Class 9 · Mathematics · Chapter 8
Sequences & Progressions
Important Questions Bank
15 Short Answer + 10 Long Answer Questions with complete solutions
📝 15 Short Questions
📖 10 Long Questions
✅ Full Solutions
🎯 AP · GP · Sequences
📘 Ganita Manjari Grade 9
📖 10 Long Questions
✅ Full Solutions
🎯 AP · GP · Sequences
📘 Ganita Manjari Grade 9
Part A — Short Answer Questions (2–3 Marks)
15 questions covering Theory, Application, and Computation
1
Define a sequence. Give two examples of finite sequences and two examples of infinite sequences.
Answer
A sequence is an ordered list of numbers where each number is called a term of the sequence.
Finite sequences:
1. 6, 12, 24, 48, 96 — a sequence of 5 terms.
2. 1, 2, 3, 4, 5 — the first five natural numbers.
Infinite sequences:
1. 1, 3, 5, 7, 9, 11, … — all odd numbers (continues indefinitely).
2. 1, 4, 9, 16, 25, … — all square numbers (continues indefinitely).
The three dots (…) indicate that an infinite sequence continues without end.
2
What is an explicit formula for a sequence? Write the explicit formula for the nth term of the sequence of odd numbers and verify it for the first three terms.
Answer
An explicit formula (or explicit rule) uses the term’s position number n to directly calculate the value of the term, without needing to know previous terms.
For the sequence of odd numbers 1, 3, 5, 7, 9, …:
Explicit formula: un = 2n − 1
Verification:
u₁ = 2(1) − 1 = 1 ✓
u₂ = 2(2) − 1 = 3 ✓
u₃ = 2(3) − 1 = 5 ✓
3
Using the explicit rule un = 2n − 1, find the 53rd and 108th terms of the sequence of odd numbers.
Answer
Using un = 2n − 1
53rd term:
u₅₃ = 2(53) − 1 = 106 − 1 = 105108th term:
u₁₀₈ = 2(108) − 1 = 216 − 1 = 215
u₅₃ = 2(53) − 1 = 106 − 1 = 105108th term:
u₁₀₈ = 2(108) − 1 = 216 − 1 = 215
u₅₃ = 105 | u₁₀₈ = 215
4
What is a recursive formula? Write the recursive rule for the sequence 1, 4, 7, 10, 13, …
Answer
A recursive formula (or recursive rule) defines each term of a sequence by relating it to one or more previous terms. Unlike an explicit formula, you need to know earlier terms to use it.
For the sequence 1, 4, 7, 10, 13, …, each term is 3 more than the previous term.
Recursive rule: t₁ = 1, tₙ = tₙ₋₁ + 3 for n ≥ 2
Verification:
t₂ = t₁ + 3 = 1 + 3 = 4 ✓
t₃ = t₂ + 3 = 4 + 3 = 7 ✓
5
Find the first four terms of the sequence given by the recursive rule: u₁ = 1, un = 2un−1 + 3 for n ≥ 2.
Answer
Substituting values of n successively:
u₁ = 1 (given)
u₂ = 2u₁ + 3 = 2(1) + 3 = 5
u₃ = 2u₂ + 3 = 2(5) + 3 = 13
u₄ = 2u₃ + 3 = 2(13) + 3 = 29
First four terms: 1, 5, 13, 29
6
For the sequence tn = 3n − 7: (i) find the first, second, and third terms, (ii) find which term of the sequence is 332, and (iii) is 557 a term of this sequence?
Answer
Using tₙ = 3n − 7
(i) First three terms:
t₁ = 3(1) − 7 = −4
t₂ = 3(2) − 7 = −1
t₃ = 3(3) − 7 = 2(ii) Which term is 332?
Set tₙ = 332: 3n − 7 = 332 → 3n = 339 → n = 113
So 332 is the 113th term.(iii) Is 557 a term?
Set tₙ = 557: 3n − 7 = 557 → 3n = 564 → n = 188
Since 188 is a natural number, yes, 557 is the 188th term.
t₁ = 3(1) − 7 = −4
t₂ = 3(2) − 7 = −1
t₃ = 3(3) − 7 = 2(ii) Which term is 332?
Set tₙ = 332: 3n − 7 = 332 → 3n = 339 → n = 113
So 332 is the 113th term.(iii) Is 557 a term?
Set tₙ = 557: 3n − 7 = 557 → 3n = 564 → n = 188
Since 188 is a natural number, yes, 557 is the 188th term.
7
Define an Arithmetic Progression (AP). State its general form and the formula for the nth term.
Answer
An Arithmetic Progression (AP) is a sequence in which the difference between any two consecutive terms is always the same constant value, called the common difference (d).
General form of an AP:
a, a + d, a + 2d, a + 3d, …, a + (n−1)d
where a = first term, d = common difference.
Formula for the nth term:
tₙ = a + (n − 1) × d
Example: The sequence 1, 5, 9, 13, … is an AP with a = 1 and d = 4. Here tₙ = 4n − 3.
8
Find the 10th term of the AP: 3, 8, 13, 18, … Also write its recursive rule.
Answer
Given AP: 3, 8, 13, 18, …
First term: a = 3, Common difference: d = 8 − 3 = 5
First term: a = 3, Common difference: d = 8 − 3 = 5
tₙ = a + (n − 1)d = 3 + (n − 1) × 5
10th term:
t₁₀ = 3 + (10 − 1) × 5 = 3 + 45 = 48
Recursive rule:
t₁ = 3, tₙ = tₙ₋₁ + 5 for n ≥ 2
t₁₀ = 48
9
Which term of the AP: 21, 18, 15, … is −81? Also, is 0 a term of this AP? Give reasons.
Answer
Given AP: 21, 18, 15, …
a = 21, d = 18 − 21 = −3For which n is tₙ = −81?
−81 = 21 + (n − 1)(−3)
−81 − 21 = −3(n − 1)
−102 = −3(n − 1)
n − 1 = 34
n = 35
a = 21, d = 18 − 21 = −3For which n is tₙ = −81?
−81 = 21 + (n − 1)(−3)
−81 − 21 = −3(n − 1)
−102 = −3(n − 1)
n − 1 = 34
n = 35
−81 is the 35th term of the AP.
Is 0 a term?
0 = 21 + (n − 1)(−3)
−21 = −3(n − 1)
n − 1 = 7 → n = 8
Since 8 is a natural number, yes, 0 is the 8th term.
0 = 21 + (n − 1)(−3)
−21 = −3(n − 1)
n − 1 = 7 → n = 8
Since 8 is a natural number, yes, 0 is the 8th term.
10
Define a Geometric Progression (GP). Write its general form and the formula for the nth term.
Answer
A Geometric Progression (GP) is a sequence in which each term after the first is obtained by multiplying the previous term by a fixed constant called the common ratio (r).
General form of a GP:
a, ar, ar², ar³, …, arⁿ⁻¹
where a = first term, r = common ratio.
Formula for the nth term:
tₙ = a × rⁿ⁻¹
Example: 3, 6, 12, 24, … is a GP with a = 3 and r = 2. Here tₙ = 3 × 2ⁿ⁻¹.
Recursive rule: t₁ = a, tₙ = r × tₙ₋₁ for n ≥ 2.
11
Check whether the sequence 3, 6, 12, 24, 48, … is a GP. If yes, find the common ratio and the 8th term.
Answer
Checking ratio of consecutive terms:
6/3 = 2, 12/6 = 2, 24/12 = 2, 48/24 = 2Since the ratio is constant, this is a GP with a = 3 and r = 2.8th term:
6/3 = 2, 12/6 = 2, 24/12 = 2, 48/24 = 2Since the ratio is constant, this is a GP with a = 3 and r = 2.8th term:
t₈ = a × r⁷ = 3 × 2⁷ = 3 × 128
t₈ = 384
12
What are triangular numbers? Write the formula for the nth triangular number and find the 10th triangular number.
Answer
Triangular numbers are numbers represented by triangular arrays of dots:
1, 3, 6, 10, 15, 21, …Each triangular number is the sum of natural numbers up to its position number. For example:
3rd triangular number = 1 + 2 + 3 = 6
5th triangular number = 1 + 2 + 3 + 4 + 5 = 15Formula for the nth triangular number:
1, 3, 6, 10, 15, 21, …Each triangular number is the sum of natural numbers up to its position number. For example:
3rd triangular number = 1 + 2 + 3 = 6
5th triangular number = 1 + 2 + 3 + 4 + 5 = 15Formula for the nth triangular number:
tₙ = n(n + 1) / 2
10th triangular number:
t₁₀ = 10 × 11 / 2 = 110 / 2 = 55
10th triangular number = 55
13
State the formula for the sum of the first n natural numbers. Derive it briefly and use it to find the sum of the first 50 natural numbers.
Answer
Derivation:
Let S = 1 + 2 + 3 + … + n
Also S = n + (n−1) + … + 1
Adding: 2S = (n+1) + (n+1) + … n times = n(n+1)
Let S = 1 + 2 + 3 + … + n
Also S = n + (n−1) + … + 1
Adding: 2S = (n+1) + (n+1) + … n times = n(n+1)
Sₙ = n(n + 1) / 2
Sum of first 50 natural numbers:
S₅₀ = 50 × 51 / 2 = 2550 / 2
S₅₀ = 1275
14
Describe the Virahānka–Fibonacci sequence. State who first studied it and write its recursive rule. Write the first eight terms.
Answer
The Virahānka–Fibonacci sequence is a sequence where each term is the sum of the two preceding terms. It was first studied by the Indian scholar Virahānka in the 7th century CE in his work Vṛttajātisamuchaya, in the context of Prakrit meter and poetry. It was later also studied by Fibonacci around 1200 CE.
Recursive rule:
V₁ = 1, V₂ = 2, Vₙ = Vₙ₋₁ + Vₙ₋₂ for n ≥ 3
First eight terms:
V₃ = 2 + 1 = 3
V₄ = 3 + 2 = 5
V₅ = 5 + 3 = 8
V₆ = 8 + 5 = 13
V₇ = 13 + 8 = 21
V₈ = 21 + 13 = 34
Sequence: 1, 2, 3, 5, 8, 13, 21, 34, …
15
When points (stage number, number of squares in a GP) are plotted on a graph, do they lie on a straight line? Compare this with what happens for an AP. Explain why.
Answer
For an AP:
The nth term tₙ = a + (n−1)d = dn + (a−d) is a linear expression in n. Therefore when we plot the pairs (n, tₙ), they lie on a straight line. This is why the graph of an AP is linear.Example: AP 1, 5, 9, 13, 17 → points (1,1), (2,5), (3,9), (4,13), (5,17) all lie on a straight line.For a GP:
The nth term tₙ = arⁿ⁻¹ is an exponential expression in n. Therefore when we plot (n, tₙ), the points do NOT lie on a straight line — they form a curved (exponential) pattern.
The nth term tₙ = a + (n−1)d = dn + (a−d) is a linear expression in n. Therefore when we plot the pairs (n, tₙ), they lie on a straight line. This is why the graph of an AP is linear.Example: AP 1, 5, 9, 13, 17 → points (1,1), (2,5), (3,9), (4,13), (5,17) all lie on a straight line.For a GP:
The nth term tₙ = arⁿ⁻¹ is an exponential expression in n. Therefore when we plot (n, tₙ), the points do NOT lie on a straight line — they form a curved (exponential) pattern.
Example: GP 3, 6, 12, 24, 48 → points (1,3), (2,6), (3,12), (4,24), (5,48) lie on a rapidly rising curve.
Key difference: AP → linear graph; GP → exponential (curved) graph.
Part B — Long Answer Questions (5 Marks)
10 questions with detailed step-by-step solutions
1
A person books a taxi that charges a fixed booking fee of ₹200 plus ₹40 per kilometre travelled. (a) Write the sequence of total fares for journeys of 1 km, 2 km, 3 km, etc. (b) Show this sequence is an AP and identify its first term and common difference. (c) Write the general formula for the fare after n km. (d) Find the total fare for a 15 km journey.
Step-by-step Solution
- 1Listing fares: After 1 km: 200 + 40 = ₹240. After 2 km: 200 + 80 = ₹280. After 3 km: 200 + 120 = ₹320. Sequence: 240, 280, 320, 360, …
- 2Is it an AP? Difference between consecutive terms: 280 − 240 = 40, 320 − 280 = 40, 360 − 320 = 40. The difference is constant, so this is an AP.
- 3First term: a = 240 | Common difference: d = 40
- 4General formula: tₙ = a + (n−1)d = 240 + (n−1)(40) = 240 + 40n − 40 = 200 + 40n, where n = distance in km.
- 5Fare for 15 km: t₁₅ = 200 + 40(15) = 200 + 600 = ₹800.
Total fare for 15 km = ₹800
2
An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find: (a) the first term and common difference, (b) the 29th term, and (c) write the recursive rule for this AP.
Step-by-step Solution
- 1Let first term = a and common difference = d. Using tₙ = a + (n−1)d:
- 2From 3rd term: a + 2d = 12 … (i)
- 3From last term (50th): a + 49d = 106 … (ii)
- 4Subtracting (i) from (ii): 47d = 94 → d = 2
- 5Finding a: From (i): a + 2(2) = 12 → a = 12 − 4 = 8
- 629th term: t₂₉ = 8 + (29−1)(2) = 8 + 56 = 64
- 7Recursive rule: t₁ = 8, tₙ = tₙ₋₁ + 2 for n ≥ 2
a = 8 | d = 2 | 29th term = 64
3
Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. (a) Write his salary sequence as an AP. (b) After how many years did his income reach ₹7,00,000? (c) Write the explicit formula for his salary in the nth year.
Step-by-step Solution
- 1Salary sequence: 5,00,000; 5,20,000; 5,40,000; 5,60,000; … This is an AP with a = 5,00,000 and d = 20,000.
- 2nth year salary: tₙ = 5,00,000 + (n−1)(20,000)
- 3When does salary reach ₹7,00,000? Set tₙ = 7,00,000:
- 47,00,000 = 5,00,000 + (n−1)(20,000)
- 52,00,000 = (n−1)(20,000) → n−1 = 10 → n = 11
- 6His salary reaches ₹7,00,000 in the 11th year, i.e., after 10 years of increments.
Harish’s salary reaches ₹7,00,000 after 10 increments (in year 11).
4
How many 2-digit numbers are divisible by 3? Find the sum of all these 2-digit numbers.
Step-by-step Solution
- 1Identify the AP: 2-digit numbers divisible by 3 are: 12, 15, 18, …, 99. This is an AP with a = 12, d = 3, and last term l = 99.
- 2Find the number of terms (n): Using tₙ = a + (n−1)d: 99 = 12 + (n−1)(3) → 87 = 3(n−1) → n−1 = 29 → n = 30
- 3There are 30 two-digit numbers divisible by 3.
- 4Sum of the AP: Sum = n/2 × (first term + last term)
- 5Sum = 30/2 × (12 + 99) = 15 × 111 = 1665
Count = 30 | Sum = 1665
5
A child arranges marbles in rows: row 1 has 1 marble, row 2 has 2, row 3 has 3, and so on up to 25 rows. (a) How many marbles in all? (b) Find the marbles in the first 10 rows. (c) Use the formula Sₙ = n(n+1)/2.
Step-by-step Solution
- 1The number of marbles in each row: 1, 2, 3, 4, …, 25. The total marbles = 1 + 2 + 3 + … + 25 = S₂₅.
- 2Formula: Sₙ = n(n + 1) / 2
- 3Total marbles (25 rows): S₂₅ = 25 × 26 / 2 = 650 / 2 = 325
- 4Marbles in first 10 rows: S₁₀ = 10 × 11 / 2 = 110 / 2 = 55
- 5Note: Each row count is a triangular number — this formula directly gives the sum of first n natural numbers.
Total marbles (25 rows) = 325 | First 10 rows = 55
6
Find the 31st term of an AP whose 11th term is 38 and 16th term is 73. Also write the explicit and recursive formulas for this AP.
Step-by-step Solution
- 1Let first term = a and common difference = d.
- 2t₁₁ = 38: a + 10d = 38 … (i)
- 3t₁₆ = 73: a + 15d = 73 … (ii)
- 4Subtract (i) from (ii): 5d = 35 → d = 7
- 5Find a: From (i): a + 10(7) = 38 → a = 38 − 70 = −32
- 631st term: t₃₁ = −32 + (31−1)(7) = −32 + 210 = 178
- 7Explicit formula: tₙ = −32 + (n−1)(7) = 7n − 39
- 8Recursive rule: t₁ = −32, tₙ = tₙ₋₁ + 7 for n ≥ 2
a = −32 | d = 7 | 31st term = 178
7
Find the sum of all natural numbers from 25 to 58 (inclusive) using the formula Sₙ = n(n+1)/2.
Step-by-step Solution
- 1Strategy: Sum from 25 to 58 = (Sum from 1 to 58) − (Sum from 1 to 24)
- 2= S₅₈ − S₂₄
- 3S₅₈ = 58 × 59 / 2 = 3422 / 2 = 1711
- 4S₂₄ = 24 × 25 / 2 = 600 / 2 = 300
- 5Required sum = 1711 − 300 = 1411
- 6Verification: Number of terms from 25 to 58 = 34. Average = (25+58)/2 = 41.5. Sum = 34 × 41.5 = 1411 ✓
Sum from 25 to 58 = 1411
8
A ball is dropped from a height of 24 feet. Each time it bounces, it rises to 3/4 of its previous height. (a) Write the sequence of heights as a GP. (b) Find the height after the 5th bounce. (c) After how many bounces does the height fall below 4 feet?
Step-by-step Solution
- 1Heights after each bounce: 1st: 24 × 3/4 = 18 ft; 2nd: 18 × 3/4 = 13.5 ft; 3rd: 13.5 × 3/4 = 10.125 ft; …
- 2This is a GP with first term a = 18 and common ratio r = 3/4 = 0.75.
- 3nth term: tₙ = 18 × (3/4)ⁿ⁻¹
- 4Height after 5th bounce: t₅ = 18 × (3/4)⁴ = 18 × 81/256 = 1458/256 ≈ 5.70 feet
- 5When does height fall below 4 feet? Need 18 × (3/4)ⁿ⁻¹ < 4 → (3/4)ⁿ⁻¹ < 4/18 ≈ 0.222
- 6Check: (3/4)⁵ ≈ 0.237 > 0.222 (not yet below); (3/4)⁶ ≈ 0.178 < 0.222 (below!)
- 7So n − 1 = 6 → n = 7. After the 7th bounce, the ball stays below 4 feet.
GP: 18, 13.5, 10.125, … | After 5th bounce ≈ 5.70 ft | Below 4 ft after 7th bounce
9
The number of black triangles in successive stages of the Sierpiński triangle is 1, 3, 9, 27, … (a) Verify this is a GP. (b) Write the explicit formula for the number of black triangles at the nth stage. (c) Find the number at stage 6. (d) The area of the black region at stage n is (3/4)ⁿ — explain this pattern and write the recursive rule for area.
Step-by-step Solution
- 1Verify GP: Ratios: 3/1 = 3, 9/3 = 3, 27/9 = 3. Common ratio r = 3 (constant), so this is a GP with a = 1.
- 2Explicit formula: Since stage numbers start at 0, the number of black triangles at stage n = 3ⁿ. (Alternatively, at the nth term of the GP: tₙ = 1 × 3ⁿ⁻¹ = 3ⁿ⁻¹ for n starting at 1.)
- 3Stage 6: Black triangles = 3⁶ = 729
- 4Area pattern: At Stage 0, area = 1. At Stage 1, the central triangle (area 1/4) is removed from each triangle, leaving 3/4 of the previous area.
- 5So: Stage 0 = 1, Stage 1 = 3/4, Stage 2 = (3/4)², Stage 3 = (3/4)³, … Area at stage n = (3/4)ⁿ.
- 6Recursive rule for area: s₀ = 1, sₙ = (3/4) × sₙ₋₁ for n ≥ 1. As n increases, area → 0 (the black region shrinks to nothing).
Stage 6 black triangles = 729 | Area at stage n = (3/4)ⁿ → 0
10
Find the first four terms of the sequence given by s₁ = 3, sₙ = sₙ₋₁(sₙ₋₁ − 1) for n ≥ 2. Then separately find the first four terms of the Virahānka–Fibonacci sequence defined by V₁ = 1, V₂ = 2, Vₙ = Vₙ₋₁ + Vₙ₋₂ for n ≥ 3. Compare the rates at which both sequences grow.
Step-by-step Solution
- 1Sequence 1: s₁ = 3, sₙ = sₙ₋₁(sₙ₋₁ − 1)
- 2s₂ = s₁(s₁ − 1) = 3(3 − 1) = 3 × 2 = 6
- 3s₃ = s₂(s₂ − 1) = 6(6 − 1) = 6 × 5 = 30
- 4s₄ = s₃(s₃ − 1) = 30(30 − 1) = 30 × 29 = 870
- 5First four terms of Sequence 1: 3, 6, 30, 870 — grows extremely rapidly (faster than exponential).
- 6Virahānka–Fibonacci: V₁ = 1, V₂ = 2, Vₙ = Vₙ₋₁ + Vₙ₋₂
- 7V₃ = V₂ + V₁ = 2 + 1 = 3; V₄ = V₃ + V₂ = 3 + 2 = 5
- 8First four terms: 1, 2, 3, 5 — grows much more slowly (approximately exponential with ratio ≈ 1.618).
- 9Comparison: Sequence 1 (3, 6, 30, 870) grows far faster because each term involves multiplication of the previous term by itself minus one. The Virahānka sequence grows by addition, so it grows much more slowly.
Sequence 1: 3, 6, 30, 870 (very fast) | Virahānka: 1, 2, 3, 5 (slower)
Chapter-wise Topic Coverage
Topics covered in this question bank
Sequences — Basics
Explicit Formula
Recursive Formula
Arithmetic Progressions
Geometric Progressions
Sum of Natural Numbers
Triangular Numbers
Virahānka–Fibonacci
Sierpiński Triangle
Word Problems
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