Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Arithmetic Progression
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams You are here
Question 8 Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 10 Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 12 Deleted for CBSE Board 2024 Exams
Question 13 Deleted for CBSE Board 2024 Exams
Question 14 Important Deleted for CBSE Board 2024 Exams
Question 15 Important Deleted for CBSE Board 2024 Exams
Question 16 Important Deleted for CBSE Board 2024 Exams
Question 17 Deleted for CBSE Board 2024 Exams
Question 18 Important Deleted for CBSE Board 2024 Exams
Arithmetic Progression
Last updated at May 29, 2023 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]