# Sequences and series

Canonical: https://duckyhelper.com/learn/algebra-2/sequences-and-series/
Updated: 2026-10-01

An arithmetic sequence adds the same number d each time, so \(a_n = a_1 + (n - 1)d\). A geometric sequence multiplies by the same ratio r, so \(a_n = a_1 r^{n - 1}\). A series is the sum of the terms. Arithmetic sums use \(S_n = \frac{n}{2}(a_1 + a_n)\), geometric sums use \(S_n = a_1 \cdot \frac{1 - r^n}{1 - r}\), and an infinite geometric series adds up to \(\frac{a_1}{1 - r}\) when \(|r| < 1\).

## The key idea

Look at how you get from one term to the next. If you add the same amount every time, the sequence is arithmetic. If you multiply by the same amount, it is geometric.

**The four formulas**

|  | Arithmetic (add d) | Geometric (multiply by r) |
| --- | --- | --- |
| Example | \(4, 7, 10, 13, \ldots\) with \(d = 3\) | \(4, 12, 36, 108, \ldots\) with \(r = 3\) |
| nth term | \(a_n = a_1 + (n - 1)d\) | \(a_n = a_1 r^{n - 1}\) |
| Sum of n terms | \(S_n = \frac{n}{2}(a_1 + a_n)\) | \(S_n = a_1 \cdot \frac{1 - r^n}{1 - r}\) |
| Sum of all terms | No finite sum (unless every term is 0) | \(S = \frac{a_1}{1 - r}\) if \(\|r\| < 1\) |

The \(n - 1\) is there because the first term has had no steps yet. The 40th term is 39 steps after the first.

$$
S_\infty = \frac{a_1}{1 - r}, \quad |r| < 1
$$

## Worked examples

**Example 1: a term far down an arithmetic sequence**

Problem: Find the 40th term of \(7, 11, 15, 19, \ldots\)

1. The common difference is the gap between terms.

   $$
   11 - 7 = 4
   $$
2. Use \(a_n = a_1 + (n - 1)d\) with \(a_1 = 7\), \(n = 40\), \(d = 4\).

   $$
   7 + (40 - 1)(4) = 7 + 156 = 163
   $$

Answer: \(a_{40} = 163\)

**Example 2: an arithmetic series**

Problem: Find the sum \(3 + 8 + 13 + \cdots + 98\).

1. The difference is 5. First count the terms: how many steps of 5 from 3 to 98, plus 1 for the first term.

   $$
   n = \frac{98 - 3}{5} + 1 = 20
   $$
2. Average of the first and last term, times the number of terms.

   $$
   \frac{20}{2}(3 + 98) = 10 \cdot 101 = 1010
   $$

Answer: The sum is 1,010.

**Example 3: a geometric term and sum**

Problem: For \(5, 15, 45, \ldots\), find the 8th term and the sum of the first 8 terms.

1. The ratio is \(15 \div 5 = 3\). Use \(a_n = a_1 r^{n - 1}\).

   $$
   5 \cdot 3^{8 - 1} = 5 \cdot 2187 = 10935
   $$
2. Use \(S_n = a_1 \cdot \frac{1 - r^n}{1 - r}\) with \(n = 8\).

   $$
   5 \cdot \frac{1 - 3^8}{1 - 3} = 5 \cdot \frac{-6560}{-2} = 16400
   $$

Answer: The 8th term is 10,935 and the sum of the first 8 terms is 16,400.

**Example 4 (test-hard): a repeating decimal as a fraction**

Problem: Write \(0.\overline{27} = 0.272727\ldots\) as a fraction using a geometric series.

1. Split it into blocks: \(0.27 + 0.0027 + 0.000027 + \cdots\). That is geometric with \(a_1 = \frac{27}{100}\) and \(r = \frac{1}{100}\).
2. Since \(|r| < 1\), use the infinite sum formula.

   $$
   \frac{\frac{27}{100}}{1 - \frac{1}{100}} = \frac{27}{99} = \frac{3}{11}
   $$
3. Check: \(3 \div 11 = 0.272727\ldots\)

Answer: \(\frac{3}{11}\)

## Common mistakes

- **Using n instead of n - 1.** The 40th term of \(7, 11, 15, \ldots\) is \(7 + 39 \cdot 4\), not \(7 + 40 \cdot 4\). Fix: count steps between terms, not terms.
- **Miscounting the number of terms.** From 3 to 98 by 5s is 20 terms, not 19. Fix: \(n = \frac{\text{last} - \text{first}}{d} + 1\).
- **Calling a sequence geometric because it grows fast.** Check the ratios: \(2, 6, 12, 20\) has ratios 3, 2, \(\frac{5}{3}\), so it is neither type.
- **Using the infinite sum formula when \(|r| \ge 1\).** \(2 + 4 + 8 + \cdots\) has no finite sum. Fix: check that r is between \(-1\) and 1 first.
- **Getting the ratio upside down.** For \(81, -27, 9, \ldots\), r is \(\frac{-27}{81} = -\frac{1}{3}\). Fix: divide a term by the one before it.

## Quick methods

> **Tip: Arithmetic sum = average times count**
>
> Any arithmetic series equals the average of the first and last terms times the number of terms. That is all \(\frac{n}{2}(a_1 + a_n)\) says. It also gives the famous \(1 + 2 + \cdots + 100 = 50 \cdot 101 = 5050\).

> **Tip: Two terms give you everything**
>
> For an arithmetic sequence, if you know \(a_5\) and \(a_{12}\), then \(d = \frac{a_{12} - a_5}{12 - 5}\). From there, go back to \(a_1\) or forward to any term.

## Practice

**5 practice questions**

1. What is the 10th term of the arithmetic sequence \(50, 46, 42, \ldots\)?
   A. \(10\)
   B. \(14\)
   C. \(18\)
   D. \(86\)

   Answer: \(14\). \(d = -4\), so \(a_{10} = 50 + 9(-4) = 14\). 10 uses 10 steps instead of 9, and 86 adds 4 each time instead of subtracting.

2. What is the common ratio of the geometric sequence \(81, -27, 9, \ldots\)?
   A. \(-3\)
   B. \(-\frac{1}{3}\)
   C. \(\frac{1}{3}\)
   D. \(3\)

   Answer: \(-\frac{1}{3}\). Divide a term by the one before it: \(\frac{-27}{81} = -\frac{1}{3}\). The signs alternate, so r is negative, and the terms shrink, so \(|r| < 1\).

3. What is the sum of the first 30 positive odd numbers, \(1 + 3 + 5 + \cdots + 59\)?
   A. \(450\)
   B. \(870\)
   C. \(900\)
   D. \(930\)

   Answer: \(900\). There are 30 terms with average \(\frac{1 + 59}{2} = 30\), so the sum is \(30 \cdot 30 = 900\). In general, the first n odd numbers add to \(n^2\).

4. What is the sum of the infinite series \(9 + 6 + 4 + \cdots\)?
   A. \(13.5\)
   B. \(19\)
   C. \(27\)
   D. The series has no finite sum.

   Answer: \(27\). \(r = \frac{6}{9} = \frac{2}{3}\), which is less than 1, so the sum is \(\frac{9}{1 - \frac{2}{3}} = 27\). 13.5 uses \(r = \frac{1}{3}\), and 19 adds only the three terms shown.

5. In an arithmetic sequence, \(a_5 = 17\) and \(a_{12} = 45\). What is \(a_1\)?

   Answer: 1. \(d = \frac{45 - 17}{12 - 5} = 4\). Going back 4 steps from \(a_5\): \(a_1 = 17 - 4 \cdot 4 = 1\).

## Frequently asked questions

### What is the difference between a sequence and a series?

A sequence is a list of numbers in order, like \(2, 5, 8, 11\). A series is what you get when you add those numbers: \(2 + 5 + 8 + 11 = 26\). Sequence questions ask for a term; series questions ask for a sum.

### How do I tell if a sequence is arithmetic or geometric?

Subtract neighbors: if the differences are all equal, it is arithmetic. Divide neighbors: if the ratios are all equal, it is geometric. If neither is constant, it is some other kind of sequence, and these formulas do not apply.

### When does an infinite geometric series have a sum?

Only when the ratio is between \(-1\) and 1, so \(|r| < 1\). Then the terms shrink toward 0 fast enough for the total to settle at \(\frac{a_1}{1 - r}\). If \(|r| \ge 1\), the terms do not shrink, and the sum has no finite value.

### What does sigma notation mean?

The symbol \(\sum\) means "add up". \(\sum_{k=1}^{4} 2k\) means put \(k = 1, 2, 3, 4\) into \(2k\) and add: \(2 + 4 + 6 + 8 = 20\). The numbers below and above the sigma tell you where k starts and stops.

## Related

- [Exponential growth and decay on the SAT](https://duckyhelper.com/learn/sat-math/exponential-functions/)
- [Logarithms](https://duckyhelper.com/learn/algebra-2/logarithms/)
- [Functions, function notation, domain and range](https://duckyhelper.com/learn/algebra-1/functions/)
- [Linear functions and slope on the SAT](https://duckyhelper.com/learn/sat-math/linear-functions/)
- [Algebra 2 study guides](https://duckyhelper.com/learn/algebra-2/)

## Try asking Ducky

- "Is this sequence arithmetic, geometric, or neither?"
- "Why is it n - 1 and not n in the formula?"
- "Show me how 0.272727... turns into 3/11 one more time, slowly."

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