 7^{8} ⟹ S_{n} = a_{1} \frac{1 - r ^{n}}{1 - r}$
4. Substitute. 0.7⁸ = 0.05764801, so the numerator is 20 × 0.94235199 and the divisor is 1 − 0.7 = 0.3.
  
  $S_{8} = 20 \times \frac{1 - 0.05764801}{0.3} = \frac{18.8470398}{0.3} = 62.823466$
5. Now let the list run on. Each extra term multiplies rⁿ by 0.7, so it heads to 0 and the numerator settles at 1. The sum has a ceiling, which is the one thing an arithmetic series can never have.
  
  $r^{n} \to 0 ⟹ S_{\infty} = \frac{a _{1}}{1 - r} = \frac{20}{0.3} = 66.666667$
6. The gap between the 8-term sum and that ceiling is 3.843201, and it is exactly 66.666667 × 0.7⁸. So the fraction of the total still missing after n terms is rⁿ — the same number that measures how fast the terms shrink also measures how much is left.
  
  $S_{\infty} - S_{8} = 3.843201 = 66.666667 \times 0. 7^{8}$
Answer

The panel prints a₈ = 1.647086, S₈ = 62.823466 and a ceiling of S_∞ = 66.666667. The missing-fraction result is what to keep, because it turns “converges” from a promise into a schedule: 0.7ⁿ ≤ 0.01 needs n ≥ 12.9114, so 13 terms puts you within 1% of everything this series will ever be. Move the ratio and the schedule moves violently while the ceiling stays finite — at r = 0.9 that same 1% costs 44 terms, at r = 0.99 it costs 459. And the 86.588344 the pairing produced was refutable before any of this: a partial sum of positive terms cannot beat the total of all of them, so a method that says otherwise is wrong on sight.

$0. 7^{n} \leq 0.01 ⟺ n \geq \frac{ln 0.01}{ln 0.7} = 12.9114 ⟹ n = 13$

Learning path

Money over time

Leads to Simple vs compound the closed form for a geometric sum.

References (1)

On the Gauss schoolroom story, which the derivation deliberately hedges: B. Hayes, "Gauss's Day of Reckoning." American Scientist 94(3), 200, 2006.

By the 42Math team · Updated 2026-05-21

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

arithmetic growth - Arithmetic sequence with constant difference d=2.
geometric growth - Geometric growth sequence with ratio r>1.
geometric decay - Geometric decay sequence with ratio between 0 and 1.