Algebra 2

Sequences and series

An arithmetic sequence adds the same number d each time, so an=a1+(n−1)da_n = a_1 + (n - 1)d. A geometric sequence multiplies by the same ratio r, so an=a1rn−1a_n = a_1 r^{n - 1}. A series is the sum of the terms. Arithmetic sums use Sn=n2(a1+an)S_n = \frac{n}{2}(a_1 + a_n), geometric sums use Sn=a1⋅1−rn1−rS_n = a_1 \cdot \frac{1 - r^n}{1 - r}, and an infinite geometric series adds up to a11−r\frac{a_1}{1 - r} when ∣r∣<1|r| < 1.

Updated

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)
Example4,7,10,13,…4, 7, 10, 13, \ldots with d=3d = 34,12,36,108,…4, 12, 36, 108, \ldots with r=3r = 3
nth terman=a1+(n−1)da_n = a_1 + (n - 1)dan=a1rn−1a_n = a_1 r^{n - 1}
Sum of n termsSn=n2(a1+an)S_n = \frac{n}{2}(a_1 + a_n)Sn=a1⋅1−rn1−rS_n = a_1 \cdot \frac{1 - r^n}{1 - r}
Sum of all termsNo finite sum (unless every term is 0)S=a11−rS = \frac{a_1}{1 - r} if ∣r∣<1|r| < 1

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

S∞=a11−r,∣r∣<1S_\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,…7, 11, 15, 19, \ldots

  1. The common difference is the gap between terms.
    11−7=411 - 7 = 4
  2. Use an=a1+(n−1)da_n = a_1 + (n - 1)d with a1=7a_1 = 7, n=40n = 40, d=4d = 4.
    7+(40−1)(4)=7+156=1637 + (40 - 1)(4) = 7 + 156 = 163

Answer a40=163a_{40} = 163

Example 2: an arithmetic series

Problem Find the sum 3+8+13+⋯+983 + 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=98−35+1=20n = \frac{98 - 3}{5} + 1 = 20
  2. Average of the first and last term, times the number of terms.
    202(3+98)=10⋅101=1010\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,…5, 15, 45, \ldots, find the 8th term and the sum of the first 8 terms.

  1. The ratio is 15÷5=315 \div 5 = 3. Use an=a1rn−1a_n = a_1 r^{n - 1}.
    5⋅38−1=5⋅2187=109355 \cdot 3^{8 - 1} = 5 \cdot 2187 = 10935
  2. Use Sn=a1⋅1−rn1−rS_n = a_1 \cdot \frac{1 - r^n}{1 - r} with n=8n = 8.
    5⋅1−381−3=5⋅−6560−2=164005 \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 0.27‾=0.272727…0.\overline{27} = 0.272727\ldots as a fraction using a geometric series.

  1. Split it into blocks: 0.27+0.0027+0.000027+⋯0.27 + 0.0027 + 0.000027 + \cdots. That is geometric with a1=27100a_1 = \frac{27}{100} and r=1100r = \frac{1}{100}.
  2. Since ∣r∣<1|r| < 1, use the infinite sum formula.
    271001−1100=2799=311\frac{\frac{27}{100}}{1 - \frac{1}{100}} = \frac{27}{99} = \frac{3}{11}
  3. Check: 3÷11=0.272727…3 \div 11 = 0.272727\ldots

Answer 311\frac{3}{11}

Common mistakes

  • Using n instead of n - 1. The 40th term of 7,11,15,…7, 11, 15, \ldots is 7+39⋅47 + 39 \cdot 4, not 7+40⋅47 + 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=last−firstd+1n = \frac{\text{last} - \text{first}}{d} + 1.
  • Calling a sequence geometric because it grows fast. Check the ratios: 2,6,12,202, 6, 12, 20 has ratios 3, 2, 53\frac{5}{3}, so it is neither type.
  • Using the infinite sum formula when ∣r∣≥1|r| \ge 1. 2+4+8+⋯2 + 4 + 8 + \cdots has no finite sum. Fix: check that r is between −1-1 and 1 first.
  • Getting the ratio upside down. For 81,−27,9,…81, -27, 9, \ldots, r is −2781=−13\frac{-27}{81} = -\frac{1}{3}. Fix: divide a term by the one before it.

Quick methods

Practice

5 practice questions

  1. What is the 10th term of the arithmetic sequence 50,46,42,…50, 46, 42, \ldots?

    1. 1010
    2. 1414
    3. 1818
    4. 8686
    Show answer

    Answer: 1414

    d=−4d = -4, so a10=50+9(−4)=14a_{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,…81, -27, 9, \ldots?

    1. −3-3
    2. −13-\frac{1}{3}
    3. 13\frac{1}{3}
    4. 33
    Show answer

    Answer: −13-\frac{1}{3}

    Divide a term by the one before it: −2781=−13\frac{-27}{81} = -\frac{1}{3}. The signs alternate, so r is negative, and the terms shrink, so ∣r∣<1|r| < 1.

  3. What is the sum of the first 30 positive odd numbers, 1+3+5+⋯+591 + 3 + 5 + \cdots + 59?

    1. 450450
    2. 870870
    3. 900900
    4. 930930
    Show answer

    Answer: 900900

    There are 30 terms with average 1+592=30\frac{1 + 59}{2} = 30, so the sum is 30⋅30=90030 \cdot 30 = 900. In general, the first n odd numbers add to n2n^2.

  4. What is the sum of the infinite series 9+6+4+⋯9 + 6 + 4 + \cdots?

    1. 13.513.5
    2. 1919
    3. 2727
    4. The series has no finite sum.
    Show answer

    Answer: 2727

    r=69=23r = \frac{6}{9} = \frac{2}{3}, which is less than 1, so the sum is 91−23=27\frac{9}{1 - \frac{2}{3}} = 27. 13.5 uses r=13r = \frac{1}{3}, and 19 adds only the three terms shown.

  5. In an arithmetic sequence, a5=17a_5 = 17 and a12=45a_{12} = 45. What is a1a_1?

    Show answer

    Answer: 1

    d=45−1712−5=4d = \frac{45 - 17}{12 - 5} = 4. Going back 4 steps from a5a_5: a1=17−4⋅4=1a_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 2,5,8,112, 5, 8, 11. A series is what you get when you add those numbers: 2+5+8+11=262 + 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 −1-1 and 1, so ∣r∣<1|r| < 1. Then the terms shrink toward 0 fast enough for the total to settle at a11−r\frac{a_1}{1 - r}. If ∣r∣≥1|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". ∑k=142k\sum_{k=1}^{4} 2k means put k=1,2,3,4k = 1, 2, 3, 4 into 2k2k and add: 2+4+6+8=202 + 4 + 6 + 8 = 20. The numbers below and above the sigma tell you where k starts and stops.

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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