The Auditorium — Teacher Answers

🔐 Student Activity Unlock Codes (case-insensitive)

Phase 1 → 2

STALLS

Phase 2 → 3

CIRCLE

Phase 3 → 4

BALCONY

Phase 4 → Done

CURTAIN

🎨 Oral Launch — Read Aloud to Class

"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."

Students work entirely on their whiteboard. Display phase content on student devices only after each gate is unlocked.

Reading the Pattern

Arithmetic sequences · Deriving \(u_n\)

🗣 Phase 1 Prompt — Read Aloud

"The architect shows you the first four rows: Row 1 has 12 seats, Row 2 has 15, Row 3 has 18, Row 4 has 21."

Task 1a — Row 10 and Row 20

The constant difference is \(d = 3\) seats per row.

\(u_{10} = 12 + 9 \times 3 = 39\) seats
\(u_{20} = 12 + 19 \times 3 = 69\) seats

u₁₀ = 39 seats · u₂₀ = 69 seats

🔍 What to look for

Most groups will build a table. Some jump straight to multiplying. Both valid — reward groups who articulate why it works.

Task 1b — General term \(u_n\)

\(u_n = 12 + 3(n - 1)\)

Also acceptable simplified: \(u_n = 3n + 9\).

u_n = 12 + 3(n − 1)

🔍 Common error

Students sometimes write \(u_n = 12 + 3n\) (off by one). Ask: does your formula give 12 when \(n = 1\)?

Task 1c — Which row has 63 seats?

\(12 + 3(n-1) = 63 \implies 3(n-1) = 51 \implies n = 18\)

Row 18

✓ Gate check before giving STALLS

Verify a correct general expression for \(u_n\) and \(n = 18\) with working. Both must be present.

A New Theatre

Extracting \(u_1\) and \(d\) · Solving for \(n\)

🗣 Phase 2 Prompt — Read Aloud

"The architect has a second design — only two rows shown: Row 5 has 38 seats, and Row 12 has 59 seats. Figure out everything else."

Task 2a — Find \(d\) and \(u_1\)

\(d = \dfrac{59 - 38}{7} = 3\)

\(u_5 = u_1 + 4d \implies 38 = u_1 + 12 \implies u_1 = 26\)

d = 3 · u₁ = 26 seats

🔍 Key insight

From Row 5 to Row 12 spans exactly 7 differences. Ask groups to justify the step count, not just the division.

Task 2b — General term

\(u_n = 26 + 3(n - 1)\)

u_n = 26 + 3(n − 1)

Task 2c — How many rows fit within 150 seats?

\(26 + 3(n-1) \leq 150 \implies n \leq 42.33\ldots \implies n = 42\)
Check: \(u_{42} = 26 + 123 = 149 \leq 150\) ✓

42 rows

✓ Gate check before giving CIRCLE

Verify \(d = 3\), \(u_1 = 26\), general term, and \(n = 42\) with a clear inequality argument.

The Sum Problem

Deriving \(S_n\) · Gauss pairing

🗣 Phase 3 Prompt — Read Aloud

"Third design: Row 1 has 20 seats, each row adds 6. The client needs total seat counts."

Task 3a — Total seats, first 5 rows

20 + 26 + 32 + 38 + 44 = 160

S₅ = 160 seats

Task 3b — Total seats, first 15 rows

\(u_{15} = 20 + 14 \times 6 = 104\)

\(S_{15} = \dfrac{15}{2}(20 + 104) = 15 \times 62 = 930\)

S₁₅ = 930 seats

Task 3c — Total seats, 40 rows

\(u_{40} = 20 + 39 \times 6 = 254\)

\(S_{40} = \dfrac{40}{2}(20 + 254) = 20 \times 274 = 5\,480\)

S₄₀ = 5 480 seats

🗣 Teacher verbal prompt (if no group finds the pairing)

Say: "Write the sum from Row 1 to Row 40. Now write it again directly underneath, but in reverse. What do you notice about the columns?" Every column sums to \(20 + 254 = 274\), giving \(40 \times 274\) — written twice, so divide by 2.

📐 Formula students should derive

\(S_n = \dfrac{n}{2}(u_1 + u_n) = \dfrac{n}{2}(2u_1 + (n-1)d)\)

Task 3d — Verify

\(u_1 = 8,\ d = 5,\ n = 30\). \(u_{30} = 8 + 145 = 153\)

\(S_{30} = \dfrac{30}{2}(8 + 153) = 15 \times 161 = 2\,415\)

S₃₀ = 2 415 seats

✓ Gate check before giving BALCONY

The formula \(S_n = \tfrac{n}{2}(u_1 + u_n)\) must be written explicitly. Ask a group member to explain the Gauss pairing — not just recite the formula.

Sigma Notation & Reverse Problems

Sigma notation · Solving for unknowns · HL system

Task 4 Intro — Deriving the second form of \(S_n\)

Students substitute \(u_n = u_1 + (n-1)d\) into \(S_n = \dfrac{n}{2}(u_1 + u_n)\):

\(S_n = \dfrac{n}{2}\bigl(u_1 + u_1 + (n-1)d\bigr) = \dfrac{n}{2}\bigl(2u_1 + (n-1)d\bigr)\)

S_n = n/2 · (2u₁ + (n−1)d)

🔍 Why this matters

This form is essential for Task 4b where \(u_n\) is not given. Make sure every group has both forms on their board before moving on.

Task 4a — Sigma notation

The \(\Sigma\)/S hint should lead groups to understand: sum the terms \(7 + 4(k-1)\) for \(k = 1\) to \(25\). This is an arithmetic sequence with \(u_1 = 7\), \(d = 4\), \(n = 25\).

\(S_{25} = \dfrac{25}{2}(7 + 103) = 25 \times 55 = 1\,375\)

∑ = 1 375

Phase 3c in sigma notation: \(\displaystyle\sum_{k=1}^{40}(20 + 6(k-1))\)

🔍 What to look for

Students should identify that \(k\) is the row index and the bounds give the number of terms. Do not pre-explain sigma notation — let groups reason it out.

Task 4b — Working backwards (SL)

Given: \(S_{25} = 1\,350\), \(n = 25\), \(d = 4\). Find \(u_1\).

\(\dfrac{25}{2}(2u_1 + 96) = 1\,350 \implies 25(2u_1 + 96) = 2\,700 \implies 2u_1 + 96 = 108 \implies u_1 = 6\)

u₁ = 6 seats

HL Extension — Full worked solution

Given: \(d = 3\), last row \(u_n = 49\), total \(S_n = 424\). Find \(n\) and \(u_1\).

(1) \(u_1 + 3(n-1) = 49 \implies u_1 = 52 - 3n\)

(2) \(\dfrac{n}{2}(u_1 + 49) = 424\)

Substitute (1) into (2):
\(\dfrac{n}{2}(52 - 3n + 49) = 424 \implies n(101 - 3n) = 848\)

\(3n^2 - 101n + 848 = 0\)

\(\Delta = 10201 - 10176 = 25\)

\(n = \dfrac{101 \pm 5}{6} \implies n = 16 \text{ (integer, valid)} \text{ or } n = \tfrac{106}{6} \text{ (rejected)}\)

\(u_1 = 52 - 48 = 4\)

Verify: \(u_{16} = 4 + 45 = 49\) ✓ · \(S_{16} = 8 \times 53 = 424\) ✓

n = 16 rows · u₁ = 4 seats

Note: Students should check both roots and reject the non-integer with a clear reason.

✓ Gate check before giving CURTAIN

SL: Task 4a value 1 375, Phase 3c in sigma notation, Task 4b with \(u_1 = 6\) and full working.

HL: additionally, quadratic formed and solved correctly, non-integer root explicitly rejected.

✦

Consolidation from the Boards

Select 2–3 groups · 10–12 min debrief

Suggested discussion sequence

One board showing the table approach for Phase 1 — connect to the general term: why is it \((n-1)\) steps, not \(n\)?
One board showing the Gauss pairing — write both the forward and backward sums explicitly. Derive the formula from it as a class.
Sigma notation — have a group explain what \(k\) represents and why the bounds are 1 to 25. Ask: could you write the same sum starting from \(k = 0\)?
HL only: show how the two conditions become a system leading to a quadratic. Emphasise: always verify both roots — context eliminates non-integer answers.