Solve all your doubts with Teachoo Black (new monthly pack available now!)
Are you in school? Do you love Teachoo?
We would love to talk to you! Please fill this form so that we can contact you
Ex 9.2
Ex 9.2, 2 Deleted for CBSE Board 2023 Exams
Ex 9.2, 3 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 4 Deleted for CBSE Board 2023 Exams
Ex 9.2, 5 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 6 Deleted for CBSE Board 2023 Exams
Ex 9.2, 7 Important Deleted for CBSE Board 2023 Exams You are here
Ex 9.2, 8 Deleted for CBSE Board 2023 Exams
Ex 9.2, 9 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 10 Deleted for CBSE Board 2023 Exams
Ex 9.2, 11 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 12 Deleted for CBSE Board 2023 Exams
Ex 9.2, 13 Deleted for CBSE Board 2023 Exams
Ex 9.2, 14 Important
Ex 9.2, 15 Important
Ex 9.2, 16 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 17 Deleted for CBSE Board 2023 Exams
Ex 9.2, 18 Important Deleted for CBSE Board 2023 Exams
Ex 9.2
Last updated at Sept. 3, 2021 by Teachoo
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]