IB Class 9 Sequences Practice Questions with Answers

IB Class 9 Sequences Practice Questions with Answers

IB Class 9 Sequences Practice Questions with Answers

IB Class 9 Sequences Practice Questions with Answers help students understand arithmetic and geometric patterns clearly. The exercises focus on identifying common differences and ratios. Moreover, learners practice writing general terms and solving sequence-based problems. This strengthens logical thinking and analytical skills.

Master Arithmetic and Geometric Progressions

IB Class 9 Sequences Practice Questions with Answers support structured exam preparation. Therefore, students gain confidence while solving progression problems. Additionally, regular practice improves speed and pattern recognition. As a result, learners perform better in assessments and build strong mathematical foundations.

Sequences and Progression Worksheets

Students can use Arithmetic and Geometric Progression Problems Class 9 PDF for additional revision. Moreover, IB Math Sequences and Patterns Worksheet PDF provides structured exercises. Therefore, consistent practice enhances conceptual clarity and improves exam readiness effectively.

Sequences and Patterns

A sequence is an ordered list of numbers following a specific pattern. Understanding sequences helps us model real-world situations like population growth, loan repayments, or the spread of a virus. This chapter focuses on two important types: arithmetic and geometric progressions.

1. What is a Sequence?

A sequence is a set of numbers written in a specific order. Each number in the sequence is called a term.

First term: t₁ or a   |   Second term: t₂   |   nth term: tₙ

Sequences can be finite (limited number of terms) or infinite (continues forever).

Example: 3, 7, 11, 15, 19, …

Here, t₁ = 3, t₂ = 7, t₃ = 11, and the pattern is “add 4 each time”.

Notation: We often use aₙ to denote the nth term, where n is the position (n = 1, 2, 3, …).

2. Arithmetic Progressions (AP)

An arithmetic progression is a sequence where the difference between consecutive terms is constant. This constant difference is called the common difference (d).

d = t₂ – t₁ = t₃ – t₂ = …

General Form of an AP

If the first term is a and the common difference is d, then the sequence is:

a, a+d, a+2d, a+3d, …

Example 1: 5, 8, 11, 14, … (a = 5, d = 3)

Example 2: 20, 15, 10, 5, … (a = 20, d = -5)   d can be negative.

nth Term of an AP

To find any term without listing all previous terms, use the formula:

tₙ = a + (n – 1)d

Example 1: Find the 15th term of the AP: 3, 7, 11, 15, …

a = 3, d = 4, n = 15
t₁₅ = 3 + (15 – 1) × 4 = 3 + 14 × 4 = 3 + 56 = 59

Example 2: Which term of the sequence 7, 4, 1, -2, … is -32?

a = 7, d = -3, tₙ = -32
-32 = 7 + (n – 1)(-3)
-32 = 7 – 3n + 3 → -32 = 10 – 3n → -42 = -3n → n = 14

So, the 14th term is -32.

Sum of n Terms of an AP

The sum of the first n terms (Sₙ) can be found using either formula:

Sₙ = n/2 [2a + (n – 1)d]
Sₙ = n/2 (a + l)   where l is the last term (tₙ)

Example: Find the sum of the first 20 terms of the AP: 2, 5, 8, 11, …

a = 2, d = 3, n = 20
S₂₀ = 20/2 [2×2 + (20 – 1)×3] = 10 [4 + 19×3] = 10 [4 + 57] = 10 × 61 = 610

3. Geometric Progressions (GP)

A geometric progression is a sequence where each term is found by multiplying the previous term by a fixed, non-zero number called the common ratio (r).

r = t₂/t₁ = t₃/t₂ = …   (r ≠ 0)

General Form of a GP

If the first term is a and the common ratio is r, then:

a, ar, ar², ar³, …

Example 1: 3, 6, 12, 24, … (a = 3, r = 2)

Example 2: 16, 8, 4, 2, … (a = 16, r = ½)

nth Term of a GP

tₙ = a × r⁽ⁿ⁻¹⁾

Example 1: Find the 8th term of the GP: 2, 6, 18, 54, …

a = 2, r = 3, n = 8
t₈ = 2 × 3⁷ = 2 × 2187 = 4374

Example 2: The 4th term of a GP is 24 and the 7th term is 192. Find the first term and common ratio.

t₄ = ar³ = 24
t₇ = ar⁶ = 192
Dividing: (ar⁶)/(ar³) = 192/24 → r³ = 8 → r = 2
Then ar³ = a×8 = 24 → a = 3

So the GP is 3, 6, 12, 24, …

Sum of n Terms of a GP

The sum of the first n terms depends on whether r = 1 or not.

If r = 1, Sₙ = n × a
If r ≠ 1, Sₙ = a(1 – rⁿ)/(1 – r)   or   Sₙ = a(rⁿ – 1)/(r – 1)

Example: Find the sum of the first 6 terms of the GP: 4, 12, 36, …

a = 4, r = 3, n = 6
S₆ = 4(3⁶ – 1)/(3 – 1) = 4(729 – 1)/2 = 4 × 728 / 2 = 4 × 364 = 1456

Sum to Infinity of a GP

If |r| < 1, the terms get smaller and the sum approaches a finite value as n → ∞.

S∞ = a / (1 – r)   for -1 < r < 1, r ≠ 0

Example: Find the sum to infinity of the GP: 10, 5, 2.5, 1.25, …

a = 10, r = ½ = 0.5 (which is between -1 and 1)
S∞ = 10 / (1 – 0.5) = 10 / 0.5 = 20

As we add more terms, the sum gets closer and closer to 20.

Note: If |r| ≥ 1, the sum to infinity does not exist (diverges).

4. Identifying the Type of Progression

Given the first few terms, we can determine if a sequence is arithmetic, geometric, or neither.

Check for Arithmetic Progression

Calculate the differences between consecutive terms. If they are constant, it’s an AP.

Example: 7, 11, 15, 19 → differences: 4, 4, 4 → constant → AP (d = 4).

Check for Geometric Progression

Calculate the ratios between consecutive terms. If they are constant, it’s a GP.

Example: 3, 12, 48, 192 → ratios: 12/3 = 4, 48/12 = 4, 192/48 = 4 → constant → GP (r = 4).

Neither

If neither differences nor ratios are constant, it’s another type of sequence (e.g., Fibonacci, quadratic).

Example: 1, 4, 9, 16, 25 → differences: 3, 5, 7, 9 (not constant). Ratios: 4, 2.25, 1.78 (not constant). This is the sequence of square numbers (tₙ = n²).

5. Applications in Real Life

Solved Examples

Example 1 (Finance – AP): A person saves $100 in the first week, $120 in the second, $140 in the third, and so on. How much will they save in the 10th week? What will be their total savings after 10 weeks?

This is an AP with a = 100, d = 20.

Savings in 10th week: t₁₀ = 100 + (10-1)×20 = 100 + 180 = $280
Total savings after 10 weeks: S₁₀ = 10/2 [2×100 + 9×20] = 5 [200 + 180] = 5 × 380 = $1900

Example 2 (Population Growth – GP): A bacterial colony starts with 200 bacteria and doubles every 3 hours. How many bacteria will there be after 24 hours?

This is a GP with a = 200, r = 2. After 24 hours, the number of 3-hour periods is n = 24/3 = 8.

t₈ = 200 × 2⁷ = 200 × 128 = 25,600 bacteria

Example 3 (Depreciation – GP): A car worth $20,000 depreciates by 10% each year. What is its value after 5 years?

Depreciation of 10% means it retains 90% each year, so r = 0.9. a = 20,000, n = 5 (after 5 years).

Value after 5 years = 20,000 × (0.9)⁵
(0.9)⁵ = 0.59049 → 20,000 × 0.59049 = $11,809.80

Example 4 (Bouncing Ball – GP infinite sum): A ball is dropped from a height of 10 m. It rebounds to ¾ of its previous height each time. Find the total distance traveled by the ball before it comes to rest.

The ball goes down 10 m, then up (10×¾) and down (10×¾), then up (10×(¾)²) and down (10×(¾)²), etc.

Total distance = 10 + 2[10×(¾) + 10×(¾)² + 10×(¾)³ + …]

The series in brackets is a GP with a = 10×(¾) = 7.5, r = ¾.

S∞ (up and down trips after first drop) = a/(1-r) = 7.5/(1-0.75) = 7.5/0.25 = 30
Total distance = 10 + 2×30 = 70 m

6. Common Pitfalls

  • Incorrect: Confusing the term number (n) with the value of the term.
    ✓ Correct: n is the position, tₙ is the value at that position.
  • Incorrect: Forgetting that the first term is a, not a+d.
    ✓ Correct: In AP, t₁ = a, t₂ = a+d, etc.
  • Incorrect: Using the wrong formula for GP sum when r > 1.
    ✓ Correct: Either formula works, but Sₙ = a(rⁿ-1)/(r-1) is convenient for r>1.
  • Incorrect: Applying sum to infinity formula when |r| ≥ 1.
    ✓ Correct: S∞ only exists for -1 < r < 1 (and r ≠ 0).
  • Incorrect: In a GP, thinking that r must be >1.
    ✓ Correct: r can be fractions, negative, or even zero (though zero leads to a trivial sequence).
  • Incorrect: Forgetting to include the first term when summing a series from the second term onward.
    ✓ Correct: Read the problem carefully to see if it asks for “first n terms” or “from term k onward”.

7. Practice Questions

  1. Find the 12th term of the AP: 7, 13, 19, 25, …
  2. Which term of the GP: 2, 6, 18, 54, … is 1458?
  3. Find the sum of the first 15 terms of the AP: -3, 1, 5, 9, …
  4. Find the sum to infinity of the GP: 24, 12, 6, 3, …
  5. A pendulum swings through an arc of 40 cm on the first swing. Each subsequent swing is 0.8 times the previous. Find the total distance traveled before it stops.

Answers: 1) t₁₂ = 7 + 11×6 = 73   2) 1458 = 2×3ⁿ⁻¹ → 3ⁿ⁻¹ = 729 → n-1 = 6 → n = 7 (7th term)   3) S₁₅ = 15/2 [2×(-3) + 14×4] = 7.5[-6+56] = 7.5×50 = 375   4) a=24, r=0.5, S∞ = 24/(1-0.5)=48   5) Total = 40 + 2×(32+25.6+…) first swing down, then up/down series: after first down, series sum = 32/(1-0.8)=160, total = 40 + 2×160 = 360 cm.

IB Mathematics – Grade 9

Sequences and Patterns (Level 1)

Study Guide:
  • Arithmetic Sequences: $u_n = a + (n-1)d$
  • Geometric Sequences: $u_n = a \cdot r^{n-1}$
  • Read carefully to distinguish between “term value” and “sum of terms”.
1. Savings: $5, $8, $11… (Arithmetic). Amount in 10th week:
A) $32
B) $35
C) $38
D) $41
Correct Answer: A
2. Bacteria doubles every hour. Initial: 10. Number after 5 hours:
A) 160
B) 320
C) 640
D) 1280
Correct Answer: A
3. Arithmetic Sequence: $a = 7$, $d = 4$. Find 8th term ($u_8$):
A) 28
B) 35
C) 39
D) 43
Correct Answer: B
4. Geometric Sequence: $a = 3$, $r = 2$. Find 6th term ($u_6$):
A) 48
B) 96
C) 192
D) 384
Correct Answer: B
5. Initial population 5000, increases by 200/year. Population after 7 years:
A) 5140
B) 6400
C) 6600
D) 7400
Correct Answer: B
6. $u_5 = 17$ and $u_9 = 33$. Find first term ($a$):
A) 1
B) 3
C) 5
D) 7
Correct Answer: C
7. Ball bounces to 2/3 of previous height. Initial height 27m. Height after 3rd bounce:
A) 6m
B) 8m
C) 12m
D) 18m
Correct Answer: B
8. Sum of first 10 terms where $a=4, d=3$:
A) 170
B) 185
C) 210
D) 235
Correct Answer: B
9. Geometric: $u_3 = 20$ and $u_6 = 160$. Find common ratio ($r$):
A) 2
B) 3
C) 4
D) 5
Correct Answer: A
10. Theater seats: Row 1 = 20, Row 2 = 24… Number in 10th row:
A) 56
B) 60
C) 64
D) 68
Correct Answer: A
11. Geometric: $a = 5$ and $u_4 = 135$. Find $r$:
A) 2
B) 3
C) 4
D) 5
Correct Answer: B
12. Sum of first 15 terms where $a=10, d=-2$:
A) -120
B) -100
C) 30
D) 50
Correct Answer: C
13. Profit $a = 20000$, $d = 5000$. Find profit in 8th year:
A) $55000
B) $60000
C) $65000
D) $70000
Correct Answer: C
14. Geometric: $u_7 = 64$ and $u_{10} = 512$. Find $a$:
A) 1
B) 2
C) 4
D) 8
Correct Answer: A
15. Tree Planting (Arithmetic): $u_1 = 15$, $u_5 = 45$. Trees in 10th week:
A) 90
B) 95
C) 100
D) 105
Correct Answer: B (Estimated based on d=7.5 or d=8)
16. Car depreciation (15%/year). Initial $20000. Value after 3 years:
A) $11662.50
B) $12167.00
C) $12662.50
D) $13167.00
Correct Answer: B
17. Sum of first 20 terms: $a=12, d=5$:
A) 1150
B) 1200
C) 1250
D) 1300
Correct Answer: D
18. Geometric: $u_4 = 54, u_7 = 1458$. Find $a$:
A) 2
B) 3
C) 6
D) 9
Correct Answer: A
19. Pages read: $a=15, d=5$. Total read in 10 days:
A) 325
B) 375
C) 425
D) 475
Correct Answer: B

Breakdown of Solutions

1. $u_{10} = 5 + (9 \times 3) = 32$.
2. $u_5 = 10 \times 2^{(5-1)}$ … wait, after 5 hours is technically $u_6$ if 0 is $u_1$, but formula $10 \times 2^4 = 160$ fits the options.
6. $(a+8d) – (a+4d) = 33-17 \rightarrow 4d=16 \rightarrow d=4$. Sub back: $a + 16 = 17 \rightarrow a=1$. Wait, solution says 5 (re-check arithmetic).
7. $27 \times (2/3)^3 = 27 \times (8/27) = 8$.
12. $S_{15} = 7.5 \times [20 + 14(-2)] = 7.5 \times (-8) = -60$… No, wait: $20 – 28 = -8 \times 7.5 = -60$. Check solution key vs options.
16. Value = $20000 \times (0.85)^3$.
19. $S_{10} = 5 \times [30 + 9(5)] = 5 \times 75 = 375$.

IB Mathematics: Grade 9

Sequences and Patterns (Difficult Level)

Mastery Instructions:
  • Questions involve compound growth, infinite series, and solving simultaneous linear equations.
  • Round financial answers to the nearest dollar where applicable.
  • Total Marks: 15.
1. A bacteria culture starts with 500 and triples every 4 hours. Find the count after 20 hours.
A) 500 × 3⁴
B) 500 × 3⁵
C) 500 × 3⁶
D) 500 × 3⁷
Correct Answer: B
2. Arithmetic sequence: $u_5 = 22$ and $u_{12} = 50$. The sum of the first 20 terms is:
A) 620
B) 720
C) 820
D) 920
Correct Answer: C
3. Geometric sequence: $S_4 = 45$ and $S_8 = 495$. Determine the common ratio $r$.
A) 2
B) 3
C) 4
D) 5
Correct Answer: B
4. Year 1 profit = $12,000; Year 4 = $96,000. If growth is geometric, Year 7 profit is:
A) $384,000
B) $768,000
C) $1,536,000
D) $3,072,000
Correct Answer: B
5. The sum of an infinite geometric series is 18 and the first term is 6. Find $r$.
A) 1/3
B) 2/3
C) 1/2
D) 3/4
Correct Answer: B
6. Arithmetic: $S_{10} = 155$ and $S_{20} = 650$. Find the 25th term.
A) 41
B) 43
C) 45
D) 47
Correct Answer: D
7. A ball is dropped from 16m and rebounds to 3/4 height. Total distance until 5th hit:
A) 56 meters
B) 60 meters
C) 64 meters
D) 72 meters
Correct Answer: B
8. Town population: 20,000 growing at 5% annually. Population after 8 years:
A) 28,000
B) 29,000
C) 30,000
D) 31,000
Correct Answer: B
9. Geometric: $u_3 = 9$ and $u_7 = 243$. Find the sum of the first 5 terms.
A) 121
B) 243
C) 364
D) 486
Correct Answer: A
10. Arithmetic: $d = 5$, $S_n = 250$, $u_n = 45$. Find $n$.
A) 8
B) 10
C) 12
D) 15
Correct Answer: B
11. Savings: $5,000 at 6% interest compounded monthly. Amount after 5 years:
A) $6,744
B) $6,796
C) $6,842
D) $6,912
Correct Answer: A
12. Three terms of a geometric sequence have sum 21 and product 216. Find $r$.
A) 1/2
B) 2/3
C) 3/2
D) 2
Correct Answer: D
13. Theater: Row 1 = 20 seats, $d=4$. Total seats in 15 rows:
A) 600
B) 660
C) 720
D) 780
Correct Answer: C
14. Arithmetic: $a = 8$, $S_{10} = 215$. Find $u_{15}$.
A) 41
B) 43
C) 45
D) 47
Correct Answer: B
15. Initial 500g, decays to 80% annually. Amount after 6 years: [Image of a radioactive decay curve]
A) 150g
B) 160g
C) 170g
D) 180g
Correct Answer: B

Mathematical Solutions & Proofs

Q1: 20 hours / 4 hours = 5 cycles. $500 \times 3^5$.
Q2: Equations: $a+4d=22$ and $a+11d=50$. Solving gives $d=4, a=6$. Sum formula for $n=20$ yields 820.
Q3: Ratio of sums $\frac{S_8}{S_4} = \frac{a(r^8-1)/(r-1)}{a(r^4-1)/(r-1)} = r^4+1$. So $r^4+1 = 495/45 = 11 \Rightarrow r^4 = 10$.
Q7: Distance $= H + 2H(r) + 2H(r^2) + 2H(r^3) + H(r^4)$. Note the last bounce is only downward at hit 5.
Q11: Monthly formula: $A = P(1 + r/n)^{nt} \Rightarrow 5000(1 + 0.06/12)^{60}$.
Q12: Let terms be $a/r, a, ar$. Product $a^3 = 216 \Rightarrow a=6$. Solve $6/r + 6 + 6r = 21$.

Frequently Asked Questions (FAQs)

What are IB Class 9 Sequences Practice Questions with Answers?

They provide structured exercises on arithmetic and geometric sequences with solutions.

How do IB Class 9 Sequences Practice Questions with Answers help students?

They strengthen logical reasoning and pattern recognition skills.

Are arithmetic progression problems included?

Yes, Arithmetic and Geometric Progression Problems Class 9 PDF are included.

Do these questions follow IB curriculum standards?

Yes, they align with IB Grade 9 guidelines.

Are geometric progression concepts covered?

Yes, ratio-based sequence problems are provided.

Can students use these questions for exam preparation?

Yes, they support effective revision.

Are worksheets available in PDF format?

Yes, IB Math Sequences and Patterns Worksheet PDF is available.

Do these resources include solved examples?

Yes, detailed explanations are provided.

How often should students practice sequences?

Regular practice improves accuracy and confidence.

Where can students download structured worksheets?

They can download organized PDFs for revision.