આ MCQ મોડ્યુલ આના પર આધારિત છે: 5.1 Introduction — Patterns in Life
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આ મૂલ્યાંકન આના પર આધારિત હશે: 5.1 Introduction — Patterns in Life Class 10 સ્તર, Algebra માં, Intermediate કઠિનતા સાથે લક્ષ્ય.
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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.
Fig 5.1: Number of seats in successive rows — an AP with common difference 2
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
Definition — Arithmetic Progression
An arithmetic progression (AP) is a list of numbers \(a_1, a_2, a_3, \ldots\) in which every term (except the first) is obtained by adding a fixed number \(d\) to the preceding term. That is, \(a_{n+1} - a_n = d\) for all \(n \ge 1\).
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 \(a=4,\ d=6\).
For the AP \(3,1,-1,-3,\ldots\): \(a=3\), \(d = 1-3 = -2\).
Activity: Build Your Own AP with Stair Steps
L3 Apply
Materials: Graph paper, pencil, ruler.
Predict: If you start at height 2 cm and each step rises by 1.5 cm, what will the height be at step 10?
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.
Derivation
Let the AP be \(a,\ a+d,\ a+2d,\ a+3d,\ \ldots\)
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: \(a_n = a + (n-1)d\).
Key Formula
\[\boxed{\,a_n = a + (n-1)d\,}\]
where \(a\) = first term, \(d\) = common difference, \(n\) = term number, \(a_n\) = \(n\)th term (also called general term).
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 \(= -62 + (11-1)\cdot 3 = -62 + 30 = -32\).
In-text: Why does reversing the AP change the sign of \(d\)? Because walking from the last term towards the first, each step subtracts \(d\), which is equivalent to adding \(-d\).
Exercise 5.1 (Selected)
Q1. In which of the following situations does the list of numbers form an AP? (i) Taxi fare: ₹15 for 1st km, ₹8 for each additional km. (ii) Amount of air in a cylinder if a vacuum pump removes \(\tfrac14\) of the air at a time. (iii) Cost of digging a well: ₹150 for 1st metre, and it rises by ₹50 for each subsequent metre. (iv) Amount at compound interest of 8% on ₹10000 year after year.
(i) 15, 23, 31, 39, ... — AP (\(d=8\)). (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.
Q2. Write the first four terms for each: (i) \(a=10,\ d=10\); (ii) \(a=-2,\ d=0\); (iii) \(a=4,\ d=-3\); (iv) \(a=-1,\ d=\tfrac12\); (v) \(a=-1.25,\ d=-0.25\).
Q2. Analyse: which row has exactly 81 seats, and what does this tell us about the hall's layout?
L4 Analyse
Solve \(81 = 25 + (n-1)\cdot 2\Rightarrow n=29\). The 29th row has 81 seats. Since there are 30 rows, only one row (29) has 81 seats; the last row has 83 seats. The pattern tells us the hall is fan-shaped — widening steadily by 2 seats per row.
Q3. A rival designer claims that if the pattern began instead at 20 seats with the same \(d=2\), the 30th row would still have the same number of seats. Evaluate this claim.
L5 Evaluate
False. Original: \(a_{30}=25+29\cdot 2=83\). Rival: \(a_{30}=20+29\cdot 2=78\). Shifting \(a\) by −5 shifts every term by −5, including the 30th. The claim is wrong.
Q4. Design an alternative seating AP for a 30-row hall where Row 1 has 30 seats and Row 30 has 88 seats. Give \(a\), \(d\), and verify.
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.
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