Last updated at May 29, 2018 by Teachoo

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Ex 5.3 ,10 Show that a1, a2 , an , form an AP where an is defined as below (i) an = 3 + 4n .Also find the sum of first 15 terms an = 3 + 4n Hence series is 7, 11, 15, .. Since difference is same, it is an AP Common difference = d = 4 Now we have to find sum of first 15 terms i.e. S15 We know that Sn = /2 (2 +( 1) ) Here, n = 15, a = 7, d = 4 Putting the values S15 = 15/2 (2 7+(15 1) 4) = 15/2 (14 + 14 4) = 15/2 (14+46) = 15/2 70 = 15 35 = 525 Ex 5.3 ,10 Show that a1, a2 , an , form an AP where an is defined as below (ii) an = 9 5n . Also find the sum of first 15 terms in each case an = 9 5n Hence , the series is 4, 1, 6, Since difference is same, it is an AP Common difference = d = 1 4 = 5 Now we have to find sum of first 15 terms i.e. S15 We know that Sn = /2 (2 +( 1) ) Here, n = 15, a = 7, d = 4 Putting the values Sn = /2 (2 +( 1) ) S15 = 15/2 (2 4+(15 1) 5) S15 = 15/2 (8+14 5) S15 = 15/2 (8 70) S15 = 15/2 62 S15 = 465

Chapter 5 Class 10 Arithmetic Progressions

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.