# Ex 9.2, 5

Last updated at March 9, 2017 by Teachoo

Last updated at March 9, 2017 by Teachoo

Transcript

Ex 9.2,5 In an A.P., if pth term is 1/q and qth term is 1/p , prove that the sum of first pq terms is 1/2 (pq + 1) where p ≠ q. We know that an = a + (n – 1)d Where an is nth term of AP, n is the number of terms, a be the first term & d be the common difference of the A.P. It is given that pth term is 1/𝑞 i.e. ap = 1/q a + (p – 1)d = 1/q Also, qth term of A.P = 1/p i.e. aq = 1/p i.e. a + (q – 1)d = 1/p Now subtracting (1) from (2) i.e. (1) – (2) [a + (p – 1)d] – [a + (q – 1)d] = 1/𝑞 – 1/𝑝 a + pd – d – [a + qd – d] = 1/𝑞 – 1/𝑝 a + pd – d – a – qd + d = 1/𝑞 – 1/𝑝 a – a – d + d + pd – qd = 1/𝑞 – 1/𝑝 0 + 0 + pd – qd = 1/𝑞 – 1/𝑝 d(p – q) = 1/𝑞 – 1/𝑝 d(p – q) = (𝑝 − 𝑞)/𝑝𝑞 d = (𝑝 − 𝑞)/(𝑝𝑞(𝑝 −𝑞)) d = 1/𝑝𝑞 Now finding first term i.e. a Putting d = 1/𝑝𝑞 in (1) 1/𝑞 = a + (p – 1)d 1/𝑞 = a + (p – 1)1/𝑝𝑞 1/𝑞 = a + 𝑝/𝑝𝑞 – 1/𝑝𝑞 1/𝑞 = a + 1/𝑞 – 1/𝑝𝑞 1/𝑞 – 1/𝑞 = a – 1/𝑝𝑞 0 = a – 1/𝑝𝑞 1/𝑝𝑞 = a a = 1/𝑝𝑞 Therefore, a = 1/𝑝𝑞 Thus, a = 1/𝑝𝑞 & d = 1/𝑝𝑞 We need to show sum of first pq term is 1/2(pq + 1) i.e. Spq = 1/2 (pq + 1) We know that Sn = n/2 ( 2a + (n – 1)d ) Where, Sn = sum of n terms of A.P. n = number of terms a = first term and d = common difference For sum of first pq terms, Putting n = pq , a = 1/𝑝𝑞 , d = 1/𝑝𝑞 Spq = pq/2 ["2 × " 1/𝑝𝑞 " + (pq – 1)" 1/𝑝𝑞] = pq/2 [2/𝑝𝑞 " + " 𝑝𝑞/𝑝𝑞 −1/𝑝𝑞] = pq/2 [(2 + 𝑝𝑞 − 1 )/𝑝𝑞] = pq/2 [(1 + 𝑝𝑞 )/𝑝𝑞] = 1/2 [pq + 1] Thus, sum of first pq terms is 1/2 (pq+1) Hence proved

Example 6
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Ex 9.2, 5 Important You are here

Ex 9.2, 9 Important

Ex 9.2, 15 Important

Ex 9.2, 17 Important

Example 14 Important

Example 15 Important

Ex 9.3, 3 Important

Ex 9.3, 11 Important

Ex 9.3, 17 Important

Ex 9.3, 18 Important

Ex 9.3, 22 Important

Ex 9.3, 28 Important

Ex 9.3, 29 Important

Ex 9.4.4 Important

Ex 9.4, 7 Important

Ex 9.4, 9 Important

Ex 9.4, 10 Important

Example 23 Important

Misc 16 Important

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Misc 25 Important

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Misc 32 Important

Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.