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Ex 9.2, 4 - How many terms of AP -6, -11/2, -5 ... are needed - Ex 9.2

Ex 9.2, 4 - Chapter 9 Class 11 Sequences and Series - Part 2
Ex 9.2, 4 - Chapter 9 Class 11 Sequences and Series - Part 3

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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.

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Davneet Singh

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.