Last updated at Dec. 8, 2016 by Teachoo

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Ex9.2 ,7 (Method 1) Find the sum to n terms of the A.P., whose kth term is 5k + 1. It is given that kth term = 5k + 1 ak = 5k + 1. Putting k = 1 a1 = 5 × 1 + 1 = 5 + 1 = 6 Similarly, Putting k = 2 a2 = 5 × 2 + 1 = 10 + 1 = 11 Putting k = 3 a3 = 5 × 3 + 1 = 15 + 1 = 16 Thus, our A.P. becomes 6 , 11 , 16 , …… Here, First term = a = 6 d = common difference = 11 – 6 = 5 We need to find sum of n term i.e. Sn Sn = n/2 [2a + (n – 1)d] putting values a = 6 , d = 5 Sn = n/2 [2 × 6 + (n – 1)5] = n/2 [12 + (n – 1)5] = n/2 [12 + 5n – 5] = n/2 [5n + 7] Hence, the sum of n terms of the A.P. is n/2 [5n + 7] Ex9.2 ,7 (Method 2) Find the sum to n terms of the A.P., whose kth term is 5k + 1. It is given that kth term = 5k + 1 ak = 5k + 1 To find sum of n terms, we use the formula Sn = n/2 [a + l] where Sn = sum of n terms of AP n = number of terms a = first term of A.P l = last term of A.P. Last term of this sequence will be an Finding first term i.e. a , Putting k = 1 in (1) a1 = 5 × 1 + 1 = 5 + 1 = 6 Finding last term , i.e. an Putting k = n in (1) an = 5n + 1 i.e. l = 5n + 1 Now, Sn = 𝑛/2[a + l] putting values a = 6 , l = 5n + 1 = 𝑛/2[6 + (5n + 1)] = 𝑛/2[6 + 5n + 1] = 𝑛/2[5n + 7] Hence, the sum of n terms of the A.P. is n/2 [5n + 7]

Chapter 9 Class 11 Sequences and Series

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.