Sequences & Series

What you need to know

We write $a_n$ for the $n$th term of a sequence, so $a_1$ is the first term.

• Arithmetic sequences go up by the same amount $d$ each time (the common difference). The $n$th term is $a_n=a_1+(n-1)d$. The sum of the first $n$ terms is $S_n=\dfrac n2(a_1+a_n)$: the number of terms times the average of the first and last. Example: for $5,8,11,\dots$, $a_{20}=5+19\cdot3=62$ and $S_{20}=20\cdot\frac{5+62}{2}=670$.
• Geometric sequences multiply by the same number $r$ each time (the common ratio). Then $a_n=a_1r^{n-1}$ and $S_n=a_1\dfrac{r^n-1}{r-1}$. Example: for $3,6,12,\dots$, $a_6=3\cdot2^5=96$ and $S_6=3\cdot\frac{2^6-1}{2-1}=189$.
• Infinite geometric series. If $r$ is between $-1$ and $1$, the sum that goes on forever is $\dfrac{a_1}{1-r}$. Example: $8+4+2+1+\dots=\frac{8}{1-\frac12}=16$.
• Sums to memorize. $1+2+\dots+n=\dfrac{n(n+1)}2$. $1^2+2^2+\dots+n^2=\dfrac{n(n+1)(2n+1)}6$. $1^3+2^3+\dots+n^3=\left(\dfrac{n(n+1)}2\right)^2$. Example: with $n=10$ these are $55$, $385$, and $55^2=3025$.
• Telescoping. Write each term as a difference so that neighbors cancel and only the ends survive. The key split is $\dfrac1{k(k+1)}=\dfrac1k-\dfrac1{k+1}$ (this is partial fractions). Example: $\frac1{1\cdot2}+\frac1{2\cdot3}+\dots+\frac1{99\cdot100}=1-\frac1{100}=\frac{99}{100}$.
• Recursions often repeat. When each term is made from the one before, compute terms until one repeats. From there the sequence cycles. If the cycle goes all the way back to $a_1$, only the remainder of the index divided by the cycle length (the period) matters. If some terms come before the cycle starts, count from the start of the cycle instead: adding the squares of the digits, starting from $2$, gives $2, 4, 16, 37, 58, 89, 145, 42, 20, 4, \dots$, a cycle of length $8$ that starts at $a_2$. Example: $a_1=2$ and $a_{n+1}=\frac1{1-a_n}$ give $2,-1,\frac12,2,\dots$, with period $3$. Since $100=3\cdot33+1$, $a_{100}=a_1=2$.
• Three-term tests. $a,b,c$ is arithmetic exactly when $2b=a+c$. For numbers that are not zero, $a,b,c$ is geometric exactly when $b^2=ac$. Example: $4,7,10$ is arithmetic ($14=4+10$) and $4,6,9$ is geometric ($36=4\cdot9$).

How AMC 10 tests it

• Problems 3–10: a direct question about an arithmetic sequence (one term, a sum, or the number of terms).
• Problems 8–15: "$a_1=\dots$, $a_{n+1}=\dots$; find $a_{2025}$." Compute a few terms, spot the cycle, and use the remainder of the index.
• Problems 10–18: an arithmetic or geometric sequence hidden in a story (rows of seats, a bouncing ball), sometimes an infinite geometric series.
• Problems 14–20: a telescoping sum of fractions, or of terms like $k\cdot k!$. Rewrite each term as a difference, such as $k\cdot k!=(k+1)!-k!$.
• Mixed conditions ("$a,b,c$ is arithmetic and $a,c,b$ is geometric") that turn into a small system of equations.

Standard approaches

1. Decide the type: arithmetic, geometric, telescoping, or recursive. For a recursion, write out the first five or six terms.
2. For an arithmetic sum, use (number of terms) $\times$ (average term). Count the terms as $\frac{\text{last}-\text{first}}d+1$.
3. For a sum of fractions, or a product, try to write each term as a difference of neighboring pieces, like $\frac1k-\frac1{k+1}$.
4. For a term with a huge index, find the period or a formula, then reduce the index.
5. Use the answer choices as a check: whole number or not, even or odd, or the pattern from small cases.

Worked example

Define $a_1=3$ and $a_{n+1}=\dfrac{a_n-1}{a_n+1}$ for $n\ge1$. What is the product $a_1a_2a_3\cdots a_{2026}$?

(A) $-2$ (B) $-\frac13$ (C) $\frac12$ (D) $1$ (E) $\frac32$

Compute terms until one repeats.
$$a_2=\frac{3-1}{3+1}=\frac12,\quad a_3=\frac{\frac12-1}{\frac12+1}=\frac{-\frac12}{\frac32}=-\frac13,\quad a_4=\frac{-\frac13-1}{-\frac13+1}=\frac{-\frac43}{\frac23}=-2,\quad a_5=\frac{-2-1}{-2+1}=3.$$
Since $a_5=a_1$, the terms repeat in blocks of $4$: $3,\frac12,-\frac13,-2$.

Multiply one block. $3\cdot\frac12\cdot\left(-\frac13\right)\cdot(-2)=1$.

Count the blocks. $2026=4\cdot506+2$. So the $2026$ terms are $506$ full blocks, each multiplying to $1$, then two extra terms: $a_{2025}=a_1=3$ and $a_{2026}=a_2=\frac12$.

The whole product is $1\cdot3\cdot\frac12=\frac32$. The answer is $\boxed{\textbf{(E)}\ \frac32}$.

Pitfalls

• Off-by-one when counting terms: $a_n$ is $n-1$ steps past $a_1$. Example: for $5,8,11,\dots$, $a_{20}=5+19\cdot3$, not $5+20\cdot3$.
• Using $\frac{a_1}{1-r}$ when $r$ is not between $-1$ and $1$. Example: $1+2+4+\dots$ has no sum, but the formula would give $-1$.
• Reducing the index wrongly. With period $4$ and a cycle that starts at $a_1$, an index that is a multiple of $4$ (remainder $0$, like $2024$) matches $a_4$, not a nonexistent $a_0$.
• Telescoping without tracking which terms survive at both ends, or dropping the factor in front: $\dfrac1{k(k+2)}=\dfrac12\left(\dfrac1k-\dfrac1{k+2}\right)$, not $\dfrac1k-\dfrac1{k+2}$.

Traps that recur

• Trying to compute terms exactly, or bounding with only one side of the ratio and landing in the wrong interval. (2019 AMC 10B #24)
• Reading 'the average drops by 1' as 'each term drops by 1' and running the sequence down from around 2025 to 1, which points at choice (E). (2025 AMC 10B #17)
• Assuming the ratio must be an integer (b = 1440, digit sum 9) or grabbing 9/8 (b = 810) without checking that 15 and 16 both divide 720. (2024 AMC 10A #19)
• Assuming every row has the same common difference (a linear function Bi + Cj + A), which cannot fit the four given values; the row differences themselves form an arithmetic progression. (2024 AMC 10A #21)