"A concert hall is being designed. The architect has one rule: every row must have a fixed number of extra seats compared to the row in front. You are the venue consultants. Your job is to answer the client's increasingly difficult questions — starting with the simplest design and working up to the full hall."
Do not hand out any printed materials. Students work entirely on their VNPS. Display only the phase content on student devices when each phase unlocks.
SL 1.2 · Arithmetic sequences · Deriving \(u_n\)
"The architect shows you the first four rows of the design: Row 1 has 12 seats, Row 2 has 15, Row 3 has 18, Row 4 has 21."
The constant difference is \(d = 3\) seats per row.
Students should arrive at these by extending their table or by reasoning "each row adds 3, so Row 10 is 9 steps from Row 1." Accept any correct method.
Most groups will build a table. Some will jump straight to multiplying by 3 and adding 12. Both are valid. Reward groups who articulate why it works — the constant difference is the key idea.
Students should notice: Row 1 needs \(0\) extra steps, Row 2 needs \(1\) extra step, Row \(n\) needs \((n-1)\) extra steps of 3.
Also acceptable simplified: \(u_n = 3n + 9\). Both forms are correct.
Students sometimes write \(u_n = 12 + 3n\) (off by one). Ask them to check: does their formula give 12 when \(n = 1\)?
Verify: (1) a correct general expression for \(u_n\) on the board, and (2) \(n = 18\) with working shown. Both must be present before releasing the code.
SL 1.2 · Extracting \(u_1\) and \(d\) from given terms · Solving for \(n\)
"The architect has a second design — but this time they've only shown you two rows: Row 5 has 38 seats, and Row 12 has 59 seats. You need to figure out everything else."
Between Row 5 and Row 12 there are \(12 - 5 = 7\) steps.
The key insight is that going from Row 5 to Row 12 spans exactly 7 differences. Students who just divide \(59 - 38\) by \(12 - 5\) without explanation should be asked to justify the step count.
Also acceptable: \(u_n = 3n + 23\).
Since \(n\) must be a whole number: \(n = 42\). Check: \(u_{42} = 26 + 3(41) = 26 + 123 = 149 \leq 150\) ✓
Verify: correct \(d = 3\), \(u_1 = 26\), general term, and \(n = 42\) with a clear inequality argument (not just trial and error).
SL 1.2 · Deriving \(S_n\) · Gauss pairing
"Third design: Row 1 has 20 seats, each row adds 6. The client needs to know total seat counts — for planning purposes."
Row values: 20, 26, 32, 38, 44.
\(u_{15} = 20 + 14 \times 6 = 104\)
If students are still adding term by term: let them. The pain at 15 rows is mild. The jump to 40 rows is where the Gauss moment should happen.
\(u_{40} = 20 + 39 \times 6 = 254\)
Say to the group (do not write this on anything): "Write the sum from Row 1 to Row 40. Now write it again directly underneath, but in reverse order — starting from Row 40. What do you notice about the columns?"
Students should see that every column sums to \(20 + 254 = 274\), giving \(40 \times 274\) — but they've written the sum twice, so divide by 2.
\(S_n + S_n = n(u_1 + u_n)\)
\(\therefore S_n = \dfrac{n}{2}(u_1 + u_n) = \dfrac{n}{2}\bigl(2u_1 + (n-1)d\bigr)\)
\(u_1 = 8,\ d = 5,\ n = 30\).
\(u_{30} = 8 + 29 \times 5 = 8 + 145 = 153\)
Verify: the formula \(S_n = \tfrac{n}{2}(u_1 + u_n)\) or equivalent is written explicitly on the board, \(S_{40} = 5\,480\), and \(S_{30} = 2\,415\). Ask a group member to explain the Gauss pairing — not just recite the formula.
SL 1.2 · Sigma notation · Solving for unknowns · HL system
Students substitute \(u_n = u_1 + (n-1)d\) into \(S_n = \dfrac{n}{2}(u_1 + u_n)\):
This form is essential for Task 4b, where \(u_n\) is not given directly. Make sure every group has both forms written on their board before moving on.
The \(\Sigma\) hint (S for sum) should be enough for groups to understand the expression is asking them to sum the terms \(7 + 4(k-1)\) as \(k\) runs from 1 to 25. This is an arithmetic sequence with \(u_1 = 7\), \(d = 4\), \(n = 25\).
Phase 3c (Task 3c) written in sigma notation:
Students should decode that \(k\) is the row index, the expression inside the sum is \(u_k\), and the bounds give the first and last row. The key is getting from the notation to "this is just an arithmetic sum." Do not pre-explain sigma notation — let groups reason it out together.
Given: \(S_{25} = 1\,350\), \(n = 25\), \(d = 4\). Find \(u_1\).
Given: \(d = 3\), last row \(u_n = 49\), total \(S_n = 424\). Find \(n\) and \(u_1\).
Set up two equations:
From equation (1):
Substitute into equation (2):
Solve the quadratic:
Find \(u_1\):
Verify:
n = 16 rows · u₁ = 4 seats
SL minimum: Task 4a decoded correctly with sigma value 1 375, and Phase 3c written in sigma notation. Task 4b with \(u_1 = 6\) and full working.
HL: additionally, the quadratic formed and solved correctly, with the non-integer root explicitly rejected.
Select 2–3 boards · 10–12 min debrief