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Algebra · Sequences

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The nth term of a linear sequence

Why a constant difference makes a sequence "dn + something", how the zeroth term pins down the constant, and how the nth term answers "is 122 in this sequence?" with an equation instead of a hunch.

Algebra · Sequences

The nth term of a linear sequence

Why a constant difference makes a sequence "dn + something", how the zeroth term pins down the constant, and how the nth term answers "is 122 in this sequence?" with an equation instead of a hunch.

Why it works

A linear (arithmetic) sequence climbs by the same amount each step: 7,11,15,19,7, 11, 15, 19, \ldots adds 4 every time. Two different kinds of rule describe it:
  • the term-to-term rule — "add 4" — tells you how to get the NEXT term;
  • the nth term — a formula in nn — hands you ANY term directly.
"Add 4" and "n+4n + 4" are not the same thing: n+4n + 4 is a formula whose terms go 5,6,7,5, 6, 7, \ldots — a sequence that adds 1. The add-4-ness of a sequence lives in the coefficient: the nth term starts 4n4n.

Why dndn? Because the sequence is the times table, shifted. The 4× table is 4,8,12,164, 8, 12, 16; our sequence 7,11,15,197, 11, 15, 19 sits exactly 33 above it. So the nth term is

4n+3.4n + 3.

The quick way to the shift: step BACK once from the first term — the "zeroth term" is 74=37 - 4 = 3, and that's the constant. (Using the first term as the constant, 4n+74n + 7, is the classic slip — its first term would be 11.) Check by substituting n=1n = 1: 4(1)+3=74(1) + 3 = 7. ✓

Descending sequences have a negative dd. 30,26,22,1830, 26, 22, 18 subtracts 4, so the nth term is 4n+(zeroth term)=344n-4n + \text{(zeroth term)} = 34 - 4n. Check: n=1n = 1 gives 30. ✓

"Is 122 a term?" is an equation, not a feeling. For 3,8,13,18,3, 8, 13, 18, \ldots (nth term 5n25n - 2): set 5n2=1225n - 2 = 122, so 5n=1245n = 124 and n=24.8n = 24.8. Positions are whole numbers — there is no 24.8th term — so 122 is NOT in the sequence, and that little calculation IS the required "explain why".

Building the rule from further-apart clues. If the 3rd term is 13 and the 7th is 29, the sequence climbed 2913=1629 - 13 = 16 in 44 steps, so d=4d = 4; step back from the 3rd term to the zeroth: 133×4=113 - 3 \times 4 = 1. The nth term is 4n+14n + 1 — check both givens. ✓