Arithmetic and Geometric Sequence Calculator
- Free
- No sign-up
- Works on phones
- Shows the working
By Yasir Hafeez · Updated
This sequence calculator handles both arithmetic sequences (add the same number each time) and geometric sequences (multiply by the same number each time). Give it the first term, the common difference or ratio and a term number n, and it returns the explicit formula for \( a_n \), the value of the nth term, the sum of the first n terms \( S_n \), and a list of the opening terms. For geometric sequences with \( |r| < 1 \) it also gives the infinite sum.
It works as an arithmetic sequence calculator and a geometric sequence calculator, with exact fractions where possible. It’s free and needs no sign-up.
How to use the sequence calculator
- Choose Arithmetic or Geometric in the Type menu.
- Enter the first term \( a_1 \).
- Enter d or r: the common difference for arithmetic, or the common ratio for geometric. Fractions such as
1/2and negative numbers are fine. - Enter n, the term you want, and press Calculate.
Formulas
| Arithmetic | Geometric | |
|---|---|---|
| Rule | add \( d \) | multiply by \( r \) |
| nth term | \( a_n = a_1 + (n - 1)d \) | \( a_n = a_1 r^{n-1} \) |
| Sum of n terms | \( S_n = \frac{n}{2}(a_1 + a_n) \) | \( S_n = a_1\frac{1 - r^n}{1 - r} \), \( r \neq 1 \) |
| Infinite sum | diverges (unless all terms are 0) | \( S_\infty = \frac{a_1}{1 - r} \) if \( \lvert r \rvert < 1 \) |
Worked example: arithmetic
The sequence 3, 7, 11, 15, … has \( a_1 = 3 \) and \( d = 4 \), as loaded above.
Formula. \( a_n = 3 + 4(n - 1) \).
10th term. \( a_{10} = 3 + 4 \cdot 9 = 39 \).
Sum of 10 terms. \( S_{10} = \frac{10}{2}(3 + 39) = 5 \cdot 42 = 210 \).
Worked example: geometric
The sequence 2, 6, 18, 54, … has \( a_1 = 2 \) and \( r = 3 \).
$$a_8 = 2 \cdot 3^7 = 4374, \qquad S_8 = 2 \cdot \frac{1 - 3^8}{1 - 3} = \frac{2(1 - 6561)}{-2} = 6560$$
Because \( r = 3 > 1 \), the infinite series diverges.
Worked example: a converging geometric series
16, 8, 4, 2, … has \( r = \frac12 \). The sixth term is \( 16 \cdot \left(\frac12\right)^5 = \frac12 \) and \( S_6 = \frac{63}{2} = 31.5 \). Since \( |r| < 1 \), the infinite sum exists:
$$S_\infty = \frac{16}{1 - \frac12} = 32$$
The partial sums 16, 24, 28, 30, 31, 31.5, … creep up toward 32 without passing it.
Tips and common mistakes
- Off-by-one errors. The exponent is \( n - 1 \), not n, because the first term has no factor of r.
- Finding d or r. Subtract consecutive terms for d. Divide consecutive terms for r.
- Negative ratios. With \( r = -2 \), terms alternate in sign: 1, −2, 4, −8, …
- Decreasing arithmetic sequences. Use a negative d: 50, 43, 36, … has \( d = -7 \) and \( a_{12} = -27 \).
From sequences to series in calculus
A sum of infinitely many terms is a series, and deciding whether it converges is a central topic in Calculus 2. The geometric series is the model case; the geometric series formula guide covers it in depth, the ratio test compares other series to it, and Taylor series build functions out of infinite sums. For general series, use the series convergence calculator.
Related guides
Further reading
- Arithmetic progression (Wikipedia) — the nth-term and sum formulas, with Gauss’s pairing argument.
- Geometric progression (Wikipedia) — terms, partial sums and convergence of geometric series.
FAQ
How do I find the nth term of an arithmetic sequence?
Use \( a_n = a_1 + (n - 1)d \). For 3, 7, 11, … the 10th term is \( 3 + 9 \cdot 4 = 39 \).
How do I find the sum of a geometric sequence?
Use \( S_n = a_1\frac{1 - r^n}{1 - r} \). For an infinite sum with \( |r| < 1 \), use \( \frac{a_1}{1 - r} \).
How can I tell if a sequence is arithmetic or geometric?
If the differences between terms are constant, it’s arithmetic. If the ratios are constant, it’s geometric. Many sequences are neither.
What if r = 1?
Every term equals \( a_1 \), so \( S_n = n a_1 \). The calculator handles this case separately.