**Ex 5.3, 10**

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

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 = 𝑛/2 (2𝑎+(𝑛−1)𝑑) 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 = 𝑛/2 (2𝑎+(𝑛−1)𝑑) Here, n = 15, a = 7, d = 4 Putting the values Sn = 𝑛/2 (2𝑎+(𝑛−1)𝑑) 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

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can contact him here.