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

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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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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