Arithmetic progression ex no. 4
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1. The sum of the first 20 natural numbers is 210. 2. The sum of all odd natural numbers from 1 to 150 is 5625. 3. The sum of the terms of an AP with a = 6 and d = 3 up to the 10th term is 195.
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![1. Find the sum of the first n natural numbers and
hence find the sum of first 20 natural numbers.
Sol. The first n natural numbers are 1, 2, 3, 4, ......n
These natural number form an A.P. with a = 1, d = 1
t1 = 1 and tn = n
Alternative method :
Sn = n/2 [t1 + tn]
S20 = 20/2[t1 + t20]
= 10 [1 + 20]
= 10 [21]
S20 = 210
The sum of first twenty terms is 210.](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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1) An arithmetic progression (AP) is a sequence of numbers where the difference between consecutive terms is constant.
2) The general formula for the nth term of an AP is an = a + (n-1)d, where a is the first term and d is the common difference.
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- The sum of the first n terms (Sn) can be calculated using the formula n/2[2a + (n - 1)×d]
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This document provides an introduction to sequences and series. It begins with definitions of sequences, finite and infinite sequences, and series. It then covers topics like arithmetic progressions, geometric progressions, and harmonic progressions. It provides formulas for the nth term and sum of terms for arithmetic and geometric progressions. It also defines arithmetic mean and geometric mean between terms in progressions. The document aims to help secondary students understand key concepts related to sequences and series.
Arithmetic series
The document discusses arithmetic series and provides examples of calculating the sum of arithmetic series (Sn) given various inputs like the first term (a1), the common difference (d), and the number of terms (n). The key formulas explained are an=a1+(n-1)d to find subsequent terms, and Sn=n/2(a1+an) or Sn=n/2[2a1+(n-1)d] to calculate the sum. Several examples are worked through step-by-step to demonstrate applying the formulas to problems involving finding individual terms, the last term, or the entire sum of an arithmetic series.
(677528443) neethu text (2)
This document provides an introduction to sequences and series. It begins with definitions of sequences, series, arithmetic progressions, geometric progressions and their key terms. It then presents several examples of finding terms in arithmetic and geometric progressions. The document also defines arithmetic mean and geometric mean, and discusses how to find the sum of terms in an arithmetic progression using formulas. Overall, the document serves as an introductory guide to common concepts involving sequences and progressions.
Arithmetic progressions
The document defines arithmetic progressions and provides examples. Some key points:
- An arithmetic progression is a list of numbers where each term is obtained by adding a fixed number (called the common difference) to the preceding term.
- The general formula for the nth term of an AP is an = a + (n-1)d, where a is the first term and d is the common difference.
- Examples show how to determine if a list of numbers forms an AP, write the next terms, and find a specific term by using the general formula.
- There are finite APs, which have a last term, and infinite APs, which go on forever without a last term.
Starr pvt. ltd. rachit's group ppt (1)
The document discusses arithmetic progressions (AP), which are sequences where the difference between consecutive terms is constant. It provides examples of AP sequences and formulas to find the nth term, the sum of terms, and other properties. Some key points:
- The nth term of an AP is given by an = a + (n-1)d, where a is the first term and d is the common difference.
- The sum of the first n terms is given by Sn = n/2 * [2a + (n-1)d].
- Examples show how to use the formulas to find individual terms, sums, number of terms, etc. given information about an AP sequence.
- Questions at
2e. Pedagogy of Mathematics - Part II (Numbers and Sequence - Ex 2.5)
Pedagogy of Mathematics - Part II (Numbers and Sequence - Ex 2.5, Numbers and Sequence, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, Arithmetic progression, definition of arithmetic progression, terms and common difference of an A.P., In an Arithmetic progression, conditions for three numbers to be in A.P.,
arithmatic progression.pptx
The document defines and provides examples of arithmetic progressions (AP). It states that an AP is a sequence where the difference between consecutive terms is constant. The key characteristics of an AP include:
- The first term is denoted as a
- The common difference is denoted as d
- The nth term is calculated as an = a + (n-1)d
Several examples are provided to illustrate how to determine if a sequence is an AP and how to calculate individual terms and the sum of terms in an AP using the above formulae.
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2e. Pedagogy of Mathematics - Part II (Numbers and Sequence - Ex 2.5)
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Arithmetic progression ex no. 4
- 1. EX. NO. 4
- 2. 1. Find the sum of the first n natural numbers and hence find the sum of first 20 natural numbers. Sol. The first n natural numbers are 1, 2, 3, 4, ......n These natural number form an A.P. with a = 1, d = 1 t1 = 1 and tn = n Alternative method : Sn = n/2 [t1 + tn] S20 = 20/2[t1 + t20] = 10 [1 + 20] = 10 [21] S20 = 210 The sum of first twenty terms is 210.
- 3. 2. Find the sum of all odd natural numbers from 1 to 150. Sol. The odd natural numbers from 1 to 150 are 1, 3, 5, 7, 9, .........., 149. These numbers form an A.P. with a = 1, d = 2 Let, 149 be nth term of an A.P. tn = a + (n – 1) d 149 = 1 + (n – 1) 2 = 1 + 2n – 2 = 2n – 1 149 + 1 = 2n 2n = 150 n = 75 ∴ 149 is 75th term of A.P.
- 4. Sn = n/ 2 [t1 + tn] S75 = 75/2[t1 + t75] = 75/ 2[ [1 + 149] = 75/2[ [150] S75 = 75 x 75 =5625
- 5. 3. Find S10 if a = 6 and d = 3. Sol. For an A.P. a = 6, d = 3 n/ [2a + (n – 1)d] Sn = 2 S10 = 10/2 [2(6) + (10 – 1 ) 3] S10 = 5 [2 (6) + 9 (3)] S10 = 5 (12 + 27) S10 = 5 (39) S10 = 195
- 6. 4. Find the sum of all numbers from 1 to 140 which are divisible by 4. Sol. The natural numbers from 1 to 140 that are divisible by 4 are 4, 8, 12, 16, .............., 140 Here, a = 4, d = t2 – t1 = 8 – 4 = 4 and tn = 140 tn = a + (n – 1) d ∴ 140 = 4 + (n – 1) 4 ∴ 140 = 4 + 4n – 4 ∴ 140 = 4n ∴ n = 140 /4 ∴ n = 35
- 7. n/ [2a + (n – 1)d] Sn = 2 35/ [2a + (35 – 1 ) d] ∴ S35 = 2 35/ [2 (4) + 34 (4)] ∴ S35 = 2 35/ (8 + 136) ∴ S35 = 2 35/ (144) ∴ S35 = 2 ∴ S35 = 35 x 72 ∴ S35 = 2520
- 8. 5. Find the sum of the first n odd natural numbers. Hence find 1 + 3 + 5 + ... + 101. Sol. The first n odd natural numbers are 1, 3, 5, 7, ............., n Here, a = 1, d = t2 – t1 = 3 – 1 = 2 Sn = n/2 [2a + (n – 1)d] ∴ Sn = n/2 [2(1) + (n – 1)(2)] ∴ Sn = n/2 (2 + 2n – 2) ∴ Sn = n/2 (2n) ∴ Sn = n2 ........ eq. (1)
- 9. Now, we have 1 + 3 + 5 + ........ + 101 Let, tn = 101 tn = a + (n – 1) d ∴ 101 = a + (n – 1) d ∴ 101 = 1 + (n – 1) 2 ∴ 101 = 1 + 2n – 2 ∴ 101 = 2n – 1 ∴ 101 + 1 = 2n ∴ 2n = 102 ∴ n = 102/2 ∴ n = 51
- 10. Now, 101 is the 51st term of A.P., We have to find sum of 51 terms i.e. S51, 2 [From (i)] Sn = n ∴ S51 = (51)2 ∴ S51 = 2601
- 11. 6. Obtain the sum of the 56 terms of an A. P. whose 19th and 38th terms are 52 and 148 respectively. Sol. t19 = 52, t38 = 148 tn = a + (n – 1) d ∴ t19 = a + (19 – 1) d ∴ 52 = a + 18d ......(i) ∴ t38 = a + (38 – 1) d ∴ 148 = a + 37d ......(ii)
- 12. Adding (i) and (ii) we get, a + 18d = 52 a + 37d = 148 . 2a + 55d = 200 . ..........Eq. no. (iii) Sn = n/2[2a + (n – 1)d] ∴ S56 = 56/2 [2a + (56 – 1) d] ∴ S56 = 28[2a + 55d] ∴ S56 = 28[200] [from eq. no. (iii) ∴ S56 = 5600 ∴ Sum of first 56 terms of A.P. is 5600.
- 13. 7. The sum of the first 55 terms of an A. P. is 3300. Find the 28th term. Sol. S55 = 3300 [Given] Sn = n/2[2a + (n – 1)d] ∴ S55 = 55/2[2a + (55 – 1) d] ∴ 3300 = 55/2[2a + 54d] 55/ × (2)[a + 27d] ∴ 3300 = 2 ∴ 3300 = 55[a + 27d] 3300/ = a + 27d 55 ∴ a + 27d= 60 ......eq. no. (1)
- 14. But, tn = a + (n – 1) d ∴ t28 = a + (28 – 1) d ∴ t28 = a + 27d ∴ t28 = 60 [From (i)] ∴ Twenty eighth term of A.P. is 60.
- 15. 8. Find the sum of the first n even natural numbers. Hence find the sum of first 20 even natural numbers. Sol. The first n even natural numbers are 2, 4, 6, 8, ..... Here, a = 2, d = t2 – t1 = 4 – 2 = 2 ∴ Sn = n/2[2a + (n – 1)d] ∴ Sn = n/2[2 (2) + (n – 1) 2] ∴ Sn = n/2[4 + 2n – 2] ∴ Sn = n/2[2n + 2] ∴ Sn =n/2 ×(2) (n + 1) ∴ Sn = n (n + 1)
- 16. ∴ S20 = 20 (20 + 1) ∴ S20 = 20 (21) ∴ S20 = 420 ∴ Sum of first twenty even natural numbers is 420