Predicting What Comes Next: Exploring Sequences and Progressions — Advanced Question Bank
A difficult, source-aligned practice resource covering number patterns, finite and infinite sequences, triangular and square numbers, term notation, explicit and recursive rules, Virahānka–Fibonacci patterns, arithmetic progressions, linear visualisation of APs, sum of the first n natural numbers, geometric progressions, fractals, Sierpiński patterns, GP graphs, bouncing-ball models and real-life growth applications.
Formula & Concept Revision
Explicit vs Recursive
Explicit: tₙ=f(n)
Recursive: term determined from earlier term(s).
Triangular Numbers
Tₙ=1+2+…+n=n(n+1)/2
Arithmetic Progression
tₙ=a+(n−1)d
Recursive: t₁=a, tₙ=tₙ₋₁+d
Geometric Progression
tₙ=arⁿ⁻¹
Recursive: t₁=a, tₙ=rtₙ₋₁
Virahānka–Fibonacci
V₁=1,V₂=2
Vₙ=Vₙ₋₁+Vₙ₋₂
Fractal GPs
Sierpiński triangle count=3ⁿ
black area=(3/4)ⁿ
carpet count=8ⁿ, area=(8/9)ⁿ
Visual Learning — HTML Canvas
JavaScript is limited to drawing these four diagrams. All answer accordions use native HTML <details>.
Section A — 30 MCQs
Conceptual, computational, visual and higher-order questions across the complete chapter.
A sequence is:
View answer / solution
Answer: B
The chapter defines a sequence as an ordered list of numbers, with each number called a term.
The nth triangular number is:
View answer / solution
Answer: B
A triangular number is the sum 1+2+…+n, giving n(n+1)/2.
The sum of the first n odd numbers is:
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Answer: C
The square-number pattern shows 1+3+5+…+(2n−1)=n².
Which statement correctly distinguishes explicit and recursive rules?
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Answer: C
An explicit formula uses n directly; a recursive formula relates a term to one or more earlier terms.
In the odd-number sequence uₙ=2n−1, the number 137 occurs at position:
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Answer: B
2n−1=137 gives 2n=138, so n=69.
For sₙ=5n−2, which number is NOT a term of the sequence?
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Answer: D
5n−2=471 gives n=473/5=94.6, not a natural-number position.
If u₁=1 and uₙ=2uₙ₋₁+3, then u₅ is:
View answer / solution
Answer: C
The sequence is 1,5,13,29,61,…
The next two terms after 1,2,3,5,8,13,21,34 in the Virahānka–Fibonacci sequence are:
View answer / solution
Answer: B
Each term is the sum of the previous two: 21+34=55 and 34+55=89.
An arithmetic progression is characterised by a constant:
View answer / solution
Answer: C
In an AP the common difference d is fixed.
The common difference of 11,7,3,−1,−5,… is:
View answer / solution
Answer: B
Each term is obtained by subtracting 4.
The nth term of an AP with first term a and common difference d is:
View answer / solution
Answer: B
The general AP is a,a+d,a+2d,…,a+(n−1)d.
When points (n,tₙ) of a non-constant AP are plotted, they lie on:
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Answer: C
The chapter visualises APs as linear patterns.
A taxi charges ₹200 fixed plus ₹40 per kilometre. The fare for 10 km is:
View answer / solution
Answer: D
Fare=200+40(10)=₹600.
An AP has t₁₁=38 and t₁₆=73. Its 31st term is:
View answer / solution
Answer: C
5d=35⇒d=7. Then a=38−70=−32 and t₃₁=−32+30×7=178.
An AP has t₃=16 and t₇−t₅=12. Its first term and common difference are:
View answer / solution
Answer: A
t₇−t₅=2d=12⇒d=6. Then a+2d=16⇒a=4.
The number of three-digit multiples of 7 is:
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Answer: C
They form the AP 105,112,…,994. Number of terms=(994−105)/7+1=128.
The number of multiples of 4 strictly between 10 and 250 is:
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Answer: C
The sequence is 12,16,…,248. Count=(248−12)/4+1=60.
1+2+3+…+n equals:
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Answer: C
Pairing the sum with its reverse gives 2S=n(n+1).
The smallest n for which 1+2+…+n>1000 is:
View answer / solution
Answer: B
T₄₄=990 and T₄₅=1035, so the least n is 45.
A geometric progression is characterised by a constant:
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Answer: B
Each term after the first is obtained by multiplying by the fixed common ratio r.
The nth term of 3,6,12,24,… is:
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Answer: C
a=3,r=2, so tₙ=3·2ⁿ⁻¹.
The sequence 1,−1,1,−1,… is a GP with common ratio:
View answer / solution
Answer: C
Every term is the previous term multiplied by −1.
Which term of 2,2√2,4,… is 128?
View answer / solution
Answer: C
Here r=√2. 2(√2)ⁿ⁻¹=128=2⁷, so (n−1)/2=6 and n=13.
Which term of 2,8,32,… is 131072?
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Answer: B
2·4ⁿ⁻¹=2¹⁷ gives 2n−1=17, hence n=9.
At Stage n of the Sierpiński triangle, the number of black triangles is:
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Answer: B
The number triples at each stage: 1,3,9,27,…
If Stage 0 of the Sierpiński triangle has black area 1, the black area at Stage n is:
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Answer: C
At each stage 3/4 of the previous black area remains.
For the Sierpiński square carpet, the number of retained red squares at Stage n follows:
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Answer: C
Each retained square generates 8 retained smaller squares at the next stage.
A ball is dropped from 80 m and rebounds to 60% of its previous height. Its height after the 5th bounce is:
View answer / solution
Answer: A
The bounce heights are 80(0.6)ⁿ; for n=5, 80(0.6)⁵=6.2208 m.
Unlike a non-constant AP, points (n,tₙ) of a typical GP:
View answer / solution
Answer: B
The chapter contrasts the straight-line AP plot with the curved GP plot.
The non-zero constant sequence 5,5,5,5,… is:
View answer / solution
Answer: C
It is an AP with d=0 and a GP with r=1.
Section B — 15 One-Mark Questions
Fast recall, formulas, term notation and short calculations.
Define a sequence.
View answer / solution
An ordered list of numbers in which each number is called a term.
What is an explicit rule?
View answer / solution
A rule that uses the position n directly to calculate the nth term.
What is a recursive rule?
View answer / solution
A rule that gives a term using one or more earlier terms.
Write the nth term of the odd-number sequence.
View answer / solution
tₙ=2n−1.
Write the nth triangular number.
View answer / solution
tₙ=n(n+1)/2.
Find the 17th triangular number.
View answer / solution
17×18/2=153.
Find the next term of 1,2,3,5,8,13,…
View answer / solution
21.
State the common difference of 2,5,8,11,…
View answer / solution
3.
Write the explicit nth-term formula for an AP.
View answer / solution
tₙ=a+(n−1)d.
Write the recursive form of an AP.
View answer / solution
t₁=a and tₙ=tₙ₋₁+d for n≥2.
Find 1+2+…+20.
View answer / solution
20×21/2=210.
State the common ratio of 3,30,300,3000,…
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10.
Write the nth-term formula for a GP.
View answer / solution
tₙ=arⁿ⁻¹.
Write a recursive rule for 3,30,300,3000,…
View answer / solution
t₁=3 and tₙ=10tₙ₋₁ for n≥2.
What is the number of black triangles at Stage 5 of the Sierpiński triangle?
View answer / solution
3⁵=243.
Section C — 15 Two-Mark Questions
Explicit/recursive rules, AP reconstruction, GP terms and applications.
Find the first five terms of tₙ=3n−4.
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For n=1,2,3,4,5: −1,2,5,8,11.
Find the first five terms of tₙ=2−5n.
View answer / solution
−3,−8,−13,−18,−23.
Find the first five terms of tₙ=n²−2n+3.
View answer / solution
2,3,6,11,18.
Determine whether 97 and 172 are terms of tₙ=5n−3.
View answer / solution
5n−3=97⇒n=20, so yes. 5n−3=172⇒n=35, so yes.
For t₁=−5, tₙ₊₁=tₙ+3, find the first five terms and decide whether 52 occurs.
View answer / solution
Terms: −5,−2,1,4,7,… Explicitly tₙ=−5+3(n−1)=3n−8. Set 3n−8=52⇒n=20, so 52 is the 20th term.
Given T₁=1,T₂=2,T₃=4 and Tₙ=Tₙ₋₁+Tₙ₋₂+Tₙ₋₃, find T₄ to T₈.
View answer / solution
T₄=7,T₅=13,T₆=24,T₇=44,T₈=81.
Find the 10th and 26th terms of the AP 3,8,13,18,…
View answer / solution
a=3,d=5. t₁₀=3+9×5=48; t₂₆=3+25×5=128.
For the AP 21,18,15,… find the term equal to −81 and decide whether 0 occurs.
View answer / solution
tₙ=21−3(n−1). For −81: n=35. For 0: n=8. Thus −81 is the 35th term and 0 is the 8th term.
An AP has 50 terms, t₃=12 and t₅₀=106. Find t₂₉.
View answer / solution
a+2d=12 and a+49d=106⇒47d=94⇒d=2,a=8. Thus t₂₉=8+28×2=64.
How many two-digit numbers are divisible by 3, and what is their sum?
View answer / solution
They are 12,15,…,99. Number=30. Sum=30(12+99)/2=1665.
Harish starts at ₹5,00,000 and gets ₹20,000 annual increments. When does the salary reach ₹7,00,000?
View answer / solution
5,00,000+20,000(n−1)=7,00,000 gives n=11. It is reached in the 11th salary year, i.e. after 10 increments.
A GP has r=2 and t₈=192. Find t₁₂.
View answer / solution
t₁₂=t₈·2⁴=192×16=3072.
Find the 10th and nth terms of 5,25,125,…
View answer / solution
a=5,r=5. Thus tₙ=5·5ⁿ⁻¹=5ⁿ, so t₁₀=5¹⁰=9,765,625.
For t₁=2,tₙ₊₁=3tₙ−2, which term is 730?
View answer / solution
Let vₙ=tₙ−1. Then vₙ₊₁=3vₙ and v₁=1, so tₙ=3ⁿ⁻¹+1. 730=3⁶+1, hence n=7.
Which term of 2,6,18,… is 4374? Give explicit and recursive rules.
View answer / solution
2·3ⁿ⁻¹=4374⇒3ⁿ⁻¹=2187=3⁷⇒n=8. Explicit: tₙ=2·3ⁿ⁻¹. Recursive: t₁=2,tₙ=3tₙ₋₁.
Section D — 15 Three-Mark Questions
Derivations, end-exercise themes, fractals and multi-step reasoning.
Derive the formula 1+2+…+n=n(n+1)/2 by pairing the sum with its reverse.
View answer / solution
Let S=1+2+…+n and also S=n+(n−1)+…+1. Adding termwise gives 2S=(n+1)+(n+1)+…+(n+1), with n copies. Thus 2S=n(n+1), so S=n(n+1)/2.
Find 25+26+…+58 using the first-n-natural-numbers formula.
View answer / solution
S₅₈−S₂₄=58·59/2−24·25/2=1711−300=1411.
Find all ways of writing 100 as a sum of consecutive positive natural numbers.
View answer / solution
For k consecutive terms starting at a: k(2a+k−1)/2=100. Testing divisors of 200 that give positive integer a yields k=1,a=100; k=5,a=18; k=8,a=9. Thus 100; 18+19+20+21+22; and 9+10+11+12+13+14+15+16.
How many three-digit numbers are divisible by 7?
View answer / solution
Smallest is 105 and largest is 994. They form an AP with d=7. If n terms, 994=105+(n−1)7⇒889=7(n−1)⇒n=128.
An AP satisfies t₄+t₈=24 and t₆+t₁₀=44. Find its first three terms.
View answer / solution
2a+10d=24⇒a+5d=12. Also 2a+14d=44⇒a+7d=22. Subtract:2d=10⇒d=5, then a=−13. First three terms: −13,−8,−3.
Find all GPs whose first two terms sum to −4 and whose fifth term is four times the third term.
View answer / solution
Let first term a and ratio r. ar⁴=4ar². For non-zero a,r, r²=4⇒r=±2. Also a(1+r)=−4. If r=2,a=−4/3. If r=−2,a=4. Thus both GPs are possible.
The sum of the first three terms of a GP is 13/12 and their product is −1. Find the terms.
View answer / solution
Let the three terms be a,ar,ar². Their product=(ar)³=−1, so ar=−1. Hence terms are −1/r,−1,−r and r+1+1/r=−13/12. This gives 12r²+25r+12=0, so r=−3/4 or −4/3. The terms are 4/3,−1,3/4 or the reverse.
The first three terms of a GP sum to 26 and the sum of their squares is 364. Find the terms.
View answer / solution
Let terms be x,y,z with y²=xz. From (x+y+z)²=364+2(xy+yz+xz), we get xy+yz+xz=156. But xy+yz+xz=y(x+z)+y²=y(x+y+z)=26y. Hence y=6. Then x+z=20 and xz=36, giving x,z=2,18. Terms: 2,6,18 or 18,6,2.
Prove that if the 4th,10th,16th terms of a GP are x,y,z, then x,y,z are in GP.
View answer / solution
For GP tₙ=arⁿ⁻¹: x=ar³,y=ar⁹,z=ar¹⁵. Then y²=a²r¹⁸=xz. Hence y/x=z/y wherever the ratios are defined, so x,y,z form a GP.
For P₁=1,P₂=2 and Pₙ=P₁+…+Pₙ₋₁+1 (n>2), find P₁ to P₈ and a simpler rule.
View answer / solution
P₃=4,P₄=8,P₅=16, so P₁…P₈=1,2,4,8,16,32,64,128. Since Pₙ equals twice Pₙ₋₁ for n≥2, the simpler recurrence is Pₙ=2Pₙ₋₁; explicit rule Pₙ=2ⁿ⁻¹.
For W₁=1,W₂=2 and Wₙ=W₁+…+Wₙ₋₂+2, find W₁ to W₈ and identify the sequence.
View answer / solution
The terms are 1,2,3,5,8,13,21,34. Subtracting consecutive defining sums shows Wₙ=Wₙ₋₁+Wₙ₋₂. It is the Virahānka–Fibonacci sequence used in the chapter.
A bacteria culture starts with 30 bacteria and doubles every hour. Find the counts after 2 hours,4 hours and n hours.
View answer / solution
Treat the starting count as hour 0. After n hours the count is 30·2ⁿ. Thus after 2 hours:120; after 4 hours:480; after n hours:30·2ⁿ.
In the Sierpiński triangle, derive formulas for the number of black triangles and total black area at Stage n.
View answer / solution
Each stage triples the number of black triangles, starting at 1, so Nₙ=3ⁿ. Each stage retains 3/4 of the previous total black area, starting at 1, so Aₙ=(3/4)ⁿ.
In the Sierpiński square carpet, obtain formulas for retained-square count and red area at Stage n.
View answer / solution
Each retained square produces 8 retained squares, so Nₙ=8ⁿ from Stage 0. Each stage retains 8/9 of the previous area, hence Aₙ=(8/9)ⁿ when Stage-0 area is 1.
A ball is dropped from 80 m and rebounds to 60% each time. Find its fifth-bounce height and the total vertical distance by the sixth ground hit.
View answer / solution
Bounce heights:48,28.8,17.28,10.368,6.2208,… so fifth height=6.2208 m. By the 6th ground hit, distance=80+2(48+28.8+17.28+10.368+6.2208)=301.3376 m.
Section E — 10 Four-Mark Questions
Advanced AP/GP reconstruction, proofs, fractals and application modelling.
Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
View answer / solution
Let tₙ=a+(n−1)d. Then a+10d=38 and a+15d=73. Subtract:5d=35⇒d=7. Hence a=38−70=−32. Therefore t₃₁=−32+30×7=178.
Determine the AP whose third term is 16 and whose seventh term exceeds the fifth term by 12.
View answer / solution
t₃=a+2d=16. Also t₇−t₅=(a+6d)−(a+4d)=2d=12, so d=6. Then a=16−12=4. AP:4,10,16,22,28,…
Find all possible ways of expressing 100 as a sum of consecutive natural numbers.
View answer / solution
Let there be k terms beginning with a≥1. Then 100=k[2a+k−1]/2, so k(2a+k−1)=200. For positive integer a, valid divisor lengths are k=1,5,8. They give a=100,18,9 respectively. Hence 100; 18+19+20+21+22; and 9+10+11+12+13+14+15+16.
Find a GP for which the sum of the first two terms is −4 and the fifth term is four times the third term.
View answer / solution
Let the GP be a,ar,ar²,… . From ar⁴=4ar², for a,r≠0, r²=4, so r=2 or −2. From a(1+r)=−4: if r=2, a=−4/3; if r=−2, a=4. Thus two GPs satisfy the conditions.
The sum of the first three terms of a GP is 13/12 and their product is −1. Find the common ratio and terms.
View answer / solution
The middle term ar satisfies (ar)³=−1, so ar=−1. Terms are −1/r,−1,−r. Their sum gives r+1+1/r=−13/12. Hence 12r²+25r+12=0=(3r+4)(4r+3). So r=−4/3 or −3/4; the terms are 3/4,−1,4/3 or 4/3,−1,3/4.
The sum of the first three terms of a GP is 26 and the sum of their squares is 364. Find the GP.
View answer / solution
Let terms x,y,z in GP, so y²=xz. The identity 26²=364+2(xy+yz+xz) gives xy+yz+xz=156. But xy+yz+xz=y(x+z)+xz=y(x+z)+y²=y(x+y+z)=26y, so y=6. Thus x+z=20 and xz=36. Therefore x,z are 2 and18, giving 2,6,18 or 18,6,2.
If the 4th,10th and16th terms of a GP are x,y,z, prove x,y,z themselves form a GP.
View answer / solution
Write tₙ=arⁿ⁻¹. Then x=ar³,y=ar⁹,z=ar¹⁵. We have y²=a²r¹⁸ and xz=(ar³)(ar¹⁵)=a²r¹⁸. Therefore y²=xz, the characteristic relation for three consecutive GP terms; equivalently y/x=z/y when non-zero.
For the Sierpiński triangle, suppose Stage 0 has area 1. Find explicit and recursive formulas for both the number of black triangles and total black area.
View answer / solution
Count: N₀=1 and each stage triples, so Nₙ=3ⁿ; recursively Nₙ=3Nₙ₋₁. Total area: A₀=1 and each stage retains 3/4, so Aₙ=(3/4)ⁿ; recursively Aₙ=(3/4)Aₙ₋₁. Thus count grows rapidly while total black area decreases toward 0.
For the Sierpiński square carpet, derive corresponding count and area formulas.
View answer / solution
Stage 0 has one red square. Each retained square is replaced by 8 retained squares, so Nₙ=8ⁿ and Nₙ=8Nₙ₋₁. Each step retains 8/9 of the area, so Aₙ=(8/9)ⁿ and Aₙ=(8/9)Aₙ₋₁, with N₀=A₀=1.
A ball is dropped from 80 m and rebounds to 60% of the previous height. Find its height after the fifth bounce and total distance by the sixth ground hit.
View answer / solution
The bounce heights form the GP 48,28.8,17.28,10.368,6.2208,…, so fifth-bounce height=6.2208 m. The initial fall is 80 m. Before the sixth ground hit there are five complete up-and-down bounce journeys, so total distance=80+2(48+28.8+17.28+10.368+6.2208)=301.3376 m.
Section F — 5 Case Studies
Taxi fares, visual APs, Sierpiński patterns, bouncing balls and bacteria growth.
Case Study 1 — Taxi Fare as an AP
A taxi company charges a fixed booking fee of ₹200 plus ₹40 for every kilometre travelled.
- a) Write the fares for 1 km,2 km and3 km.
- b) Why do these fares form an AP?
- c) Write the nth-term formula.
- d) Find the fare for 25 km.
View case-study solutions
a) Write the fares for 1 km,2 km and3 km. ₹240,₹280,₹320.
b) Why do these fares form an AP? The difference between consecutive fares is constantly ₹40.
c) Write the nth-term formula. tₙ=200+40n.
d) Find the fare for 25 km. ₹200+₹40×25=₹1200.
Case Study 2 — Growing Square Pattern
A visual pattern has 1 tiny square at Stage 1 and adds 4 squares at every new stage, producing 1,5,9,13,…
- a) Find the common difference.
- b) Write the nth term.
- c) Find Stage 20.
- d) What shape do the plotted points (n,tₙ) form?
View case-study solutions
a) Find the common difference. 4.
b) Write the nth term. tₙ=1+4(n−1)=4n−3.
c) Find Stage 20. 77 squares.
d) What shape do the plotted points (n,tₙ) form? A straight-line pattern.
Case Study 3 — Sierpiński Triangle
At Stage 0 there is one black equilateral triangle of area 1. At every stage, each black triangle is replaced by three smaller black triangles.
- a) Find the black-triangle counts at Stages 0–4.
- b) Write the explicit count rule.
- c) Write the total black-area rule.
- d) Explain the contrasting behaviour.
View case-study solutions
a) Find the black-triangle counts at Stages 0–4. 1,3,9,27,81.
b) Write the explicit count rule. Nₙ=3ⁿ.
c) Write the total black-area rule. Aₙ=(3/4)ⁿ.
d) Explain the contrasting behaviour. The number of pieces grows geometrically while the retained total area decreases geometrically.
Case Study 4 — Bouncing Ball
A ball is dropped from 24 ft. Every bounce reaches 3/4 of the previous height.
- a) Find the first three bounce heights.
- b) State a and r for the bounce-height GP.
- c) Write the nth bounce-height formula.
- d) Why is its graph not a straight line?
View case-study solutions
a) Find the first three bounce heights. 18 ft,13.5 ft,10.125 ft.
b) State a and r for the bounce-height GP. a=18,r=3/4.
c) Write the nth bounce-height formula. hₙ=18(3/4)ⁿ⁻¹.
d) Why is its graph not a straight line? The heights change by a constant ratio, not a constant difference.
Case Study 5 — Bacteria Growth
A culture begins with 30 bacteria and the population doubles every hour.
- a) Write the counts at hours 0,1,2,3.
- b) Identify the progression.
- c) Write the count after n hours.
- d) Find the count after 8 hours.
View case-study solutions
a) Write the counts at hours 0,1,2,3. 30,60,120,240.
b) Identify the progression. A GP with ratio 2.
c) Write the count after n hours. 30·2ⁿ.
d) Find the count after 8 hours. 30×256=7680.
Special Exemplar Challenge — 15 Questions
Non-routine term finding, reconstruction, patterns, AP/GP reasoning and exact calculations.
An AP has t₇=19 and t₁₅=51. Find a,d and t₃₀.
View worked solution
8d=32⇒d=4. a+6d=19⇒a=−5. t₃₀=−5+29×4=111.
An AP has t₄=7 and t₁₂=−17. Find the nth term.
View worked solution
8d=−24⇒d=−3. a+3d=7⇒a=16. Thus tₙ=16−3(n−1)=19−3n.
How many three-digit numbers are divisible by 7?
View worked solution
From 105 to994 with d=7: n=(994−105)/7+1=128.
Find the smallest n for which the nth triangular number exceeds 1000.
View worked solution
n(n+1)/2>1000. T₄₄=990,T₄₅=1035, so n=45.
Is 1225 a triangular number? If yes, find its position.
View worked solution
Solve n(n+1)/2=1225⇒n(n+1)=2450=49×50, so n=49.
Find the sum of all integers from 37 to 83.
View worked solution
S₈₃−S₃₆=83×84/2−36×37/2=3486−666=2820.
A GP has t₅=48 and t₈=384. Find r and t₁.
View worked solution
t₈/t₅=r³=8⇒r=2. Then ar⁴=48⇒16a=48⇒a=3.
Which term of 2,2√2,4,… is 128?
View worked solution
a=2,r=√2. 2(√2)ⁿ⁻¹=128⇒2^{1+(n−1)/2}=2⁷⇒n=13.
For t₁=2,tₙ₊₁=3tₙ−2, derive an explicit formula.
View worked solution
Let vₙ=tₙ−1. Then vₙ₊₁=3vₙ and v₁=1, so vₙ=3ⁿ⁻¹ and tₙ=3ⁿ⁻¹+1.
A GP has first three terms x,y,z. Show that y²=xz.
View worked solution
Write x=a,y=ar,z=ar². Then y²=a²r²=xz.
Find the sum of the first 50 odd numbers using the square-number pattern.
View worked solution
1+3+…+99=50²=2500.
An AP has first term −8 and common difference 5. Which term is 102?
View worked solution
−8+5(n−1)=102⇒5(n−1)=110⇒n=23.
A non-zero GP has t₄=54 and t₇=1458. Find r.
View worked solution
t₇/t₄=r³=1458/54=27, so r=3.
The points (1,5),(2,9),(3,13),(4,17) are plotted. Identify the progression and write its rule.
View worked solution
The y-values form AP 5,9,13,17 with d=4. tₙ=5+4(n−1)=4n+1.
A stage pattern has 1,8,64,512 retained pieces. Write explicit and recursive rules.
View worked solution
This is a GP with a=1,r=8. Explicit Nₙ=8ⁿ if Stage 0 corresponds to exponent 0; recursive N₀=1,Nₙ=8Nₙ₋₁.
Extra Olympiad / HOTS Challenge — 20 Questions
Proofs, structural properties, recursion, consecutive sums, fractals and deeper progression reasoning.
In an AP, tₘ=tₙ for two different positions m≠n. Prove the AP is constant.
View worked solution
a+(m−1)d=a+(n−1)d⇒(m−n)d=0. Since m≠n,d=0, so every term equals a.
In an AP, tₚ=q and t_q=p with p≠q. Find tₚ₊q.
View worked solution
Subtracting gives (p−q)d=q−p, so d=−1. From a+(p−1)(−1)=q, a=p+q−1. Thus tₚ₊q=a+(p+q−1)(−1)=0.
Prove that for an AP, tₘ+tₙ=2t_(m+n)/2 whenever m+n is even.
View worked solution
tₘ+tₙ=2a+(m+n−2)d. The middle-index term is a+[(m+n)/2−1]d; twice it is exactly the same expression.
In a finite AP with first term a and last term l, prove that terms equidistant from the ends have constant sum a+l.
View worked solution
If the AP has N terms, t_k=a+(k−1)d and t_{N+1−k}=a+(N−k)d. Their sum=2a+(N−1)d=a+l.
Find all positive-integer lengths k for which 100 can be expressed as a sum of k consecutive positive integers.
View worked solution
Using 100=k(2a+k−1)/2, valid positive solutions occur for k=1,5,8 only.
Three numbers x,y,z are consecutive terms of an AP. Prove 2y=x+z and state the converse.
View worked solution
If x=a,y=a+d,z=a+2d, then x+z=2a+2d=2y. Conversely, if 2y=x+z, then y−x=z−y, so the three numbers are in AP.
Three non-zero numbers x,y,z are consecutive terms of a GP. Prove y²=xz and state the converse.
View worked solution
For x=a,y=ar,z=ar², y²=a²r²=xz. Conversely, if y²=xz and x,y are non-zero, y/x=z/y, so the ratios are equal.
If t₄,t₁₀,t₁₆ of a GP are x,y,z, prove y is the geometric mean of x and z.
View worked solution
t₄=ar³,t₁₀=ar⁹,t₁₆=ar¹⁵; hence y²=a²r¹⁸=xz.
A GP has first three terms whose sum is 26 and sum of squares is 364. Find them without trial.
View worked solution
Let terms x,y,z with y²=xz. From 26²−364=2(xy+yz+xz), get xy+yz+xz=156. But this equals y(x+y+z)=26y, so y=6. Then x+z=20,xz=36⇒{x,z}={2,18}.
A GP has first three terms with sum 13/12 and product −1. Find all possible ordered triples.
View worked solution
Middle term cubed equals product, so y=−1. Then xz=1 and x+z=25/12? Since x+y+z=13/12, x+z=25/12. Thus x,z are roots of 12u²−25u+12=0:4/3 and3/4. Triples: (4/3,−1,3/4) and reverse.
For P₁=1,P₂=2 and Pₙ=P₁+…+Pₙ₋₁+1, prove Pₙ=2ⁿ⁻¹.
View worked solution
From the definition, Pₙ−Pₙ₋₁=Pₙ₋₁ for n≥3, so Pₙ=2Pₙ₋₁. With P₁=1, induction gives Pₙ=2ⁿ⁻¹.
For W₁=1,W₂=2 and Wₙ=W₁+…+Wₙ₋₂+2, prove Wₙ follows the Virahānka recurrence.
View worked solution
For n≥4, subtract definitions of Wₙ and Wₙ₋₁: Wₙ−Wₙ₋₁=Wₙ₋₂. Hence Wₙ=Wₙ₋₁+Wₙ₋₂. Directly W₃=3, so the recurrence holds from the start.
In the Sierpiński triangle, find the area of one individual black triangle at Stage n.
View worked solution
The original triangle is divided by factor 4 in area at each stage. Each individual Stage-n black triangle has area (1/4)ⁿ of the original. There are 3ⁿ of them, giving total black area (3/4)ⁿ.
For the Sierpiński square carpet, prove the red area at Stage n is (8/9)ⁿ.
View worked solution
Each retained square is split into 9 equal squares and 8 remain. Thus every stage multiplies total red area by 8/9. Starting from area1 gives (8/9)ⁿ.
A bacteria population begins at30 and doubles hourly. Find the least integer n for which it exceeds100000.
View worked solution
Need 30·2ⁿ>100000. 2¹¹=2048 gives61440; 2¹²=4096 gives122880. Least n=12 hours.
A 24-ft ball rebounds to 3/4 each time. Find the first bounce number whose height is below 4 ft.
View worked solution
Heights are18(3/4)ⁿ⁻¹. Checking: 6th≈4.271 ft, 7th≈3.203 ft. Thus the 7th bounce is the first below4 ft.
Convert the recurrence tₙ₊₁=3tₙ−2,t₁=2 into an explicit rule by a change of variable.
View worked solution
The fixed point is1. Let vₙ=tₙ−1. Then vₙ₊₁=3vₙ,v₁=1, so vₙ=3ⁿ⁻¹ and tₙ=3ⁿ⁻¹+1.
Show that the triangular-number sequence has first differences 2,3,4,… and constant second difference1.
View worked solution
Tₙ=n(n+1)/2. Tₙ₊₁−Tₙ=n+1, so first differences rise by1; therefore successive first differences have difference1.
A sequence is both an AP with common difference d and a GP with non-zero terms and ratio r. Show that if it is non-constant this is impossible.
View worked solution
For consecutive terms a,a+d,a+2d, GP condition gives (a+d)²=a(a+2d), so d²=0. Thus d=0; hence any non-zero sequence that is both AP and GP must be constant.
An AP has integer first term and integer common difference. If one term is divisible by m and d is also divisible by m, prove every term has the same remainder modulo m.
View worked solution
tₙ=a+(n−1)d. Since m divides d, (n−1)d contributes remainder0 modulo m, so every tₙ has the same remainder as a. In particular, if one term is divisible by m, all are.
High-Level Thinking Laboratory — 10 Questions
Students analyse definitions, ambiguity, graph behaviour, recursion and growth versus decay.
Why can the first few terms of a sequence fail to determine a unique 'next term'?
View worked solution
Different rules can generate the same initial terms but diverge later. A prediction is meaningful only after the intended pattern or rule is specified.
Why must n in an nth-term rule represent an allowed position such as a natural number?
View worked solution
Positions are discrete labels 1,2,3,…; a solution such as n=94.6 does not correspond to a term position.
Why is the prime-number list still a sequence even though the chapter says there is no simple regular pattern like an AP or GP?
View worked solution
A sequence needs an ordering, not necessarily a constant difference, constant ratio or simple elementary formula.
When is an explicit rule more useful than a recursive rule?
View worked solution
When a distant term such as the 1000th is needed directly; an explicit rule avoids computing all preceding terms.
When can a recursive rule be more natural than an explicit rule?
View worked solution
When the defining pattern intrinsically depends on earlier terms, as in the Virahānka–Fibonacci sequence.
Why do AP points form a straight line in an (n,tₙ) graph?
View worked solution
tₙ=a+(n−1)d=(a−d)+dn is linear in n, so equal horizontal steps produce equal vertical changes.
Why do typical GP points not form a straight line?
View worked solution
tₙ=arⁿ⁻¹ changes multiplicatively rather than by a constant additive amount; its successive vertical differences are generally not constant.
How can the Sierpiński triangle have more and more black pieces but less and less total black area?
View worked solution
The number of pieces is multiplied by3, but each individual piece becomes 1/4 as large in area; total area is therefore multiplied by3/4.
Why is the non-zero constant sequence both an AP and a GP?
View worked solution
Its consecutive differences are all0 and its consecutive ratios are all1, satisfying both definitions.
Why is the formula for triangular numbers identical to the formula for the sum of the first n natural numbers?
View worked solution
The nth triangular number is defined by arranging or adding 1+2+…+n objects, so the two quantities are literally the same number viewed geometrically and arithmetically.
Exam Strategy & Common Traps
| Topic | Best first move | Common trap |
|---|---|---|
| Sequence rule | Identify exactly what changes with position n. | Assuming the first few terms force a unique rule. |
| Term membership | Set tₙ equal to the given number and solve for n. | Accepting a non-integer n as a valid position. |
| AP | Compute consecutive differences. | Using a ratio instead of a difference. |
| GP | Compute consecutive ratios, checking signs carefully. | Ignoring a negative common ratio. |
| AP reconstruction | Write tₙ=a+(n−1)d for each condition. | Using nd instead of (n−1)d. |
| Natural-number sums | Use n(n+1)/2 or subtract two triangular sums. | Off-by-one errors in consecutive ranges. |
| Recursive rules | Write the initial term(s) as well as the recurrence. | Giving a recurrence with no starting value. |
| Fractals | Separate ‘number of pieces’ from ‘total retained area’. | Assuming both quantities have the same common ratio. |
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