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Last updated at Dec. 8, 2016 by Teachoo

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Misc 12 The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms. Let the first four terms of A.P. be a, a + d, a + 2d, a + 3d Given, First term = a = 11 Let common difference = d Also given that, Sum of first four term is 56 a + (a + d) + (a + 2d) + (a + 3d) = 56 11 + (11 + d) +(11 + 2d) +(11 + 3d) = 56 11 + 11 + 11 + 11 + d + 2d +3d = 56 44 + 6d = 56 6d = 56 – 44 6d = 12 d = 12/6 d = 2 Hence common difference d = 2 Last four terms are an, an – 1, an – 2, an – 3 We know that nth term of AP = a + (n – 1)d an = a + (n – 1)d Putting values a = 11 & d = 2 an = 11 + (n – 1)2 For an – 1 Putting n = n – 1 in (1) an – 1 = 11 + (n – 1 – 1)2 = 11 + (n – 2)2 For an – 2 Putting n = n – 2 in (1) an – 2 = 11 + (n – 2 – 1)2 an – 2 = 11 + (n – 3)2 For an – 3 Putting n = n – 3 in (1) an – 3 = 11 + (n – 3 – 1)2 an – 3 = 11 + (n – 4)2 It is given that sum of last four term is 112 i.e. an + an – 1 + an – 2 + an – 3 = 112 Putting values from (1), (2), (3), & (4) 11 + (n – 1)2 + 11 + (n – 2)2 + 11 + (n – 3)2 + 11 + (n – 4)2 = 112 11 + 11 + 11 + 11 + (n – 1)2 + (n – 2)2 + 11 + (n – 3)2 + (n – 4)2 = 112 44 + 2[n – 1 + n – 2 + n – 3 + n – 4] = 112 44 + 2[n + n + n + n – 1 – 2 – 3 – 4] = 112 44 + 2 [4n – 10] = 112 2 [4n – 10] = 112 – 44 2 [4n – 10] = 68 4n – 10 = 68/2 4n = 34 + 10 4n = 44 n = 44/4 n = 11 Hence number of terms is 11

Miscellaneous

Misc 1

Misc 2

Misc 3 Important

Misc 4

Misc 5

Misc 6

Misc 7 Important

Misc 8

Misc 9

Misc 10

Misc 11

Misc 12 You are here

Misc 13

Misc 14

Misc 15

Misc 16 Important

Misc 17

Misc 18

Misc 19 Important

Misc 20

Misc 21 Not in Syllabus - CBSE Exams 2021

Misc 22

Misc 23 Not in Syllabus - CBSE Exams 2021

Misc 24 Not in Syllabus - CBSE Exams 2021

Misc 25 Important Not in Syllabus - CBSE Exams 2021

Misc 26

Misc 27

Misc 28 Important

Misc 29

Misc 30

Misc 31

Misc 32 Important

Chapter 9 Class 11 Sequences and Series

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.