IB Math AI HL Sequences & Series — Complete Cheatsheet
Every sequences-and-series skill and trap for IB Mathematics Applications & Interpretation HL — arithmetic & geometric nth term, series sums, sum to infinity, sigma notation, and the AI applications of compound growth, savings & least-n. Hand-built by an IBO-certified Singapore tutor, with a print-ready PDF to download.
Sequences and series are where IB Mathematics Applications & Interpretation HL first turns pattern-spotting into modelling. AI treats them less as abstract algebra and more as the mathematics of growth over time: a salary that rises by a fixed raise each year, savings that compound, a drug that decays between doses, a population that settles to a steady level. A calculator is always allowed, so the marks live in recognising which pattern a worded problem hides — arithmetic or geometric — choosing the right formula, and reading the GDC's answer correctly, not in grinding out terms by hand.
This cheatsheet condenses the whole of SL 1.2–1.4 — arithmetic and geometric nth terms, the two series sums, the sum to infinity, sigma notation, and the AI applications of compound growth, savings and "how many terms" least-n problems — onto one page, and flags the traps that quietly cost method marks. The print-ready PDF is at the bottom, free to download.
§1 — Two engines: arithmetic vs geometric SL 1.2–1.3
A sequence is an ordered list of numbers; add its terms and you get a series. Almost every question runs on one of two engines: the arithmetic sequence adds a fixed common difference $d$, and the geometric sequence multiplies by a fixed common ratio $r$.
Arithmetic (add $d$):$u_n=u_1+(n-1)d$
Geometric (times $r$):$u_n=u_1 r^{\,n-1}$
Sum to infinity:$S_\infty=\dfrac{u_1}{1-r}$ — only when $|r| < 1$ (geometric)
TrickTo classify a sequence, test consecutive terms: a constant difference $u_{n+1}-u_n$ means arithmetic; a constant ratio $u_{n+1}/u_n$ means geometric.
TrapIt takes $(n-1)$ steps to reach the $n$th term, not $n$ — so both formulas carry the exponent/multiplier $n-1$. Slipping to $n$ is the single biggest source of dropped marks on this topic.
§2 — Arithmetic sequences: the nth term SL 1.2
An arithmetic sequence steps up (or down) by the same amount every time. Fix $u_1$ and $d$ and you can reach any term directly.
nth term:$u_n=u_1+(n-1)d$
Common difference:$d=u_{n+1}-u_n$ — the same for every consecutive pair
From two terms:$u_j-u_i=(j-i)\,d$ — solve for $d$ first, then back-substitute for $u_1$
TrickGiven two terms, the number of steps between the $i$th and $j$th is $j-i$, not $j$. Divide the change in value by the change in position to get $d$ in one line.
TrapCount gaps, not terms: reaching the $n$th term takes $n-1$ jumps of $d$. Writing $u_n=u_1+nd$ shifts every answer by one $d$.
§3 — Arithmetic series: summing the terms SL 1.2
The series is the running total $S_n$ of the first $n$ terms. There are two versions of the same formula — pick the one that matches what the question gives you.
TrickAverage the first and last term, then multiply by how many terms there are: $S_n=\left(\dfrac{u_1+u_n}{2}\right)n$. That is all the first formula says.
NoteThe two forms are the same equation — substitute $u_n=u_1+(n-1)d$ into the first to get the second. Use the top one when you already know $u_n$, the bottom one when you only know $d$.
§4 — Geometric sequences: the nth term SL 1.3
A geometric sequence scales by the same factor $r$ each step, so its terms grow or decay exponentially — exactly the shape of compound interest and radioactive-style decay.
nth term:$u_n=u_1 r^{\,n-1}$
Common ratio:$r=\dfrac{u_{n+1}}{u_n}$ — the same for every consecutive pair
Grow or decay:$|r| > 1$ grows without bound; $0 < |r| < 1$ shrinks toward $0$
TrapThe exponent is $n-1$, so the first term uses $r^0=1$. Writing $u_1 r^{\,n}$ shifts every term one place along the sequence.
TrickFor "the first term below/above a level", scroll a GDC table of $u_n$ until it crosses — and always check the term just before, since a threshold is an inequality and the answer sits right at the boundary.
§5 — Geometric series: summing the terms SL 1.3
Adding the first $n$ geometric terms has a closed form given in the booklet in two equivalent shapes.
Sum of $n$ terms:$S_n=\dfrac{u_1(r^{\,n}-1)}{r-1}=\dfrac{u_1(1-r^{\,n})}{1-r},\quad r\neq 1$
TrickThe two forms are identical — use the left one when $r > 1$ and the right one when $|r| < 1$, so the top and bottom stay positive and you avoid sign slips.
TrapThe series sum uses $r^{\,n}$ (with $n$ = the number of terms), one power higher than the $r^{\,n-1}$ in the $n$th-term formula. Mixing the two exponents is a classic error.
§6 — Sum to infinity SL 1.3
If the terms shrink fast enough, adding infinitely many of them still gives a finite total. This happens precisely when the ratio is a proper fraction.
Converges when:$|r| < 1$, i.e. $-1 < r < 1$ — the terms must tend to $0$
Sum to infinity:$S_\infty=\dfrac{u_1}{1-r}$
Find $r$ from $S_\infty$:rearrange to $r=1-\dfrac{u_1}{S_\infty}$
Each term is a fixed fraction r of the one before, so the running total edges toward the finite limit S∞ = u₁/(1−r) — shown here for r = ½. The gap to S∞ never quite closes, it just shrinks to nothing.
TrapThe sum to infinity exists only if $|r| < 1$. If $|r|\ge 1$ the terms do not shrink, the total runs off to infinity, and $\dfrac{u_1}{1-r}$ is meaningless — check $r$ before you quote it.
TrickGiven $u_1$ and $S_\infty$, don't re-derive: divide the first term by the sum and subtract from $1$ to get $r=1-\dfrac{u_1}{S_\infty}$ in one step.
§7 — Sigma notation SL 1.2
Sigma notation packs a whole sum into one symbol. The counter $k$ runs from the bottom value up to the top value, one term added for each.
Linear sum:$\displaystyle\sum_{k=1}^{n}(pk+q)=p\,\dfrac{n(n+1)}{2}+qn$
TrickSplit the sum: $\sum(pk+q)=p\sum k+\sum q$. Here $\sum_{k=1}^{n}k=\tfrac{n(n+1)}{2}$, and the constant $q$ is added once per term, so $\sum q = qn$.
NoteRead the limits before anything else — they set how many terms you are adding. A sum from $k=0$ or $k=3$ has a different term count from one starting at $k=1$, which changes both the number of terms and $S_n$. Any sum with $\sum$ can also be typed straight into the GDC.
§8 — Compound growth & the least-n question SL 1.4
This is the headline AI application. Money, populations and investments grow geometrically, and the favourite Paper 2 question asks "after how many years / terms does it first pass a target?" — a least-$n$ problem best cracked on the GDC.
Compound growth:value after $n$ periods $=P(1+i)^{\,n}$, with rate $i$ per period
Least $n$ (growth):$P(1+i)^{\,n} > T\ \Rightarrow\ n > \dfrac{\ln(T/P)}{\ln(1+i)}$, then round up
Least $n$ (running total):smallest $n$ with $S_n=\tfrac{n}{2}\big(2u_1+(n-1)d\big)\ge$ target
TrickSkip the logs: on the GDC, tabulate the amount (or the running sum $S_n$) against $n$ and read off the first $n$ that clears the target. It is faster and sidesteps rounding slips entirely.
Trap$n$ is a whole number and the inequality is strict, so round up (take the ceiling) — never round to nearest. A value of $n>12.3$ means $n=13$, and you should confirm $S_{12}$ falls short while $S_{13}$ clears the goal.
§9 — Recurrences, steady states & bouncing balls SL 1.4
Two synthesis applications AI loves. A repeating dose or deposit that decays between steps is a linear recurrence, and a bouncing ball's total path length is a geometric series in disguise.
Linear recurrence:$A_{n+1}=rA_n+c$ — keep a fraction $r$, then add $c$ each step; iterate on the GDC for any $A_k$
Steady state:the fixed point solves $L=rL+c\ \Rightarrow\ L=\dfrac{c}{1-r}$
Dropped from height u₁, the ball rebounds to r u₁, then r²u₁, … — a geometric sequence of peak heights. The total distance sums every rise and fall: D = u₁(1+r)/(1−r).
TrapFor the long-run limit divide by $1-r$, never by $r$. And in the bouncing ball, every bounce is counted twice (up and down) while the first drop counts once — that is why the numerator carries a factor of $2$ on the bounces but not on $u_1$.
NoteThe fixed point $L=\dfrac{c}{1-r}$ is exactly the same idea as a Markov steady state in AHL 1.15 — the long-run level a repeating process settles to, independent of where it started. The scalar recurrence here scales straight up to a matrix one there.
Maya starts a job on an annual salary of $\$45\,000$. Her contract gives her a $4\%$ pay rise at the start of every year after the first, so her yearly salaries form a geometric sequence with first term $u_1=45\,000$ and ratio $r=1.04$. (a) Find her salary in year $10$. (b) Find her total earnings over the first $10$ years. (c) Find the least number of complete years after which her total earnings first exceed $\$700\,000$.
Solution
(a) Year 10 is the $10$th term, so use the exponent $n-1=9$: $u_{10}=45\,000\,(1.04)^{9}=45\,000\times1.42331=\$64\,049$ (nearest dollar). (M1)(A1)(the raise applies $9$ times by year 10, not $10$ — the $n-1$ trap.)
(b) Total of the first $10$ terms is a geometric series: $S_{10}=\dfrac{u_1(r^{10}-1)}{r-1}=\dfrac{45\,000\,(1.04^{10}-1)}{0.04}=\$540\,275$ (nearest dollar). (M1)(A1)
(c) Require $S_n=\dfrac{45\,000\,(1.04^{n}-1)}{0.04} > 700\,000$, i.e. $1.04^{n} > 1.6222$, so $n > \dfrac{\ln 1.6222}{\ln 1.04}=12.3$. Rounding up, $n=13$ years. (M1)(A1)(check: $S_{12}=\$676\,161 < 700\,000$ and $S_{13}=\$748\,208$ — the first to clear the target.)
Examiner's note: keep full calculator precision throughout — rounding $1.04^{9}$ early throws off the later parts. In part (c) the inequality is strict and $n$ counts whole years, so you must round up: reading $12.3$ as $12$ is the classic lost mark. The quickest safe route is a GDC table of $S_n$ against $n$ — scroll until the running total first passes $\$700\,000$. Give money answers with sensible units and rounding.
Common Student Questions
How do I tell whether a sequence is arithmetic or geometric?
Test consecutive terms. If the difference $u_{n+1}-u_n$ is the same each time, it is arithmetic with common difference $d$. If the ratio $u_{n+1}/u_n$ is the same each time, it is geometric with common ratio $r$. Check two or three pairs — a worded problem that says "increases by $\$200$ a year" is arithmetic, while "grows by 4% a year" or "halves each hour" is geometric.
Why is it $u_1 r^{\,n-1}$ and not $u_1 r^{\,n}$?
Because the first term has had the ratio applied zero times: $u_1=u_1 r^{0}=u_1$. By the $n$th term you have multiplied by $r$ a total of $n-1$ times, so $u_n=u_1 r^{\,n-1}$. The same logic gives $u_1+(n-1)d$ for arithmetic sequences: it takes $n-1$ steps to reach term $n$. Using $r^{\,n}$ or $nd$ shifts every answer one place along.
When does a geometric series have a sum to infinity?
Only when $|r| < 1$, that is $-1 < r < 1$. Then the terms shrink toward $0$ fast enough that the infinite total settles on the finite value $S_\infty=\dfrac{u_1}{1-r}$. If $|r|\ge 1$ the terms stay the same size or grow, the sum runs off to infinity, and the formula does not apply — always check $r$ before quoting a sum to infinity.
For "how many years" questions, should I use logs or the GDC?
Either works, and both are examinable. Logs give $n>\dfrac{\ln(T/P)}{\ln(1+i)}$ directly, but the safest method on Paper 2 is a GDC table: list the amount (or the running sum) against $n$ and read off the first $n$ that passes the target. Whichever you use, remember $n$ is a whole number and the inequality is strict, so you round up — never to the nearest integer.
There are two arithmetic-sum formulas — which one do I use?
They give the same answer, so use whichever fits your data. $S_n=\dfrac{n}{2}(u_1+u_n)$ is quickest when you already know the last term $u_n$ (just average the ends and multiply by $n$). $S_n=\dfrac{n}{2}\big(2u_1+(n-1)d\big)$ is the one to reach for when you know the common difference $d$ but not the last term. They are the same equation — the second is just the first with $u_n=u_1+(n-1)d$ substituted in.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
Notes PDF — every arithmetic and geometric formula in full, with worked compound-growth, sum-to-infinity and recurrence edge cases and the exact GDC keystrokes.
Tutorial booklet — 25+ AI HL-style questions from basic nth-term fluency to full compound-growth, savings and least-n applications.
Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations for every sequences and series method.
Predicted-paper questions — the exact sequences, series and least-n question types most likely in the next session.
