This MCQ module is based on: Sequences and Arithmetic Progressions
TOPIC 23 OF 25
Sequences and Arithmetic Progressions
🎓 Class 9
Mathematics
CBSE
Theory
Ch 8 — Predicting What Comes Next: Exploring Sequences and Progressions
⏱ ~40 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Sequences and Arithmetic Progressions
Targeting Class 9 level in Sequences Series, with Intermediate difficulty.
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8.1 Patterns That Repeat — Why Sequences?
Look at the dates 2, 9, 16, 23, 30 on a calendar — every Sunday in a month. Look at the savings 100, 200, 300, 400 — adding ₹100 each month. Look at the heights of bars in a staircase rising 6 cm at a step. These are all examples of sequences? — ordered lists of numbers that follow a pattern.
The mathematics of sequences gives us tools to predict what will come next, find any specific value far ahead in the list, and add up many terms quickly without listing them all.
Calendar dates
2, 9, 16, 23, 30 — adds 7 every step.
Monthly savings
100, 200, 300, 400 — adds 100 every step.
Staircase
6, 12, 18, 24 cm — adds 6 every step.
Odd numbers
1, 3, 5, 7, 9 — adds 2 every step.
8.2 Sequences — Definitions and Notation
Definition — Sequence
A sequence is an ordered list of numbers, each of which is called a term. The first term is denoted \(a_1\), the second \(a_2\), and the n-th term \(a_n\). The number n is the position of the term in the list.
3
→
7
→
11
→
15
→
19
→
…
Finite and Infinite Sequences
- Finite sequence: has a last term. Example: number of students in classes 1 to 10 of a school.
- Infinite sequence: goes on forever, denoted with "…". Example: the natural numbers 1, 2, 3, 4, …
Example 1 — Recognising a pattern
Identify the pattern and write the next two terms of: 5, 10, 15, 20, …
Each term is 5 more than the previous. The next two terms are 25 and 30.
Example 2 — Tabulating a sequence
| Position n | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Term \(a_n\) | 2 | 5 | 8 | 11 | 14 | 17 |
Each new term is obtained by adding 3 to the previous one. The pattern is therefore "+3".
8.3 Arithmetic Progression (AP)
Of all sequences, the simplest pattern is "add the same number every time." This special sequence has a name.
Definition — Arithmetic Progression
A sequence in which each term (after the first) is obtained by adding a fixed number to the previous term is called an Arithmetic Progression? (AP). The fixed number is the common difference?, denoted by \(d\).
If the first term is \(a\), the AP looks like: \( a,\;\; a+d,\;\; a+2d,\;\; a+3d,\;\; \ldots \)
Fig. 8.1 — A general AP with first term a and common difference d.
Quick Test for an AP
To check whether a list is an AP, compute the difference between every consecutive pair:
\( a_2 - a_1,\quad a_3 - a_2,\quad a_4 - a_3,\quad \ldots \)
If all these differences are equal, the sequence is an AP and that common value is d. If even one differs, it is not an AP.
Example 3 — Identify the APs
Which of these are APs? For those that are, write \(a\) and \(d\).
- 2, 4, 6, 8, …
- 1, 4, 9, 16, …
- 10, 7, 4, 1, –2, …
- 3, 3, 3, 3, …
- 1, 2, 4, 8, …
Solution
(1) Diffs 2, 2, 2 — AP with a = 2, d = 2. (2) Diffs 3, 5, 7 — not equal, NOT an AP. (3) Diffs −3, −3, −3, −3 — AP with a = 10, d = −3. (4) Diffs 0, 0, 0 — AP with a = 3, d = 0 (a constant sequence). (5) Diffs 1, 2, 4 — not equal, NOT an AP (it is geometric).
8.4 The n-th Term of an AP
Suppose the AP has first term a and common difference d. Then by the pattern in Fig. 8.1:
\(a_1 = a,\quad a_2 = a + d,\quad a_3 = a + 2d,\quad a_4 = a + 3d,\quad \ldots\)
So in general:
Formula — General Term of an AP
\[ a_n \;=\; a + (n-1)\,d \]
where \(a\) = first term, \(d\) = common difference, \(n\) = position number.
Derivation
From \(a_1 = a\) and \(a_{k+1} = a_k + d\), we add d a total of \((n-1)\) times to reach the n-th term:
\(a_n = \underbrace{a}_{a_1} + \underbrace{d + d + \cdots + d}_{(n-1) \text{ times}} = a + (n-1)d.\)
Example 4 — Direct use of the formula
Find the 20th term of the AP: 5, 9, 13, 17, …
Solution
Here \(a = 5,\; d = 9 - 5 = 4,\; n = 20\).
\(a_{20} = a + 19d = 5 + 19(4) = 5 + 76 = 81\).
\(a_{20} = a + 19d = 5 + 19(4) = 5 + 76 = 81\).
Example 5 — Reverse problem
Which term of the AP 8, 14, 20, … is 134?
Solution
\(a = 8,\; d = 6,\; a_n = 134\).
\(a + (n-1)d = 134 \Rightarrow 8 + (n-1)(6) = 134 \Rightarrow (n-1)(6) = 126 \Rightarrow n-1 = 21 \Rightarrow n = 22\).
So 134 is the 22nd term.
\(a + (n-1)d = 134 \Rightarrow 8 + (n-1)(6) = 134 \Rightarrow (n-1)(6) = 126 \Rightarrow n-1 = 21 \Rightarrow n = 22\).
So 134 is the 22nd term.
Example 6 — Two given terms
The 3rd term of an AP is 12 and the 8th term is 27. Find a and d.
Solution
\(a_3 = a + 2d = 12\) … (i)
\(a_8 = a + 7d = 27\) … (ii)
Subtract (i) from (ii): \(5d = 15 \Rightarrow d = 3\).
Substitute back: \(a + 2(3) = 12 \Rightarrow a = 6\). So the AP is 6, 9, 12, 15, ….
\(a_8 = a + 7d = 27\) … (ii)
Subtract (i) from (ii): \(5d = 15 \Rightarrow d = 3\).
Substitute back: \(a + 2(3) = 12 \Rightarrow a = 6\). So the AP is 6, 9, 12, 15, ….
Example 7 — Decreasing AP
Find the 25th term of the AP 50, 47, 44, …
Solution
\(a = 50,\; d = -3,\; n = 25\). \(a_{25} = 50 + 24(-3) = 50 - 72 = -22\).
8.5 Real-World APs
Example 8 — Salary increment
Anita's starting salary is ₹25,000 per month. She gets a fixed annual increment of ₹1,500. What will her monthly salary be in her 10th year of service?
Solution
Year 1 salary = 25000, year 2 = 26500, year 3 = 28000, … forms an AP with \(a = 25000,\; d = 1500\).
\(a_{10} = 25000 + 9(1500) = 25000 + 13500 = ₹38{,}500\).
\(a_{10} = 25000 + 9(1500) = 25000 + 13500 = ₹38{,}500\).
Example 9 — Stadium seating
A stadium has 20 seats in the front row, 24 in the second, 28 in the third, and so on for 30 rows. How many seats are in the last row?
Solution
AP with \(a = 20,\; d = 4,\; n = 30\). \(a_{30} = 20 + 29(4) = 20 + 116 = 136\) seats.
Fig. 8.2 — Stadium row seats grow as an AP.
Example 10 — Plant height
A sapling is 12 cm tall when planted and grows by 0.7 cm each week. What is its height after 26 weeks?
Solution
Week 0 height = 12 cm. Heights at end of weeks 1, 2, 3, … form an AP with \(a = 12.7,\; d = 0.7\).
But the simpler approach: height after n weeks = 12 + 0.7n. After 26 weeks: 12 + 0.7(26) = 12 + 18.2 = 30.2 cm.
But the simpler approach: height after n weeks = 12 + 0.7n. After 26 weeks: 12 + 0.7(26) = 12 + 18.2 = 30.2 cm.
Historical Note
The Indian mathematician Aryabhata (476 CE) gave one of the earliest formulas for the sum of an AP in his work Aryabhatiya. Centuries earlier, the Sulba Sutras (~800 BCE) used arithmetic progressions to design fire altars in geometric brick patterns.
Activity 8.1 — Spotting APs Around You
L3 ApplyMaterials: Notebook, ruler, observation around school/home.
Predict: How many real-life APs can you find in 30 minutes? Try at least 6.
- List 6 real situations that look like APs (rows of seats, monthly bills, building floors, taxi fare, etc.).
- For each, write the first three terms and identify a and d.
- Pick any two and find the 15th term using \(a_n = a + (n-1)d\).
- Try to find one situation that is not an AP, and explain why.
Examples: (i) Auto fare 25, 35, 45, … (a = 25, d = 10). (ii) Multiples of 7: 7, 14, 21, … (a = 7, d = 7). (iii) Staircase rise 18, 36, 54 cm. Non-AP: a tree's leaves doubling each week (geometric, not AP).
Competency-Based Questions
Scenario: Riya begins a 30-day fitness plan. On day 1 she does 12 push-ups. Each day she does 3 more push-ups than the previous day.
Q1. How many push-ups will Riya do on day 15?
L3 ApplyAnswer: (b) 54. a = 12, d = 3, n = 15. \(a_{15} = 12 + 14 \times 3 = 12 + 42 = 54\).
Q2. The doctor recommends she should not exceed 90 push-ups in a single day. On which day will Riya first exceed this limit? Analyse.
L4 AnalyseAnswer: Need \(12 + (n-1)(3) > 90 \Rightarrow (n-1)(3) > 78 \Rightarrow n-1 > 26 \Rightarrow n > 27\). So on day 28, push-ups become \(12 + 27 \times 3 = 93 > 90\). Riya will first exceed the recommended limit on day 28.
Q3. Riya's friend says, "If you do 12 push-ups on day 1 and double them each day, that's much faster!" Evaluate whether the friend's plan is realistic and compare it to Riya's plan on day 10.
L5 EvaluateAnswer: The friend's plan is geometric (not AP). Day 10 of Riya's plan: \(12 + 9 \times 3 = 39\). Day 10 of doubling plan: \(12 \times 2^9 = 12 \times 512 = 6144\). Doubling is not realistic — no one can do thousands of push-ups in a day. AP growth is steady and sustainable.
Q4. Design a 4-week alternative plan starting at 10 push-ups, capped at 60 push-ups by day 28, that uses an AP. Find the daily increment d.
L6 CreateOne design: Day 1 = 10, day 28 = 60. \(a_{28} = a + 27d \Rightarrow 60 = 10 + 27d \Rightarrow d = 50/27 \approx 1.85\). Since push-ups are integers, round to \(d = 2\) — then day 28 = \(10 + 54 = 64\), close to the target. Alternative: keep d = 2 and stop at 60 on day 26.
Assertion–Reason Questions
Assertion (A): The sequence 7, 7, 7, 7, … is an AP.
Reason (R): The common difference is 0, which is a fixed number.
Reason (R): The common difference is 0, which is a fixed number.
Answer: (a) — A constant sequence is an AP with d = 0; R correctly explains A.
Assertion (A): The sequence 1, 4, 9, 16, … is an AP.
Reason (R): The differences between consecutive terms (3, 5, 7, …) are themselves in AP.
Reason (R): The differences between consecutive terms (3, 5, 7, …) are themselves in AP.
Answer: (d) — A is false: the differences are not all equal, so 1, 4, 9, 16 is NOT an AP. R is correct on its own (the differences 3, 5, 7 do form an AP).
Assertion (A): If \(a_n = 4n + 5\), the sequence is an AP with d = 4.
Reason (R): Whenever \(a_n\) is a linear expression of the form \(pn + q\), the sequence is an AP with common difference equal to p.
Reason (R): Whenever \(a_n\) is a linear expression of the form \(pn + q\), the sequence is an AP with common difference equal to p.
Answer: (a) — Both true and R is the structural reason. \(a_{n+1} - a_n = (4(n+1)+5) - (4n+5) = 4\).
Frequently Asked Questions
What is a sequence in Class 9 Maths?
A sequence is an ordered list of numbers, often written a_1, a_2, a_3, .... Each number is called a term. Sequences can be finite (with a last term) or infinite, and they can follow a rule (like an AP) or be defined by a formula a_n.
What is an arithmetic progression (AP) in Class 9 Maths?
An arithmetic progression is a sequence where each term after the first is obtained by adding a fixed number d, called the common difference, to the previous term. Examples: 5, 9, 13, 17, ... (d = 4) and 100, 90, 80, ... (d = -10).
What is the formula for the nth term of an AP?
If the first term is a and the common difference is d, the n-th term is a_n = a + (n - 1) d. For instance, in the AP 3, 7, 11, 15, ..., the 10th term is 3 + (10 - 1) x 4 = 3 + 36 = 39.
How do you find the common difference of an AP?
Subtract any term from the next: d = a_{n+1} - a_n. For example, in 8, 5, 2, -1, ..., d = 5 - 8 = -3 (or 2 - 5 = -3, the same). The common difference can be positive, negative, or zero.
If a = 7 and d = 4, what is the 15th term of the AP?
Using a_n = a + (n - 1)d with n = 15: a_15 = 7 + (15 - 1) x 4 = 7 + 56 = 63.
Who first studied arithmetic progressions in India?
The Indian mathematician and astronomer Aryabhata (476-550 CE), in his work Aryabhatiya (499 CE), gave formulas for the n-th term and the sum of an arithmetic progression, well before similar formulas appeared in European mathematics.
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