Arithmetic Sequence Formula

Algebra 1 · Axiom Academy

LESSON Arithmetic Sequence Formula Start at the first term, hop by the common difference, and count the hops — that count is the secret behind the explicit formula. 1. Each Term Is One Hop Further An arithmetic sequence moves by the same fixed step every time. Watch a marker land on 3, then hop +4 to 7, again to 11, and so on. Every hop is the same size — that constant size is the common difference d . 2, 5, 8, 11, 14… → d = 3 (each hop adds 3) 10, 7, 4, 1, −2… → d = -3 (each hop subtracts 3) 5, 5, 5, 5, 5… → d = 0 (no hop at all) 2. Build Every Term From the First Pin the marker on the first term a_1 = 3 and never move that anchor. To reach a later term, hop away from it. Watch the hop counter : reaching a_2 takes 1 hop, a_3 takes 2 hops, a_4 takes 3. Each landing reads . 3 = 3 + 0(4) — zero hops from the anchor. 3. Counting Hops to the nth Term Why one less? Because the first term costs no hop — you start there. So to reach term n you hop (n-1) times, and each hop adds d . Watch the counter climb to 5 hops as the marker reaches a_6 , landing on a_6 = a_1 + 5d . To reach term n : start at a_1 , then add d a total of (n-1) times. 4. The Explicit Formula: Jump Straight There Now we cash in. Instead of hopping 49 times to reach the 50th term, the explicit formula bundles all (n-1) hops into one move: add to a_1 in a single jump. Watch one long arc replace the whole chain of hops.

This is the written version of the interactive lesson above. See the full Algebra 1 course.