Sequences — Arithmetic and Geometric

⚡ Key Takeaways

Sequences — Arithmetic and Geometric. An arithmetic sequence adds a fixed amount (the common difference dd) at each step; a geometric sequence multiplies by a fixed factor (the common ratio rr). Key results: arithmetic terms grow linearly (an=a1+(n−1)da_n = a_1 + (n-1)d), geometric terms grow exponentially (an=a1rn−1a_n = a_1 r^{n-1}), and both have compact sum formulas. They appear everywhere from loan interest to physics free-fall to computer science. Formalized over centuries; Gauss famously summed an arithmetic series as a child (~1786).

💡 The Core Insight

There are exactly two ways to get from one term to the next using a single fixed operation: add the same number or multiply by the same number. Arithmetic sequences encode the first, geometric the second. Everything else — the nn-th term formula and the sum formula — follows mechanically from repeating that one operation n−1n-1 times.

Formal Statement

DefinitionArithmetic Sequence

A sequence (an)(a_n) is arithmetic if there is a constant dd (the common difference) such that

1
an+1−an=dfor all n≥1.\quad a_{n+1} - a_n = d \quad\text{for all } n \ge 1.

Derivation hint: This is the defining recurrence; "constant difference" is literally the definition. The closed-form (nn-th term) formula is

2
an=a1+(n−1) d,\quad a_n = a_1 + (n-1)\,d,

where a1a_1 is the first term, nn is the position (a positive integer), and dd is the common difference. Derivation hint: Starting from a1a_1, you add dd a total of (n−1)(n-1) times to reach ana_n.

DefinitionGeometric Sequence

A sequence (an)(a_n) is geometric if there is a constant r≠0r \ne 0 (the common ratio) such that

3
an+1an=rfor all n≥1, an≠0.\quad \frac{a_{n+1}}{a_n} = r \quad\text{for all } n \ge 1,\ a_n \neq 0.

Derivation hint: This is the defining recurrence; "constant ratio" is literally the definition. The closed-form (nn-th term) formula is

4
an=a1 r n−1.\quad a_n = a_1\, r^{\,n-1}.

Derivation hint: Starting from a1a_1, you multiply by rr a total of (n−1)(n-1) times to reach ana_n.

TheoremSum of a Finite Arithmetic Series

The sum of the first nn terms of an arithmetic sequence is

5
Sn=n2 (a1+an)=n2 (2a1+(n−1)d).\quad S_n = \frac{n}{2}\,(a_1 + a_n) = \frac{n}{2}\,\bigl(2a_1 + (n-1)d\bigr).

Derivation hint: Pair the first and last terms, the second and second-to-last, etc.; each pair sums to a1+ana_1 + a_n, and there are n/2n/2 pairs.

TheoremSum of a Finite Geometric Series

For r≠1r \ne 1, the sum of the first nn terms of a geometric sequence is

6
Sn=a1 1−r n1−r.\quad S_n = a_1\,\frac{1 - r^{\,n}}{1 - r}.

Derivation hint: Compute Sn−rSnS_n - r S_n; all interior terms cancel telescopically, leaving a1−a1rna_1 - a_1 r^n.

TheoremSum of an Infinite Geometric Series

If ∣r∣<1|r| < 1, the infinite series converges to

7
S∞=a11−r.\quad S_\infty = \frac{a_1}{1 - r}.

Derivation hint: As n→∞n \to \infty, rn→0r^n \to 0 because ∣r∣<1|r|<1, so the rnr^n term in [6] vanishes.

Why It Holds / Derivation

ProofClosed form for the arithmetic n-th term, Eq. 2

Step 1. By the definition [1], each term is the previous plus dd: a2=a1+da_2 = a_1 + d, a3=a2+d=a1+2da_3 = a_2 + d = a_1 + 2d, a4=a1+3da_4 = a_1 + 3d.

Step 2. Observe the pattern: the coefficient of dd is always one less than the index.

Step 3. By induction, assume ak=a1+(k−1)da_k = a_1 + (k-1)d. Then ak+1=ak+d=a1+(k−1)d+d=a1+k da_{k+1} = a_k + d = a_1 + (k-1)d + d = a_1 + k\,d, which is the formula with n=k+1n = k+1.

Step 4. The base case n=1n=1 gives a1=a1+0a_1 = a_1 + 0, true. Hence [2] holds for all n≥1n \ge 1. ■\blacksquare

∎
ProofSum formula for an arithmetic series, Eq. 5

Step 1. Write the sum forwards and backwards:

Sn=a1+a2+⋯+an−1+anS_n = a_1 + a_2 + \cdots + a_{n-1} + a_n
Sn=an+an−1+⋯+a2+a1.S_n = a_n + a_{n-1} + \cdots + a_2 + a_1.

Step 2. Add the two lines column by column. Each column sums to a1+ana_1 + a_n (since as one index rises by dd, the partner falls by dd, keeping each column constant).

Step 3. There are nn columns, so 2Sn=n(a1+an)2 S_n = n(a_1 + a_n).

Step 4. Divide by 2: Sn=n2(a1+an)S_n = \tfrac{n}{2}(a_1 + a_n). Substituting an=a1+(n−1)da_n = a_1 + (n-1)d gives the second form. ■\blacksquare

∎
ProofSum formula for a geometric series, Eq. 6

Step 1. Write the sum:

Sn=a1+a1r+a1r2+⋯+a1r n−1.S_n = a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{\,n-1}.

Step 2. Multiply both sides by rr:

rSn=a1r+a1r2+⋯+a1r n−1+a1r n.r S_n = a_1 r + a_1 r^2 + \cdots + a_1 r^{\,n-1} + a_1 r^{\,n}.

Step 3. Subtract: Sn−rSn=a1−a1r nS_n - r S_n = a_1 - a_1 r^{\,n}, because every middle term cancels.

Step 4. Factor: Sn(1−r)=a1(1−rn)S_n(1 - r) = a_1(1 - r^n). For r≠1r \ne 1, divide to get Sn=a11−rn1−rS_n = a_1\dfrac{1 - r^n}{1 - r}. ■\blacksquare

∎
mathematician · Carl Friedrich Gauss | 1777–1855

The legend (recorded by his biographer Wolfgang Sartorius) holds that a young Gauss, asked by his teacher J. G. Büttner to sum the integers from 1 to 100, produced the answer 5050 almost instantly. He had recognized the arithmetic-series pairing trick: 1+100=1011+100 = 101, 2+99=1012+99 = 101, …, fifty pairs of 101, giving 50×101=505050 \times 101 = 5050. Whether literally true or embellished, the story crystallizes the essence of [5]. Gauss went on to revolutionize number theory, statistics (the Gaussian distribution), and astronomy.

Building Intuition

🧠 Intuition

Think of an arithmetic sequence as climbing a staircase: every step is the same height dd. After n−1n-1 steps you are at height a1+(n−1)da_1 + (n-1)d — your altitude grows in a straight line. Think of a geometric sequence as repeated photocopying at a fixed zoom factor rr. Zoom in at 200% (r=2r=2) and sizes double each time: 1, 2, 4, 8, …. Zoom out at 50% (r=12r = \tfrac12) and they shrink: 1, ½, ¼, …. The shrinking case is why an infinite geometric sum can be finite — the pieces you add get vanishingly small fast enough that the total settles on a fixed number. A clean test: take differences of consecutive terms. If those differences are constant, it's arithmetic. Take ratios instead — if those are constant, it's geometric.

Figure
Arithmetic (straight line) vs. geometric (curving) growth from the same starting term
term index n value arithmetic d=2 geometric r=2 1 2 3 4 5

Worked Examples

ExampleArithmetic: theatre seating

Given. A theatre has 22 seats in the front row, and each row behind has 3 more seats. There are 18 rows. Find (a) the number of seats in the last row, and (b) the total seating capacity.

Step 1 — Identify parameters. a1=22a_1 = 22 seats, d=3d = 3 seats/row, n=18n = 18 rows.

Step 2 — Last row, use [2].

a18=22+(18−1)(3)=22+51=73.a_{18} = 22 + (18-1)(3) = 22 + 51 = 73.

Step 3 — Total, use [5].

S18=182 (22+73)=9×95=855.S_{18} = \frac{18}{2}\,(22 + 73) = 9 \times 95 = 855.

Answers.

a18=73 seats,S18=855 seats\boxed{a_{18} = 73 \text{ seats}, \qquad S_{18} = 855 \text{ seats}}
ExampleGeometric: bouncing ball and infinite sum

Given. A ball is dropped from 10 m. Each bounce returns it to 60% of the previous height (r=0.6r = 0.6). Find (a) the peak height after the 4th bounce, and (b) the total vertical distance the ball travels before coming to rest.

Step 1 — Peak heights form a geometric sequence. a1=10×0.6=6a_1 = 10\times 0.6 = 6 m (first peak), r=0.6r = 0.6. The 4th-bounce peak is a4a_4:

a4=6×(0.6)3=6×0.216=1.296 m.a_4 = 6 \times (0.6)^{3} = 6 \times 0.216 = 1.296 \text{ m}.

Step 2 — Total distance. The ball drops 10 m, then each peak height is travelled twice (up and down). The peaks 6,3.6,2.16,…6, 3.6, 2.16, \dots form an infinite geometric series with a1=6a_1 = 6, r=0.6r=0.6, ∣r∣<1|r|<1. Using [7]:

∑peaks=61−0.6=60.4=15 m.\sum \text{peaks} = \frac{6}{1 - 0.6} = \frac{6}{0.4} = 15 \text{ m}.

Step 3 — Combine. Total distance =10+2(15)=40= 10 + 2(15) = 40 m.

Answers.

a4=1.296 m,total distance=40 m\boxed{a_4 = 1.296 \text{ m}, \qquad \text{total distance} = 40 \text{ m}}

Common Misconceptions

⚠️ Common mistake

The geometric sum formula [6] requires r≠1r \ne 1. When r=1r = 1 every term equals a1a_1, so Sn=n a1S_n = n\,a_1 — you cannot use the fraction (it would divide by zero). Also, the infinite formula [7] is valid only when ∣r∣<1|r| < 1. If ∣r∣≥1|r| \ge 1 the terms do not shrink to zero and the infinite series diverges (has no finite sum).

⚠️ Common mistake

Off-by-one in the exponent and the multiplier. The nn-th term uses r n−1r^{\,n-1} and (n−1)d(n-1)d, not r nr^{\,n} or n dn\,d. The reason: from a1a_1 to ana_n there are only n−1n-1 steps. A frequent slip is computing a1rna_1 r^n for the nn-th term.

Counterexample

Consider 2,4,8,14,22,…2, 4, 8, 14, 22, \dots — is it arithmetic or geometric? Neither. Differences: 4−2=24-2=2, 8−4=48-4=4, 14−8=614-8=6 — not constant, so not arithmetic. Ratios: 4/2=24/2 = 2, 8/4=28/4 = 2, 14/8=1.7514/8 = 1.75 — not constant, so not geometric. (In fact this is a quadratic sequence whose second differences are constant.) The lesson: always check that the difference or ratio is genuinely constant across all terms, not just the first pair.

Connections

connection

Arithmetic sequences are discrete Linear Functions: the term an=a1+(n−1)da_n = a_1 + (n-1)d is a line in nn with slope dd. Geometric sequences are discrete Exponential Functions: an=a1r n−1a_n = a_1 r^{\,n-1} matches y=c bxy = c\,b^x. The infinite geometric series [7] is the first example of a convergent Infinite Series and previews the limits central to Calculus. Taking logarithms turns a geometric sequence into an arithmetic one, linking the two families through Logarithms. Finally, Compound Interest is geometric growth applied to money.

Real-World Applications

🔬 Where this appears

Historical Development

⏳ History
- c. 1850 BCE — Rhind Papyrus (Egypt): Problem 79 lists houses, cats, mice, etc. in powers of 7 — an early geometric series summation.
- c. 300 BCE — Euclid (Elements IX.35): Gives a rigorous geometric-series sum for integer ratios, essentially proving [6].
- c. 250 BCE — Archimedes: Uses the sum 1+14+116+⋯=431 + \tfrac14 + \tfrac1{16} + \cdots = \tfrac43 to find the area of a parabolic segment — a converging geometric series centuries before formal limits.
- 5th century BCE — Zeno of Elea: His paradoxes (Achilles and the tortoise) implicitly hinge on infinite geometric series, later resolved by [7].
- c. 1202 — Fibonacci (Liber Abaci): Popularizes sequence problems and arithmetic/geometric progressions in European commerce.
- 14th century — Nicole Oresme: Proves the harmonic series diverges and studies geometric series, foreshadowing convergence theory.
- c. 1786 — Carl Friedrich Gauss: The schoolroom anecdote of summing 11 to 100100 via pairing illustrates [5].
- 17th–18th century — Newton & Euler: Embed geometric series within power-series methods, making them foundational to calculus and analysis.
- 19th century — Cauchy & Weierstrass: Formalize convergence (∣r∣<1|r|<1 criterion), placing the infinite geometric sum on rigorous footing.

Subtleties & Limits

open-problem

The clean theory above covers constant difference or ratio. Most real sequences are neither: they may have constant second differences (quadratic), variable ratios, or chaotic behaviour. Determining a closed form for a general recurrence an+1=f(an)a_{n+1} = f(a_n) is often impossible — even the simple-looking logistic map an+1=r an(1−an)a_{n+1} = r\,a_n(1 - a_n) produces chaos for many rr, despite mixing arithmetic and geometric ingredients. More broadly, deciding whether an arbitrary integer sequence has any finite formula connects to deep unsolved questions in number theory, and recognizing patterns (as catalogued in the OEIS, with hundreds of thousands of sequences) remains as much art as algorithm.

See Also

Linear Functions · Exponential Functions · Infinite Series · Sigma Notation · Compound Interest · Logarithms · Mathematical Induction · Fibonacci Sequence