Generate the sequence, get the nth term and the sum.

Arithmetic: a constant difference

Each term adds a fixed amount. Starting at 3 with a difference of 5: 3, 8, 13, 18, 23… The nth term is a + (n−1)d, so the 10th is 3 + 9×5 = 48 — no counting required.

The sum has a closed form too: n/2 × (2a + (n−1)d), giving 255 for those ten terms. The trick behind it is Gauss's: pair the first term with the last, the second with the second-last, and every pair has the same total.

Geometric: a constant ratio

Each term multiplies by a fixed factor. Starting at 2 with a ratio of 3: 2, 6, 18, 54… The nth term is a · r⁽ⁿ⁻¹⁾ and the sum is a(1 − rⁿ)/(1 − r).

Geometric growth is compound interest, population growth and viral spread. It also has a property arithmetic sequences lack: when |r| < 1 the terms shrink toward zero and an infinite sum converges to a finite value, a/(1−r). Halving forever — 1 + ½ + ¼ + ⅛ … — totals exactly 2.

Fibonacci: each term from the previous two

0, 1, 1, 2, 3, 5, 8, 13, 21… Defined by aₙ = aₙ₋₁ + aₙ₋₂ rather than by a fixed step, which makes it neither arithmetic nor geometric — though it becomes almost geometric as it grows.

The ratio between consecutive terms converges on φ ≈ 1.618, the golden ratio. 13/8 is 1.625; 89/55 is 1.6182. That convergence is why Fibonacci numbers appear in phyllotaxis and spiral packing: they are the whole-number approximation to an irrational optimum.

What is the difference between arithmetic and geometric?

Arithmetic sequences add a fixed amount each step; geometric ones multiply by a fixed factor. Arithmetic grows in a straight line, geometric grows exponentially — which is why compound interest is geometric and simple interest is arithmetic.

How do I find the nth term without listing everything?

Use the closed form. Arithmetic is a + (n−1)d, geometric is a · r^(n−1). Both jump straight to the answer, which matters when n is large.

Why does the golden ratio appear in the Fibonacci sequence?

Because the ratio of consecutive terms converges to it. As the numbers grow, dividing one by its predecessor approaches φ ≈ 1.6180339887 — the positive root of x² = x + 1, which is exactly the relationship the sequence encodes.

Can a geometric sequence sum to a finite total forever?

Yes, when the ratio is between −1 and 1. The terms shrink fast enough that an infinite sum converges to a/(1−r). Adding 1 + ½ + ¼ + ⅛ and onward forever gives exactly 2.