Arithmetic sequences and series

Starting at $u_1$, reaching $u_n$ requires $n-1$ equal jumps—not $n$ jumps. Each jump contributes $d$, so the explicit formula is $u_n=u_1+(n-1)d$. This formula lets us find any term directly.

The formula is linear in $n$. Expanding gives $u_n=dn+(u_1-d)$, so the common difference is the coefficient of $n$. This connects arithmetic sequences to straight-line functions, while the constant term adjusts for the fact that sequence indexing begins at $n=1$.

To test whether a number belongs to a sequence, set the nth-term formula equal to that number and solve for $n$. It is a term only if the solution is a positive integer. A decimal or negative index does not represent a term in the usual IB sequence domain $n\in\mathbb N$.

The nth-term formula can be rearranged to find any missing quantity. If $u_n$, $n$ and $d$ are known, then $u_1=u_n-(n-1)d$. If $d\ne0$, the position of a known value is $n=1+\frac{u_n-u_1}{d}$.

Two non-consecutive terms also determine an arithmetic sequence. From $u_p=u_1+(p-1)d$ and $u_q=u_1+(q-1)d$, subtract the equations. The first term cancels, giving $d=\frac{u_q-u_p}{q-p}$. Substitute back to find $u_1$.

When finding a number of terms, the result for $n$ must be a positive integer. In an inequality such as ‘the first negative term’ or ‘at least a given total’, solve the boundary and then test the nearest permitted integers.

A sequence lists terms; a series adds them. If $u_1,u_2,\ldots,u_n$ form an arithmetic sequence, then $S_n=u_1+u_2+\cdots+u_n$ is the sum of its first $n$ terms.

Pair the first and last terms, then the second and second-last terms. Every pair has the same total $u_1+u_n$. This produces $S_n=\frac n2(u_1+u_n)$. Replacing $u_n$ with $u_1+(n-1)d$ gives the equivalent form $S_n=\frac n2[2u_1+(n-1)d]$.

Choose the form that matches the information given. Use $\frac n2(u_1+u_n)$ when the first and last terms are known. Use $\frac n2[2u_1+(n-1)d]$ when the first term and common difference are known.

A sum can be negative when sufficiently many terms are negative. The formula still works; do not replace negative terms with magnitudes. Keep the exact expression until the final step, particularly when solving a quadratic equation for $n$.

The Greek capital letter sigma, $\sum\limits$, means ‘add’. In $\sum\limits_{r=1}^{n}f(r)$, the lower value tells us where the index starts, the upper value tells us where it stops, and $f(r)$ gives each term to add.

An arithmetic series can be written as $\sum\limits_{r=1}^{n}[u_1+(r-1)d]$. The index letter is a dummy variable: changing $r$ to $k$ does not change the sum, provided the limits and term expression remain consistent.

Always inspect the starting index. The expressions $u_1+(r-1)d$ for $r=1,\ldots,n$ and $u_1+rd$ for $r=0,\ldots,n-1$ generate the same terms. Mixing their limits causes an off-by-one error.

Arithmetic sequences model constant additive change: the same amount is added or subtracted during each equal time interval or stage. Examples include rows of seats increasing by a fixed number, a weekly training distance increasing steadily, or an asset losing the same dollar amount each year.

Translate the context carefully. Identify what one term represents, what the first term is, the units of $d$, and whether the question asks for one future value $u_n$ or a cumulative total $S_n$. State the meaning of your answer in context and include units.

Watch the starting time. If a quantity is $96$ at time $t=0$ and falls by $4$ per hour, then $Q(t)=96-4t$. If you instead label the initial measurement $u_1$, then $u_n=96-4(n-1)$. Both models are correct, but their indices mean different things.

Real data may be only approximately arithmetic. A model can still be useful, but an IB interpretation should mention that a constant-difference assumption may stop being realistic beyond the observed range.

A calculator is useful for generating a term table, evaluating a large finite sum and solving an equation or inequality involving $S_n$. It does not replace the mathematical model: write the sequence formula, substitute the known values, and then use technology for the numerical step.

For a term table, enter the nth-term expression as a function of an integer variable and view values beginning at $n=1$. For a sum, use the calculator's summation template with correct lower and upper limits. For an unknown $n$, graph both sides or use a numerical solver, then apply the positive-integer constraint.

On common IB-approved calculators, a table or spreadsheet can generate $u_n$, while the catalog or calculation menu contains a sigma/sum command. Menu names vary by model, so practise on the calculator you will use in the examination.

Keep exact values during algebra and round only a contextual final answer when appropriate. Counts of people, rows or payments must be whole numbers. If technology gives $n=23.14$ for a minimum threshold, test $n=23$ and $n=24$ rather than reporting a decimal number of terms.

Attempt each question before opening the worked steps. The set moves from recognizing arithmetic structure to multi-step IB-style reasoning. Show formulas, substitutions and exact values before using a calculator.

After each solution, identify the decision that mattered most: finding $d$, counting $n-1$ gaps, choosing $u_n$ or $S_n$, handling sigma limits, or enforcing an integer constraint.