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第 14 週 · 等差與等比數列 · 第 1 節

等差數列:公差與通項公式

3、5、7、9… 每次加同一個數 —— 這就是等差數列。用「台階」把公差與通項公式 Tₙ = a + (n−1)d 從零講透,配滿圖例。

▶ 第一關: 公差 d 與識別等差數列
3579+2+2+2T₁=asame step each time = d
35795−3=27−5=29−7=2all differences equal → arithmetic, d = 2
▶ 第二關: 通項公式 Tₙ = a + (n−1)d
T₁ = a(0 steps of d)T₂ = a + d(1 step)T₃ = a + 2d(2 steps)Tₙ = a + (n−1)d(n−1) steps of d
nTₙstraight line, slope = darithmetic = linear growth
Tₙ = a + (n − 1) dfirst termsteps = n−1common difference
▶ 第三關: 反求 a、d、項數 n
a=2, T₅=14:2 + 4d = 14d = 3plug the known term into Tₙ = a + (n−1)d, then solve
is 100 in 1, 4, 7, … ?1 + (n−1)×3 = 100 → n = 34integer → YESif n came out 8.5 (not whole) → NOT in the sequence
⚔ BOSS: 給兩項求整條數列
T₃ = 7, T₇ = 19① 4 steps apart: 4d = 19−7 = 12 → d = 3② back-substitute: a + 2d = 7 → a = 1