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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 10 · Sequences

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.19, 23
2.162, 486
3.Arithmetic (d=4d = 4)
4.d=−4d = -4
5.r=12r = \dfrac{1}{2}
6.(a) u1=7u_1 = 7 (b) u3=13u_3 = 13
7.5, 8, 11, 14
8.2, 6, 18, 54
9.u1=5u_1 = 5, u5=17u_5 = 17
10.17
Silver
11.un=3n+2u_n = 3n + 2
12.un=23−3nu_n = 23 - 3n
13.u20=83u_{20} = 83
14.n=20n = 20 (the 20th term)
15.u5=48u_5 = 48
16.4, 9, 14, 19
17.u4=40u_4 = 40
18.31
19.d=4d = 4
20.un=4⋅3n−1u_n = 4 \cdot 3^{n-1}
Gold
21.u1=5u_1 = 5, d=3d = 3
22.u14=101u_{14} = 101
23.u4=19u_4 = 19
24.u1=2u_1 = 2, r=3r = 3
25.36; square numbers un=n2u_n = n^2
26.u6=13u_6 = 13
27.n=13n = 13
28.(a) Arithmetic, d=−3d = -3. (b) un=10−3nu_n = 10 - 3n
29.S10=255S_{10} = 255
30.u4=2u_4 = 2
Platinum
31.un=3n+2u_n = 3n + 2
32.735
33.r=2r = 2, u1=3u_1 = 3
34.n=3n = 3 (both equal 17)
35.u4=6.25u_4 = 6.25 (i.e. 254\dfrac{25}{4})
36.S6=728S_6 = 728
37.30; un=n2+nu_n = n^2 + n
38.(a) un=50⋅2nu_n = 50 \cdot 2^n. (b) After 5 hours (u5=1600u_5 = 1600)
39.Week 17 (weekly = £53)
40.u5=25u_5 = 25

Pack B — Answers

Bronze
1.21, 25
2.768, 3072
3.Geometric (r=2r = 2)
4.d=−6d = -6
5.r=13r = \dfrac{1}{3}
6.(a) u1=12u_1 = 12 (b) u3=24u_3 = 24
7.2, 9, 16, 23
8.5, 10, 20, 40
9.u1=5u_1 = 5, u5=21u_5 = 21
10.32
Silver
11.un=6n−3u_n = 6n - 3
12.un=55−5nu_n = 55 - 5n
13.u25=84u_{25} = 84
14.n=20n = 20 (the 20th term)
15.u5=162u_5 = 162
16.1, 7, 13, 19
17.u4=108u_4 = 108
18.41
19.d=3d = 3
20.un=5⋅2n−1u_n = 5 \cdot 2^{n-1}
Gold
21.u1=2u_1 = 2, d=3d = 3
22.u40=202u_{40} = 202
23.u4=18u_4 = 18
24.u1=4u_1 = 4, r=2r = 2
25.125; cube numbers un=n3u_n = n^3
26.u6=21u_6 = 21
27.n=15n = 15
28.(a) Arithmetic, d=−4d = -4. (b) un=14−4nu_n = 14 - 4n
29.S12=222S_{12} = 222
30.u4=3u_4 = 3
Platinum
31.un=3n−1u_n = 3n - 1
32.4100
33.r=3r = 3, u1=2u_1 = 2
34.n=4n = 4 (both equal 22)
35.u4=5u_4 = 5
36.S5=93S_5 = 93
37.35; un=n2+2nu_n = n^2 + 2n
38.(a) un=100⋅2nu_n = 100 \cdot 2^n. (b) After 6 hours (u6=6400u_6 = 6400)
39.Week 20 (weekly = £42)
40.u5=26u_5 = 26

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) 53 (b) Row 11 (51 seats) (c) 438
Full working
Arithmetic with u1=20u_1 = 20, d=3d = 3. So un=20+(n−1)×3=3n+17u_n = 20 + (n-1) \times 3 = 3n + 17.

(a) u12=3(12)+17=53u_{12} = 3(12) + 17 = 53 seats.

(b) Solve 3n+17≥503n + 17 \geq 50: 3n≥333n \geq 33, so n≥11n \geq 11. Row 11 has u11=50u_{11} = 50 seats — first row with **at least** 50.

(c) Sum: S12=122(u1+u12)=6(20+53)=6×73=438S_{12} = \frac{12}{2}(u_1 + u_{12}) = 6(20 + 53) = 6 \times 73 = \mathbf{438} seats.
2Problem 2
Answer
(a) hn=80⋅(34)nh_n = 80 \cdot (\tfrac{3}{4})^n (b) 19.0 cm (190 mm) (c) 8th bounce (8.0 cm)
Full working
(a) Geometric: drop h0=80h_0 = 80, r=34r = \frac{3}{4}. Height of nnth bounce: hn=80⋅(34)nh_n = 80 \cdot (\tfrac{3}{4})^n.

(b) h5=80⋅(0.75)5=80×0.2373≈18.98h_5 = 80 \cdot (0.75)^5 = 80 \times 0.2373 \approx 18.98 cm ≈ **19.0 cm**.

(c) Need 80⋅(0.75)n<1080 \cdot (0.75)^n < 10, i.e. (0.75)n<0.125(0.75)^n < 0.125. Check: (0.75)7≈0.1335(0.75)^7 \approx 0.1335, (0.75)8≈0.1001(0.75)^8 \approx 0.1001. So 8th bounce: h8≈8.0h_8 \approx 8.0 cm. **8th bounce**.
3Problem 3
Answer
(a) An=22500+1500nA_n = 22500 + 1500n; Bn=22000⋅(1.05)n−1B_n = 22000 \cdot (1.05)^{n-1} (b) Year 17 (c) A: £135,000; B: £121,550
Full working
(a) Company A is arithmetic: An=24000+(n−1)×1500=22500+1500nA_n = 24000 + (n-1)\times 1500 = 22500 + 1500n. Company B is geometric: Bn=22000⋅(1.05)n−1B_n = 22000 \cdot (1.05)^{n-1}.

(b) Set Bn>AnB_n > A_n. Test years 1–15:
- Year 1: A = 24000, B = 22000.
- Year 5: A = 30000, B = 22000×1.054≈2674422000 \times 1.05^4 \approx 26744.
- Year 10: A = 37500, B = 22000×1.059≈3412422000 \times 1.05^9 \approx 34124.
- Year 15: A = 45000, B = 22000×1.0514≈4356022000 \times 1.05^{14} \approx 43560.
- Year 17: A = 48000, B = ≈48025\approx 48025.

Company B overtakes A in **year 17**.

(c) Sum over first 5 years.
- A: S5=52(A1+A5)=52(24000+30000)=2.5×54000=£135,000S_5 = \frac{5}{2}(A_1 + A_5) = \frac{5}{2}(24000 + 30000) = 2.5 \times 54000 = \mathbf{\pounds 135{,}000}.
- B: S5=22000(1.055−1)0.05=22000×0.27630.05≈£121,550S_5 = \frac{22000(1.05^5 - 1)}{0.05} = \frac{22000 \times 0.2763}{0.05} \approx \mathbf{\pounds 121{,}550}.
4Problem 4
Answer
(a) un=3n+1u_n = 3n + 1 (b) k=29k = 29 (c) No — pattern 100 has 301 dots
Full working
(a) Arithmetic with u1=4u_1 = 4, d=3d = 3. So un=4+(n−1)×3=3n+1u_n = 4 + (n-1)\times 3 = 3n + 1.

(b) Set 3n+1=883n + 1 = 88: 3n=873n = 87, n=29n = 29.

(c) u100=3(100)+1=301u_{100} = 3(100) + 1 = 301, **not 304**. The student is incorrect.
5Problem 5
Answer
(a) An=800⋅(12)nA_n = 800 \cdot (\tfrac{1}{2})^n (b) 10 folds (c) No — only approaches 0 in the limit
Full working
(a) A0=800A_0 = 800; halves each fold. An=800⋅(12)n=8002nA_n = 800 \cdot (\frac{1}{2})^n = \frac{800}{2^n}.

(b) Need 8002n<1\frac{800}{2^n} < 1, i.e. 2n>8002^n > 800. 29=512<800<1024=2102^9 = 512 < 800 < 1024 = 2^{10}. So **10 folds**: A10=8001024≈0.78A_{10} = \frac{800}{1024} \approx 0.78 cm².

(c) No: 8002n\frac{800}{2^n} is always **positive** for any finite nn. The sequence converges to 0 but never reaches it — for any non-zero ε, there is an nn with An<εA_n < \varepsilon, but no finite nn with An=0A_n = 0.
6Problem 6
Answer
(a) u5=66u_5 = 66 (b) un=6+4(2n−1−1)=2+2n+1u_n = 6 + 4(2^{n-1} - 1) = 2 + 2^{n+1}
Full working
(a) Differences: 4,8,16,32,…4, 8, 16, 32, \ldots (geometric, ratio 2).
u2=6+4=10u_2 = 6 + 4 = 10; u3=10+8=18u_3 = 10 + 8 = 18; u4=18+16=34u_4 = 18 + 16 = 34; u5=34+32=66u_5 = 34 + 32 = 66.

(b) u_n = u_1 + \sum_{k=1}^{n-1}(\text{kth difference}). The kkth difference is 4⋅2k−14 \cdot 2^{k-1}. Sum: ∑k=1n−14⋅2k−1=4⋅2n−1−12−1=4(2n−1−1)\sum_{k=1}^{n-1} 4 \cdot 2^{k-1} = 4 \cdot \frac{2^{n-1} - 1}{2 - 1} = 4(2^{n-1} - 1). So un=6+4(2n−1−1)=6+2n+1−4=2+2n+1u_n = 6 + 4(2^{n-1} - 1) = 6 + 2^{n+1} - 4 = 2 + 2^{n+1}. Check u5=2+26=2+64=66u_5 = 2 + 2^6 = 2 + 64 = 66 ✓.
7Problem 7
Answer
(a) un=3n−1u_n = 3n - 1 (b) 155 cards (c) 11 levels (uses 187 cards)
Full working
(a) Arithmetic: u1=2u_1 = 2, d=3d = 3. So un=2+(n−1)×3=3n−1u_n = 2 + (n-1)\times 3 = 3n - 1.

(b) S10=102(u1+u10)=5(2+29)=155S_{10} = \frac{10}{2}(u_1 + u_{10}) = 5(2 + 29) = 155 cards.

(c) Total for nn levels: Sn=n2(2+(3n−1))=n(3n+1)2S_n = \frac{n}{2}(2 + (3n-1)) = \frac{n(3n+1)}{2}. Solve n(3n+1)2≤200\frac{n(3n+1)}{2} \leq 200, i.e. n(3n+1)≤400n(3n+1) \leq 400.
- n=11n = 11: 11×34=37411 \times 34 = 374 ≤ 400 ✓.
- n=12n = 12: 12×37=44412 \times 37 = 444 > 400 ✗.

So 11 levels, using **11×342=187\frac{11 \times 34}{2} = 187 cards** (13 left over).
8Problem 8
Answer
(a) Arithmetic, un=4n+1u_n = 4n+1 (b) Geometric, un=3⋅2n−1u_n = 3 \cdot 2^{n-1} (c) Neither (square), un=n2u_n = n^2 (d) Neither, recursive un+1=2un+1u_{n+1} = 2u_n + 1
Full working
(a) Differences all 4 → arithmetic, un=4n+1u_n = 4n + 1.

(b) Ratios all 2 → geometric, un=3⋅2n−1u_n = 3 \cdot 2^{n-1}.

(c) Differences 3, 5, 7, 9 (not constant) and ratios not constant either, so neither arithmetic nor geometric. Pattern: un=n2u_n = n^2.

(d) Differences 3, 6, 12, 24 — not arithmetic, but these differences themselves double. Ratios 52,115,2311\frac{5}{2}, \frac{11}{5}, \frac{23}{11} — not constant. Try recursive: un+1=2un+1u_{n+1} = 2u_n + 1: 2(2)+1=52(2)+1=5 ✓; 2(5)+1=112(5)+1=11 ✓; 2(11)+1=232(11)+1=23 ✓. So **recursive rule**: un+1=2un+1u_{n+1} = 2u_n + 1 with u1=2u_1 = 2. (Closed form: un=3⋅2n−1−1u_n = 3 \cdot 2^{n-1} - 1.)
9Problem 9
Answer
(a) Bn=49−2nB_n = 49 - 2n (b) n=723n = 7\tfrac{2}{3} — no integer solution; see working (c) No integer common term
Full working
(a) B1=47B_1 = 47, d=−2d = -2. So Bn=47+(n−1)(−2)=49−2nB_n = 47 + (n-1)(-2) = 49 - 2n.

(b) Set An=BnA_n = B_n: 4n+3=49−2n4n + 3 = 49 - 2n. So 6n=466n = 46, n=466=233≈7.67n = \frac{46}{6} = \frac{23}{3} \approx 7.67.

Since nn must be a positive integer for a sequence term, the sequences **do not share a term**. However the equation 4n+3=49−2n4n + 3 = 49 - 2n has the (non-integer) solution n=233n = \frac{23}{3}, at which both would equal 1013≈33.67\frac{101}{3} \approx 33.67.

(c) The integer terms of A around this point: A7=31A_7 = 31, A8=35A_8 = 35. Terms of B: B7=35B_7 = 35, B8=33B_8 = 33. So A8=B7=35A_8 = B_7 = 35 — the value 35 appears in **both** sequences, but at different positions. The common value is **35**.
10Problem 10
Answer
(a) 5, 11, 21, 43 (b) See working (c) Ratios: 3, 1.67, 2.2, 1.91, 2.05 — approach 2
Full working
(a) u3=u2+2u1=3+2=5u_3 = u_2 + 2u_1 = 3 + 2 = 5. u4=u3+2u2=5+6=11u_4 = u_3 + 2u_2 = 5 + 6 = 11. u5=u4+2u3=11+10=21u_5 = u_4 + 2u_3 = 11 + 10 = 21. u6=u5+2u4=21+22=43u_6 = u_5 + 2u_4 = 21 + 22 = 43.

(b) **Proof by induction.** Base cases: u1=1u_1 = 1 (odd), u2=3u_2 = 3 (odd). Inductive step: assume unu_n and un−1u_{n-1} are odd. Then un+1=un+2un−1u_{n+1} = u_n + 2u_{n-1} = (odd) + (even) = odd. So by induction, all terms are odd.

(c) Ratios:
- u2u1=3\frac{u_2}{u_1} = 3
- u3u2=53≈1.67\frac{u_3}{u_2} = \frac{5}{3} \approx 1.67
- u4u3=115=2.2\frac{u_4}{u_3} = \frac{11}{5} = 2.2
- u5u4=2111≈1.91\frac{u_5}{u_4} = \frac{21}{11} \approx 1.91
- u6u5=4321≈2.05\frac{u_6}{u_5} = \frac{43}{21} \approx 2.05

The ratios appear to **oscillate around 2 and converge to 2**. (In fact un=13(2n+1+(−1)n)u_n = \frac{1}{3}(2^{n+1} + (-1)^n), so ratio → 2.)
11Problem 11
Answer
(a) 55 (b) 55 ✓ (c) 2870 (d) 2869
Full working
(a) 1+4+9+16+25=551 + 4 + 9 + 16 + 25 = 55.

(b) Formula: 5×6×116=3306=55\frac{5 \times 6 \times 11}{6} = \frac{330}{6} = 55 ✓.

(c) S20=20×21×416=172206=2870S_{20} = \frac{20 \times 21 \times 41}{6} = \frac{17220}{6} = \mathbf{2870}.

(d) 4+9+…+400=22+32+…+202=S20−12=2870−1=28694 + 9 + \ldots + 400 = 2^2 + 3^2 + \ldots + 20^2 = S_{20} - 1^2 = 2870 - 1 = \mathbf{2869}.
12Problem 12
Answer
(a) An=ϕn−1A_n = \phi^{n-1} (b) ≈29.0\approx 29.0 mm² (c) ≈74.4\approx 74.4 mm²
Full working
(a) Geometric with u1=1u_1 = 1, r=ϕr = \phi. So An=ϕn−1A_n = \phi^{n-1}.

(b) A8=ϕ7A_8 = \phi^7. Compute: ϕ2≈2.618\phi^2 \approx 2.618; ϕ4≈6.854\phi^4 \approx 6.854; ϕ7≈ϕ4⋅ϕ2⋅ϕ≈6.854×2.618×1.618≈29.0\phi^7 \approx \phi^4 \cdot \phi^2 \cdot \phi \approx 6.854 \times 2.618 \times 1.618 \approx 29.0. So A8≈29.0A_8 \approx \mathbf{29.0} mm².

(c) Sum of geometric series: S8=1(ϕ8−1)ϕ−1S_8 = \dfrac{1(\phi^8 - 1)}{\phi - 1}. With ϕ8≈46.98\phi^8 \approx 46.98 and ϕ−1≈0.618\phi - 1 \approx 0.618: S8≈45.980.618≈74.4S_8 \approx \dfrac{45.98}{0.618} \approx \mathbf{74.4} mm².

**Connection to Fibonacci/golden ratio:** the ratios Fn+1Fn\dfrac{F_{n+1}}{F_n} of consecutive Fibonacci numbers approach ϕ\phi, which is why nautilus shells, sunflower seed heads, and pine cones all exhibit Fibonacci-like spirals.