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Arithmetic and geometric sequences: the nth term, the sum, and the off-by-one in n

Arithmetic sequences add the same number each step; geometric ones multiply by the same number. The formulas are short, and nearly every lost point is an n that should have been n − 1, a ratio that lost its minus sign, or a sum formula fed the wrong last term.

Short answer

Arithmetic (common difference d): an = a1 + (n − 1)d and Sn = n(a1 + an)2. Geometric (common ratio r): an = a1 · rn − 1 and Sn = a1(1 − rn)1 − r for r ≠ 1. The n − 1 is there because the first term has taken zero steps. To find d, subtract a term from the next one; to find r, divide the next term by the one before it, and keep the sign.

Why it works

In 5, 9, 13, … the first term has no 4 added yet, the second has one, the third has two. The 20th term has had 19 of them: a20 = 5 + 19 · 4. That's the whole reason for n − 1, and the same count applies to the exponent in a geometric sequence. The arithmetic sum formula comes from pairing the first and last terms: a1 + an, a2 + an−1, and so on all have the same total, and there are n/2 pairs. The geometric sum formula comes from subtracting r · Sn from Sn, which makes every middle term cancel.

To tell them apart, test both: if consecutive differences are equal, it's arithmetic; if consecutive ratios are equal, it's geometric. An arithmetic sequence is a line sampled at whole numbers, with slope d; a geometric sequence is an exponential.

Example 1: an arithmetic term and sum

For 5, 9, 13, …, find a20 and S20.

  1. d = 9 − 5 = 4Common difference. Check with the next pair: 13 − 9 = 4.
  2. a20 = 5 + (20 − 1) · 4 = 5 + 76 = 81Nineteen steps of 4 after the first term.
  3. S20 = 20(5 + 81)2 = 10 · 86 = 860The sum formula needs the actual last term, 81, from the line above.

Example 2: a geometric term and sum

For 3, 6, 12, …, find a8 and S8.

  1. r = 6 ÷ 3 = 2Common ratio. Check: 12 ÷ 6 = 2.
  2. a8 = 3 · 27 = 3 · 128 = 384Seven doublings after the first term. The exponent is n − 1 = 7, not 8.
  3. S8 = 3(1 − 28)1 − 2 = 3(−255)−1 = 765Here the exponent is n = 8, not 7. Top and bottom are both negative, so the sum is positive, as a sum of positive terms must be.

Example 3: a negative common ratio

For 64, −32, 16, …, find a6 and S6.

  1. r = −32 ÷ 64 = −12Divide a term by the one before it, sign included. The terms alternate, so r is negative.
  2. a6 = 64 · (−½)5 = 64 · (−132) = −2An odd power of a negative is negative. Listing the terms confirms it: 64, −32, 16, −8, 4, −2.
  3. S6 = 64(1 − (−½)6)1 − (−½)Sum formula with n = 6. Both the exponent and the denominator keep the negative ratio.
  4. S6 = 63 ÷ 32 = 63 · 23 = 42(−½)⁶ = 1/64, positive, so the top is 64(1 − 1/64) = 64 − 1 = 63. The bottom is 1 + ½ = 3/2. Check by adding: 64 − 32 + 16 − 8 + 4 − 2 = 42.

Example 4: which term is it?

Which term of 7, 10, 13, … is 100?

  1. 7 + (n − 1) · 3 = 100Set the nth-term formula equal to 100 with a₁ = 7 and d = 3.
  2. 3(n − 1) = 93Subtract 7.
  3. n − 1 = 31Divide by 3. This is the number of steps, not the term number.
  4. n = 32Add 1. Check: a₃₂ = 7 + 31 · 3 = 100.

Example 5: two terms given, find the first

An arithmetic sequence has a4 = 15 and a10 = 33. Find a1.

  1. 6d = 33 − 15 = 18, so d = 3From term 4 to term 10 is 10 − 4 = 6 steps.
  2. a1 = 15 − 3 · 3 = 6Walk back three steps from a₄. Check: 6 + 9 · 3 = 33.

Common mistakes, and the exact line they happen on

1. Using n instead of n − 1

  1. a20 = 5 + 20 · 4 = 85This is the mistake (Example 1). That's the 21st term. The first term has had no 4 added, so the 20th has had 19.
  2. a8 = 3 · 28 = 768The geometric version (Example 2). Eight doublings gives the 9th term. Sanity check by listing: 3, 6, 12, 24, 48, 96, 192, 384.

Quick test: plug in n = 1. The formula must return a1. With n instead of n − 1 it returns the second term.

2. Forgetting to add 1 when solving for n

Which term of 7, 10, 13, … is 100? (Example 4)

  1. n = 100 − 73 = 31This is the mistake. (100 − 7)/3 counts the steps from 7 to 100, and there's one more term than there are steps. The 31st term is 7 + 30 · 3 = 97.

3. Dropping the sign of the ratio

For 64, −32, 16, … (Example 3).

  1. r = ½, so a6 = 64 · (½)5 = 2This is the mistake. The sign was treated as decoration. The sixth term is −2, and the sum changes too: with r = ½ the formula gives 126, not 42.

If the terms alternate in sign, r is negative. Even powers of it are positive, odd powers negative, and the sum formula's denominator 1 − r becomes 1 + |r|.

4. Losing the sign in the geometric sum

  1. S8 = 3(1 − 256)1 = −765This is the mistake (Example 2). The denominator 1 − r = 1 − 2 = −1 was written as 1. A sum of positive terms can't be negative; the two minus signs cancel and the answer is 765.

If r > 1 it's often cleaner to use the equivalent form Sn = a1(rn − 1)/(r − 1), where everything stays positive: 3 · 255 / 1 = 765.

5. Using the geometric formula on an arithmetic sequence

For 5, 9, 13, …, writing r = 9/5 and a20 = 5 · (9/5)19 is a mistake. The ratios aren't constant: 13/9 ≠ 9/5. The differences are. Test both before choosing a formula, and remember that a sequence can be neither.

Practice

  1. Find a15 of 2, 7, 12, …

    Show answer

    d = 5, so a15 = 2 + 14 · 5 = 72.

  2. Find a6 of 5, 15, 45, …

    Show answer

    r = 3, so a6 = 5 · 35 = 5 · 243 = 1215.

  3. Find the sum of the first 10 terms of 4, 7, 10, …

    Show answer

    a10 = 4 + 9 · 3 = 31, so S10 = 10(4 + 31)/2 = 175.

  4. Find the sum of the first 5 terms of 2, −6, 18, …

    Show answer

    r = −3. S5 = 2(1 − (−3)5)/(1 − (−3)) = 2(1 + 243)/4 = 122. Check: 2 − 6 + 18 − 54 + 162 = 122.

  5. Which term of 11, 18, 25, … is 200?

    Show answer

    11 + 7(n − 1) = 200, so 7(n − 1) = 189, n − 1 = 27, n = 28. Check: 11 + 27 · 7 = 200.

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