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Sequences & Series 🔢

These all look similar, and that's the whole trap. The skill isn't the math — it's reading the question and knowing which of five formulas it wants. Learn the tells below and you've basically won this topic.

The whole game Match the words → the formula

Before you compute anything: read the question, find the tell, grab the formula. This is the top of your cheat sheet.

"Compute A₁₁ …" / "find the nth term" — and the terms add the same amount each step.
aₙ = a₁ + (n − 1)·d
Sheet: nth term, arithmetic. (n−1) hops, not n.
"find the nth term" — and the terms multiply by the same number (a common ratio).
aₙ = a₁ · r^(n − 1)
Sheet: nth term, geometric. Exponent is (n−1).
"Find the sum of the first n terms" — arithmetic (adding).
Sₙ = n·(a₁ + aₙ) / 2
Sheet: arithmetic sum = average of first & last, times how many.
"Find the sum of the first n terms" — geometric (multiplying).
Sₙ = a₁·(1 − rⁿ) / (1 − r)
Sheet: finite geometric sum.
"Find the sum of the infinite series" (and |r| < 1).
S = a₁ / (1 − r)
Sheet: infinite geometric — only if |r|<1.
"Simplify 105! / 102!" (factorials).
everything ≤ 102 cancels → 105 · 104 · 103
Sheet: n!/m! → cancel down, multiply what's left.

Worked Example 1 — "Compute A₁₁"

Compute A₁₁ for the arithmetic sequence with A₁ = 33 and common difference d = 6.

1Tell: terms add 6 each time → arithmetic nth term → aₙ = a₁ + (n−1)d
2Plug in: A₁₁ = 33 + (11 − 1)(6) = 33 + (10)(6)
3= 33 + 60 = 93
⚠ Your trap: it's (n − 1), not n. Going from term 1 to term 11 is 10 hops, not 11. That single "−1" is the whole problem.

Worked Example 2 — "Determine A₁"

A geometric sequence has common ratio r = 3, and its 5th term is 324. Determine A₁ and write the formula.

1Tell: "common ratio" → geometric → aₙ = a₁·r^(n−1)
25th term: 324 = a₁ · 3^(5−1) = a₁ · 3⁴
33⁴ = 81, so 324 = 81·a₁ → a₁ = 324 ÷ 81 = 4
4Formula: aₙ = 4 · 3^(n−1)
⚠ Your trap: the 5th term raises r to (5 − 1) = 4, not 5. And 3⁴ = 81 (count it: 3, 9, 27, 81) — don't fumble the power.

Worked Example 3 — finite geometric sum

Find the sum of the first 30 terms of the geometric series with first term 7 and common ratio 1.01.

1Tell: "sum of the first n terms," geometric → Sₙ = a₁(1 − rⁿ)/(1 − r)
2S₃₀ = 7 · (1 − 1.01³⁰) / (1 − 1.01)
31.01³⁰ ≈ 1.34785 → top: 1 − 1.34785 = −0.34785; bottom: 1 − 1.01 = −0.01
4(−0.34785)/(−0.01) = 34.785 → × 7 ≈ 243.49
⚠ Two traps you've hit: (1) "30 terms" means n = 30 — don't drop the first term and use 29. (2) 1 − 1.01 = −0.01; a negative top over a negative bottom is positive.

Worked Example 4 — infinite series

Evaluate the sum of the infinite geometric series with first term 4 and common ratio 2/7.

1Tell: "infinite" + |r| = 2/7 < 1 (so it converges) → S = a₁/(1 − r)
2S = 4 / (1 − 2/7) = 4 / (5/7)
3Divide by a fraction = multiply by its flip: 4 · (7/5) = 28/5 (= 5.6)
⚠ Your trap: 1 − 2/7 = 5/7 (not 5). Then keep-change-flip: 4 ÷ (5/7) = 4 × 7/5. This only works because |r| < 1.

Now you Faded — fill the blanks

Find the sum of the first 25 terms of the arithmetic sequence with a₁ = 4 and d = 3.

  1. Tell → which formula? __________________
  2. First find the 25th term: a₂₅ = 4 + (___ − 1)(3) = ______
  3. Now the sum: S₂₅ = 25·(4 + ____)/2 = ______
Check it ›
Sum of an arithmetic series → Sₙ = n(a₁+aₙ)/2. a₂₅ = 4 + (25−1)(3) = 4 + 72 = 76. S₂₅ = 25·(4 + 76)/2 = 25·80/2 = 25·40 = 1000.

Practice set (6) — name the tell first

For each: write which formula it wants before you compute. Then check the key and log any miss.

  1. Compute A₁₅ for the arithmetic sequence with A₁ = 7 and common difference d = 4.
  2. A geometric sequence has first term 5 and common ratio 2. Find the 6th term.
  3. Find the sum of the first 20 terms of the arithmetic series with a₁ = 3 and a₂₀ = 60.
  4. Find the sum of the first 8 terms of the geometric series with first term 2 and common ratio 3.
  5. Evaluate the sum of the infinite geometric series with first term 9 and common ratio 1/3.
  6. Simplify 10! / 7!.
Answer key (worked) ›

1.

Arithmetic nth term: aₙ=a₁+(n−1)d. A₁₅ = 7 + (14)(4) = 7 + 56 = 63.

2.

Geometric nth term: aₙ=a₁·r^(n−1). 6th term = 5·2⁵ = 5·32 = 160.

3.

Arithmetic sum: Sₙ=n(a₁+aₙ)/2. S₂₀ = 20·(3+60)/2 = 20·63/2 = 630.

4.

Finite geometric sum: 2·(1−3⁸)/(1−3) = 2·(1−6561)/(−2) = 2·(−6560)/(−2) = 6560.

5.

Infinite (|r|=1/3<1): 9/(1−1/3) = 9/(2/3) = 9·3/2 = 27/2 = 13.5.

6.

10!/7! → everything ≤ 7 cancels → 10·9·8 = 720.
📝 Capture before you leave: put the five formulas on your sheet with their tell next to each (nth arith / nth geo / arith sum / finite geo sum / infinite geo). Add your two reminders: "(n−1) hops" and "count the first term."
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