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Notes On Maths Arithmetic Progression (CBSE Class 10)

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Comprehensive CBSE Class 10 Maths - Arithmetic Progression PPT Unlock the complete guide to Arithmetic Progression with this detailed PPT for Class 10. Perfect for students aiming to strengthen their concepts and boost exam preparation. This presentation covers: Derivation and formula for the nth term of an Arithmetic Progression. Step-by-step breakdown of the sum of first n terms. Derivation of the sum of first n natural numbers. Real-life examples. Ideal for learners and tutors, this PPT ensures a thorough understanding of key concepts and techniques. Perfect for revision before exams!

Bhagyalakshmi S / Ajman

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  1. ARITHMETIC PROGRESSION Prepared by, BHAGYALAKSHMI S
  2. Sequence A list of numbers having specific relation between the consecutive terms is generally called a sequence. Examples : 3,5,7,... (each term is obtained by adding 2 to the previous term). 4,8, 16,32,... (each term is obtained by multiplying the previous term by 2). ,2,3,5,8,... (Fibonacci sequence). (Alternating sequence of I and-I).
  3. Arithmetic Sequence or Arithmetic Progression(AP) Terms are formed by adding a fixed number to the previous term or the difference between two successive terms is a fixed number. This fixed number is called the common difference (d). • It can be positive, negative or zero. Let the terms in the AP be al , a-2 , an. Then, — al = — = an —an-1= d Let al = a. So AP can be represented as: a, a +d, a+2d, a+3d, . a is the first term and d is the common difference.
  4. AP Example 1,2,3,4,... each term is I more than the term preceding it. each term is 30 less than the term preceding it. 3,—2,—1,0,... each term is obtained by adding I to the term preceding it. 3,3,3,3,... each term is obtained by adding (or subtracting) O to the term preceding it.
  5. Example Problem I ? Write the first four terms of the AP when the first term a and the common difference d are given as follows: a =4, d=—3. Solution: First term: al = a = 4 Second term: a-2 = a+d = 3) = I Third term: = a-2 3) = 2 Fourth term: = +d = —2+( 3) — 5 Thus, the first four terms are:
  6. nth term of an AP The terms of an AP are, a is the first term and d is the common difference. The first term is: al = a The second term is. = a +d The third term is: ag = a +2d The nth term is: an — a+(n—l)d
  7. Example Problem 2 ? Find the 10th term of the AP: 2,7,12,... Solution: Here, (1=7 2=5 The formula for the nth term of an AP is: an = Substituting the values: = = 2+45-47 Therefore, the 10th term of the given AP is 47.
  8. Example Problem 3 ? Check whether 50 is a term in the AP: 5,10,15,20,... Solution: First term a = 5 and Common difference d = 10—5 = 5 To check whether 50 is a term in the AP, we use the formula for the nth term of an AP: Substitute the values: an = a+(n—l)d 50 = 50-5 45 = 50-1) n 1=9 n Therefore, 50 is the 10th term of the given AP.
  9. Sum of first n terms of an AP Let the terms of the arithmetic progression (AP) be: a, a +d, a+2d,... The n-th term of this AP is: an = a+(n—l)d Let S denote the sum of the first n terms of the AP Rewriting the terms in reverse order: (1) (2)
  10. Sum of first n terms of an AP contd. Adding equations (l) and (2) term-wise: Simplifying each term: Since every pair adds up to 2a + (n — l)d, there are n terms of this form, so we have: Dividing both sides by 2: Sn = — [2a+(n— 2 l)dl
  11. Sum of first n terms of an AP contd. Split 2a as, a+a Substitute into the equation• S Sn = [a+a+(n—l)d] 2 We have, an = a l)d Group the terms: Sn = [a+a+(n 2 Sn = — [a + an ] 2
  12. Example problem 4 ? Find the sum of the first 22 terms of the AP: Here, a = 8, d = 3—8 =—5, and n =22. We know that the sum of the first n terms of an AP is given by the formula: 2 Substitute the values a = 8, d = —5, and n - [2(8) + = 22 = 11[16-105] S22 = Il S22 — —979 herefore, the sum of the first 22 terms of the given AP is —979.
  13. l, 2,3,...,n is an AP where, Sum of the first n natural numbers an = n a (first term), d = I (common difference). The sum of the first n terms of an AP is given by the formula: Sn = — [a + an ] Where an is the n-th term 2 Substituting a = I and an = n into the sum formula: here, Sn = — (1+n) 2 Thus, the sum of the first n natural numbers is: n(n+l) 2
  14. Example Problem 5 ? If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term. Solution: Given, S14 = 1050, n -14, a=10 We use the formula for the sum of the first n terms of an AP, 2 Substituting the known values: = [2(10) + (14 -l)d] 1050 = 7[20+13d] 1050 = 140+91d
  15. Example Problem 5 contd. Simplifying: 1050-140 = 91d 910= Now, we find the 20th term using the formula for the n-th term of an AP: an = a+(n—l)d Substituting a = 10, d = 10, and n = 20 = 10+190 = 200 Therefore, the 20th term of the AP is 200.
  16. Example Problem 6 ? Find the sum of the first 24 terms of the list of numbers whose n-th term is given by an = 3+2n Solution: We are given the formula for the n-th term as: an = 3+2n Let's compute the first few terms: al = 3+2(1) = 5 = 3+2(2) = 7 3+2(4) = 11 The list of numbers is
  17. Example Problem 6 contd. The common difference d =7—5=2 Hence, this forms an arithmetic progression (AP) with, We use the formula for the sum of the first n terms of an AP: 2 Substitute n = 24, a = 5, and d = 2 sn = [2(5) +(24 = 12(10+46)
  18. Real Life Applications Daily Savings: If you save a fixed amount of money every day, your total savings over time form an arithmetic progression. Construction: In staircases, the heights of consecutive steps often follow an AP to ensure uniformity and ease of construction. Seating Arrangements: Rows in a stadium or theater are designed so that the number of seats in consecutive rows increases by a fixed amount, forming an AP. Natural Patterns: Growth rings in trees sometimes form an AP.

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