

Get live Maths 1-on-1 Classs - Class 6 to 12
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 You are here
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
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 March 22, 2023 by Teachoo
Ex 9.2,4 How many terms of the A.P. 6, 11/2, 5 . are needed to give the sum 25? AP is of the form 6, 11/2, 5 . Here First term = a = 6 Common difference = d = 11/2 ( 6) = 11/2 + 6 = ( 11 + 12 )/2 = 1/2 & Sum of n terms = Sn = 25 We need to find n We know that Sn = n/2 [2a + (n 1)d] Here, Sn = 25 , a = 6 , d =1/2 Putting values 25 = n/2 ["2( 6) + (n 1)" (1/2)] 25 2 = n [" 12 + " n/2 " " 1/2] 50 = n [n/2 " 12 " 1/2] 50 = n [n/2 " " ((12(2) + 1)/2)] 50 = n [n/2 " " ((24 + 1)/2)] 50 = n [n/2 25/2] 50 = n [(n 25)/2] 50 2 = n(n 25) 100 = n2 25n 0 = n2 25n + 100 n2 25n + 100 = 0 n2 20n 5n + 100 = 0 n (n 20) 5 (n 20) = 0 (n 5)(n 20) = 0 Thus, n 5 = 0 , n 20 = 0 So, n = 5 , n = 20 Hence, n = 5 , 20 both values gives the required sum.