This MCQ module is based on: 5.1 Introduction — Patterns in Life
5.1 Introduction — Patterns in Life
This mathematics assessment will be based on: 5.1 Introduction — Patterns in Life
Targeting Class 10 level in Algebra, with Intermediate difficulty.
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5.1 Introduction — Patterns in Life
In everyday life, many situations follow a predictable pattern where quantities change by a fixed amount each step. Consider the following examples:
- Reena's savings: she deposits ₹1000 in a piggy bank in the first month, ₹1500 in the second, ₹2000 in the third — each month her deposit grows by ₹500.
- The length of a taxi ride's fare: ₹15 for the first km, and ₹8 for every km after that.
- The number of seats in successive rows of an auditorium: 20 in the first row, 22 in the second, 24 in the third, and so on.
- The heights of a ladder's rungs above the ground increase by the same amount as we move up rung by rung.
All of these are examples of an Arithmetic Progression? — a list of numbers in which each term is obtained by adding a fixed number to the preceding one.
5.2 Arithmetic Progressions
Consider these lists of numbers:
| List | Terms | Pattern? |
|---|---|---|
| (i) | 2, 5, 8, 11, 14, ... | Add 3 each time |
| (ii) | 100, 70, 40, 10, ... | Add −30 each time |
| (iii) | −3, −2, −1, 0, ... | Add 1 each time |
| (iv) | 3, 3, 3, 3, ... | Add 0 each time |
| (v) | −1.0, −1.5, −2.0, −2.5, ... | Add −0.5 each time |
Here \(d\) is called the common difference?. The first term is denoted \(a\) (or \(a_1\)).
A general AP: \(a,\; a+d,\; a+2d,\; a+3d,\; \ldots\)
The common difference \(d\) can be positive, negative, or zero.
Example 1 — Finding the Common Difference
For the AP 6, 9, 12, 15, 18, ..., \(d = 9 - 6 = 3\). Check: \(12 - 9 = 3\), \(15 - 12 = 3\) ✓.
Example 2 — Testing whether a list is an AP
Is the list 4, 10, 16, 22, 28 an AP?
Differences: \(10-4=6\), \(16-10=6\), \(22-16=6\), \(28-22=6\). All equal to 6, so yes, it is an AP with \(a=4,\ d=6\).
Example 3 — Write the AP given \(a\) and \(d\)
(i) \(a=10,\ d=10\): \(10, 20, 30, 40, \ldots\)
(ii) \(a=-2,\ d=0\): \(-2,-2,-2,-2,\ldots\)
(iii) \(a=4,\ d=-3\): \(4,1,-2,-5,\ldots\)
(iv) \(a=-1,\ d=\tfrac12\): \(-1,-\tfrac12,0,\tfrac12,\ldots\)
Example 4 — Find \(a\) and \(d\) from a given AP
For the AP \(3,1,-1,-3,\ldots\): \(a=3\), \(d = 1-3 = -2\).
- Mark a horizontal axis (step number 1–10) and a vertical axis (height).
- Plot \(a_1 = 2\). Then add \(d = 1.5\) repeatedly to get \(a_2 = 3.5, a_3 = 5, \ldots\).
- Join the dots. Observe that the points lie on a straight line — a key property of an AP.
- Now try with \(d = -0.5\). What happens?
Observation: The graph of \(a_n\) versus \(n\) for an AP is always a set of collinear points, because \(a_n = a + (n-1)d\) is a linear function of \(n\). If \(d>0\) the line slopes upward; if \(d<0\) it slopes downward; if \(d=0\) the line is horizontal.
5.3 nth Term of an AP
To find the hundredth term of an AP without listing all 100 terms, we derive a general formula.
Then, \(a_1 = a = a + (1-1)d\)
\(a_2 = a + d = a + (2-1)d\)
\(a_3 = a + 2d = a + (3-1)d\)
\(a_4 = a + 3d = a + (4-1)d\)
Pattern: \(a_n = a + (n-1)d\).
Example 5 — Finding the 10th term
Find the 10th term of the AP 2, 7, 12, ...
Here \(a=2,\ d=7-2=5,\ n=10\). So \(a_{10} = 2 + (10-1)\cdot 5 = 2 + 45 = 47\).
Example 6 — Which term of the AP 21, 18, 15, ... is −81?
\(a=21,\ d=-3,\ a_n=-81\). Using \(a_n = a+(n-1)d\):
\(-81 = 21 + (n-1)(-3)\)
\(-81 - 21 = -3(n-1)\)
\(-102 = -3(n-1)\), so \(n-1 = 34\), giving \(n = 35\).
Hence the 35th term is −81. Is any term zero? Set \(0 = 21 + (n-1)(-3)\Rightarrow n-1 = 7\Rightarrow n = 8\). Yes — the 8th term is 0.
Example 7 — Determining an AP when two terms are known
The 3rd term of an AP is 5 and the 7th term is 9. Find the AP.
We have \(a_3 = a + 2d = 5\) and \(a_7 = a + 6d = 9\). Subtract: \(4d = 4 \Rightarrow d = 1\). Then \(a = 5 - 2 = 3\). AP: \(3,4,5,6,7,\ldots\)
Example 8 — Checking if a number is a term of an AP
Is 301 a term of the AP 5, 11, 17, 23, ...?
\(a=5,\ d=6\). If 301 were the \(n\)th term, \(301 = 5 + (n-1)\cdot 6 \Rightarrow n-1 = \tfrac{296}{6}\), which is not an integer. Hence 301 is not a term of this AP.
Example 9 — How many two-digit numbers are divisible by 3?
First such number is 12 and last is 99. AP: 12, 15, 18, ..., 99. \(a=12,\ d=3,\ a_n = 99\).
\(99 = 12 + (n-1)\cdot 3 \Rightarrow n-1 = 29 \Rightarrow n = 30\). So there are 30 two-digit numbers divisible by 3.
Example 10 — Last term of a finite AP
Find the 11th term from the end of the AP 10, 7, 4, ..., −62.
The AP has \(a=10,\ d=-3\). Last term \(l = -62\). The 11th term from the end is the same as using the reversed AP \(-62, -59, -56, \ldots\) with \(d' = +3\). So 11th from end \(= -62 + (11-1)\cdot 3 = -62 + 30 = -32\).
Exercise 5.1 (Selected)
(ii) Each term is \(\tfrac34\) of the previous — not an AP (it is a GP).
(iii) 150, 200, 250, ... — AP (\(d=50\)).
(iv) Amounts \(10000(1.08)^n\) — not an AP.
Competency-Based Questions
Assertion–Reason Questions
Reason (R): The differences between successive terms are 3, 5, 7, 9, ...
Reason (R): \(a_n = a + (n-1)d\).
Reason (R): \(a_n = a + (n-1)d\) with \(a=10,\ d=-3\).
Frequently Asked Questions — Arithmetic Progressions
What is Part 1 — Introduction to AP & nth Term Formula | Class 10 Maths Ch 5 | MyAiSchool in NCERT Class 10 Mathematics?
Part 1 — Introduction to AP & nth Term Formula | Class 10 Maths Ch 5 | MyAiSchool is a key concept covered in NCERT Class 10 Mathematics, Chapter 5: Arithmetic Progressions. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Part 1 — Introduction to AP & nth Term Formula | Class 10 Maths Ch 5 | MyAiSchool step by step?
To solve problems on Part 1 — Introduction to AP & nth Term Formula | Class 10 Maths Ch 5 | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 10 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 5: Arithmetic Progressions?
The essential formulas of Chapter 5 (Arithmetic Progressions) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Part 1 — Introduction to AP & nth Term Formula | Class 10 Maths Ch 5 | MyAiSchool important for the Class 10 board exam?
Part 1 — Introduction to AP & nth Term Formula | Class 10 Maths Ch 5 | MyAiSchool is part of the NCERT Class 10 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Part 1 — Introduction to AP & nth Term Formula | Class 10 Maths Ch 5 | MyAiSchool?
Common mistakes in Part 1 — Introduction to AP & nth Term Formula | Class 10 Maths Ch 5 | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Part 1 — Introduction to AP & nth Term Formula | Class 10 Maths Ch 5 | MyAiSchool?
End-of-chapter NCERT exercises for Part 1 — Introduction to AP & nth Term Formula | Class 10 Maths Ch 5 | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 5, and solve at least one previous-year board paper to consolidate your understanding.