Sequences

A1.F.1Arithmetic sequences01/05

Arithmetic sequences — recursive & explicit.

The sequence we'll model

3, 7, 11, 15, 19, \dots

Step01of 05

An arithmetic sequence adds the same number to get the next term. Subtract any pair of consecutive terms to find that number — the common difference d.

d = 7 - 3 = 411 - 7 = 415 - 11 = 4

Every gap is $4$ , so this is arithmetic with $d = 4$ . The first term is $a_{1} = 3$ .

A1.F.1 — Arithmetic sequences — recursive & explicit.

The sequence we'll model

3, 7, 11, 15, 19, \dots

Step 01 of 05

An arithmetic sequence adds the same number to get the next term. Subtract any pair of consecutive terms to find that number — the common difference d.

d = 7 - 3 = 411 - 7 = 415 - 11 = 4

Every gap is $4$ , so this is arithmetic with $d = 4$ . The first term is $a_{1} = 3$ .

Step 02 of 05

The recursive formula says how to get the NEXT term from the previous one. For arithmetic, that's "add d":

a_{n} = a_{n - 1} + d

For our sequence:

a_{1} = 3, a_{n} = a_{n - 1} + 4

Recursive formulas always need TWO things: the starting value and the rule. Without $a_{1}$ the rule has nothing to start from.

Step 03 of 05

Recursive is honest but slow. Want the 100th term? You'd add 4 ninety-nine times. The explicit formula jumps straight to any term:

a_{n} = a_{1} + (n - 1) d

The logic: starting at $a_{1}$ , you add d exactly $(n - 1)$ times to reach the $n$ th term — once to get to term 2, twice to get to term 3, and so on.

Step 04 of 05

Plug in our numbers. With $a_{1} = 3$ and $d = 4$ :

a_{n} = 3 + (n - 1) (4) = 3 + 4 n - 4 = 4 n - 1

Quick check on term 5: $4 (5) - 1 = 19$ $✓$ — matches the sequence.

Step 05 of 05

Now the 100th term is one substitution.

a_{100} = 4 (100) - 1 = 399

Key insight

Recursive tells you HOW the sequence builds. Explicit lets you SKIP ahead. Both come from the same two facts: the first term

a_{1}

and the common difference

d

A1.F.2 — Geometric sequences.

The sequence we'll model

2, 6, 18, 54, 162, \dots

Step 01 of 05

A geometric sequence MULTIPLIES by the same number to get the next term. Divide any pair of consecutive terms to find it — the common ratio r.

r = \frac{6}{2} = 3 \frac{18}{6} = 3 \frac{54}{18} = 3

Every ratio is $3$ , so this is geometric with $r = 3$ and $a_{1} = 2$ .

Step 02 of 05

The recursive formula for geometric is "multiply by r":

a_{n} = a_{n - 1} \cdot r

For our sequence:

a_{1} = 2, a_{n} = a_{n - 1} \cdot 3

Same shape as arithmetic, but with multiplication where there was addition. The starting value $a_{1}$ is still required.

Step 03 of 05

The explicit formula uses an exponent because we're stacking multiplications:

a_{n} = a_{1} \cdot r^{n - 1}

Starting at $a_{1}$ , you multiply by r exactly $(n - 1)$ times. Term 2 is one multiplication, term 3 is two, and so on — same counting logic as arithmetic, just with a different operation.

Step 04 of 05

Plug in our numbers. With $a_{1} = 2$ and $r = 3$ :

a_{n} = 2 \cdot 3^{n - 1}

Quick check on term 4: $2 \cdot 3^{3} = 2 \cdot 27 = 54$ $✓$ — matches the sequence.

Term 8 jumps to $2 \cdot 3^{7} = 4, 374$ . Geometric sequences grow (or shrink, when $∣ r ∣ < 1$ ) much faster than arithmetic.

Step 05 of 05

Spotting which kind it is. Look at consecutive pairs.

Constant difference $a_{n} - a_{n - 1}$	→arithmetic, use $d$
Constant ratio $a_{n} / a_{n - 1}$	→geometric, use $r$
Neither is constant	→not arithmetic, not geometric

Key insight

Arithmetic adds d repeatedly → linear growth, formula

a_{n} = a_{1} + (n - 1) d

. Geometric multiplies by r repeatedly → exponential growth, formula

a_{n} = a_{1} \cdot r^{n - 1}

. Same skeleton, different engine.

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