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Ex 9.2, 6 - If sum of a certain number of terms of AP 25, 22, 19 - Arithmetic Progression (AP): Formulae based

Ex 9.2, 6 - Chapter 9 Class 11 Sequences and Series - Part 2
Ex 9.2, 6 - Chapter 9 Class 11 Sequences and Series - Part 3

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Ex 9.2 , 6 If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term AP is of the form 25, 22, 19, … Here First term = a = 25 Common difference = d = 22 – 25 Sum of n terms = Sn = 116. We need to find last term an First, we find n We know that Sn = n/2 [2a + (n – 1)d] Putting values 116 = n/2 [2 × 25 + (n – 1)(-3)] 116 = n/2 [50 + (n – 1)(-3)] 116 × 2 = n[50 + (n – 1)(-3)] 232 = n[50 - 3n + 3] 232 = n [53 – 3n] 232 = 53n – 3n2 3n2 – 53n + 232 = 0 3n2 – 24n – 29n + 232 = 0 3n(n – 8) – 29(n – 8) = 0 (3n – 29)(n – 8) = 0 n = 29/3 is not possible as n cannot be a fraction So, n = 8 To find last term, we use the formula an = a + (n – 1)d Here, an = last term , a = 25 , n = 8 , d = – 3 an = 25 + (8 – 1) (– 3) = 25 + (7) (– 3) = 25 – 21 = 4 Hence, last term is 4

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.