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Ex 9.2, 7 - Find sum n terms of AP, whose kth term is 5k + 1 - Ex 9.2

Ex 9.2, 7 - Chapter 9 Class 11 Sequences and Series - Part 2
Ex 9.2, 7 - Chapter 9 Class 11 Sequences and Series - Part 3

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Ex 9.2, 7 - Chapter 9 Class 11 Sequences and Series - Part 4

Ex 9.2, 7 - Chapter 9 Class 11 Sequences and Series - Part 5
Ex 9.2, 7 - Chapter 9 Class 11 Sequences and Series - Part 6

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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]

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.