Arithmetico-Geometric Sequence Calculator
Written by Thierno Sadou Diallo, formula verified per our methodology • Last checked on 9/10/2026
An arithmetico-geometric sequence is defined by u(n+1) = a×u(n) + b. Its closed form (for a≠1) is u(n) = aⁿ×(u₀−L) + L, with L=b÷(1−a). For a=2, b=1, and u₀=1, the term u(3) is exactly 15 — the well-known sequence 1, 3, 7, 15, 31...
Explanation
An arithmetico-geometric sequence combines, as its name suggests, the two simplest types of recurrence: at each step, the previous term is first multiplied by a coefficient a (as in a geometric sequence), then a constant b is added to it (as in an arithmetic sequence). This structure actually encompasses both the arithmetic sequence calculator and geometric sequence calculator already published as special cases: setting a=1 removes the multiplicative effect and gives a purely arithmetic sequence, while setting b=0 removes the constant addition and gives a purely geometric sequence. The key to solving this recurrence without calculating term by term is to identify its FIXED POINT L=b÷(1−a): the particular value that, once reached, would stay unchanged indefinitely (since a×L+b=L by definition). Subtracting this fixed point from the first term makes the sequence behave exactly like a pure geometric sequence with ratio a around that point, which gives the closed form u(n)=aⁿ×(u₀−L)+L. This fixed-point solving technique is a recurring tool in applied mathematics, used well beyond pure number sequences: it appears, for example, in the stability analysis of dynamical systems, where equilibrium states around which a system evolves are sought in the same way.
Example: the sequence 1, 3, 7, 15, 31...
Inputs
First term: u₀=1. Coefficient a=2. Coefficient b=1. Index sought: n=3.
Calculation
Fixed point L = 1÷(1−2) = −1. u(3) = 2³×(1−(−1)) + (−1) = 8×2 − 1 = 16 − 1 = 15.
Result
The fourth term of this sequence (u₃, counting u₀ as the first) is exactly 15, consistent with the sequence calculated term by term: 1 → 3 → 7 → 15.
Frequently asked questions
What happens if a=1?
The fixed point L=b÷(1−a) is then no longer defined (division by zero), which is consistent with the fact that the sequence converges to no particular value in that case: it simply degenerates into a classic arithmetic sequence, where each term is obtained by adding b to the previous one with no multiplicative effect. The formula u(n)=u₀+n×b, used automatically by this calculator in that specific case, is the direct consequence.
Does the sequence always converge to a final value?
No, only if the absolute value of a is strictly less than 1: in that case, the term aⁿ tends to zero as n increases, and the sequence gets closer and closer to its fixed point L. If |a| is greater than 1, on the contrary, the sequence moves further and further from that fixed point and diverges to infinity (or to minus infinity, depending on the sign of u₀−L); if a=−1, the sequence oscillates indefinitely without converging or diverging.
How do you check that this closed form is correct without recalculating it by hand?
The most reliable way is to compare the result of the closed form to a term-by-term calculation (literally applying u(n+1)=a×u(n)+b, n times from u₀): both methods must always give exactly the same result for any values of a, b, and n, which was verified explicitly when designing this calculator, precisely to confirm that the closed form solves the recurrence with no approximate shortcut.