Sequences are ordered lists of numbers following a pattern; series are their sums. Two types dominate competition math: arithmetic (constant difference) and geometric (constant ratio). Master their formulas and you can solve everything from bank interest to tower stacking problems.
Each term = previous term + common difference $d$.
Example: 3, 7, 11, 15, ... (d = 4)
n-th term:
\[ a_n = a_1 + (n - 1)d \]
Where $a_1$ = first term, $d$ = common difference, $n$ = term number
Alternative: $ a_n = a_k + (n - k)d $ β useful if you know some middle term
Sum of the first $n$ terms of an arithmetic sequence:
\[ S_n = \frac{n(a_1 + a_n)}{2} \]
Intuition: pair up terms from both ends β each pair sums to $a_1 + a_n$, and there are $n/2$ pairs.
Alternative form (substitute $a_n = a_1 + (n-1)d$):
$ S_n = na_1 + \frac{n(n-1)}{2}d $
\(Each term = previous term \\times common ratio\) $r$.
Example: 2, 6, 18, 54, ... (r = 3)
n-th term:
\[ a_n = a_1 \cdot r^{n-1} \]
Where $a_1$ = first term, $r$ = common ratio
Sum of the first $n$ terms of a geometric sequence:
\[ S_n = a_1 \cdot \frac{1 - r^n}{1 - r} \quad (r \neq 1) \]
Derivation trick: write $S_n$ and $r \cdot S_n$, subtract, solve for $S_n$.
If $ |r| < 1 $, the series converges to a finite value:
\[ S_\infty = \frac{a_1}{1 - r} \]
If $ |r| \geq 1 $, the series diverges (sum goes to \(\pm\) infinity or oscillates).
Common applications: repeating decimals, Zeno's paradox, bouncing ball heights
A series where most terms cancel out ("telescope").
Classic pattern: $ \sum_{k=1}^{n} \frac{1}{k(k+1)} = \sum (\frac{1}{k} - \frac{1}{k+1}) $
When expanded: $ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \cdots + (\frac{1}{n} - \frac{1}{n+1}) $
All middle terms cancel! Sum = $ 1 - \frac{1}{n+1} = \frac{n}{n+1} $
Strategy: Use partial fractions to decompose, then look for cancellation.
For sequences where neither the difference nor ratio is constant:
Useful for "find the next term" pattern problems.
Identify: Arithmetic sequence with $a_1 = 1$, $d = 2$, $n = 20$.
Step 1: Find the 20th term.
$ a_{20} = a_1 + (n-1)d = 1 + 19(2) = 1 + 38 = 39 $
Step 2: Use arithmetic series formula.
\[ S_n = \frac{n(a_1 + a_n)}{2} \]
\[ S_{20} = \frac{20(1 + 39)}{2} = \frac{20 \times 40}{2} = 10 \times 40 = 400 \]
Fun fact: The sum of the first n odd integers = $n^2$. So $20^2 = 400$. β
Answer: 400
Identify: Geometric sequence with $a_1 = 5$, $r = 2$.
After 0 hours (start): 5 bacteria (that's term 1)
After 1 hour: 10 bacteria (term 2)
After 6 hours: that's the 7th term!
Step 1: Use n-th term formula.
\[ a_n = a_1 \cdot r^{n-1} \]
Step 2: $a_1 = 5$, $r = 2$, $n = 7$ (after 6 hours = term 7)
\[ a_7 = 5 \cdot 2^{6} = 5 \times 64 = 320 \]
Check: \(5 \\to 10 \\to 20 \\to 40 \\to 80 \\to 160 \\to 320 (6 doublings).\) β
Answer: 320
Identify: Geometric series with $a_1 = 8$, $r = \frac{1}{2}$.
Since $ |r| = \frac{1}{2} < 1 $, the series converges.
Step 1: Infinite geometric series formula.
\[ S_\infty = \frac{a_1}{1 - r} \]
Step 2: Plug in values.
\[ S_\infty = \frac{8}{1 - \frac{1}{2}} = \frac{8}{\frac{1}{2}} = 8 \times 2 = 16 \]
Intuition: You keep adding half of what's left to reach 16. You get closer and closer but never exceed it β the total sum converges to exactly 16.
Answer: 16