Last updated at Dec. 8, 2016 by Teachoo

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Misc 11 A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio. Introduction Let G.P. 10,100,1000,10000,100000,1000000 Here, number of terms = 6 which is even Common ratio = 100/10 = 10 Sum of all terms = 10 + 100 + 1000 + 10000 + 100000 + 1000000 Sum of terms occupying odd places = 10 + 1000 + 100000 This is also a GP with common ratio = 1000/10 = 100 = 102 Misc 11 A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio. We know that Sn = (a(๐^๐โ 1))/(๐ โ 1) Since GP has even number of terms, We take number of terms as 2n (so that it is always even) Let G.P. be a, ar, ar2, ar3, โฆ to 2n terms Now, finding sum of all terms & sum of terms occupying odd places G.P. be a, ar, ar2, ar3, โฆ to 2n terms & Sn = (a(๐^๐โ 1))/(๐ โ 1) It is given that Sum of all the terms is 5 times the sum of terms occupying odd places i.e. S = 5S1 Putting values from (1) & (2) (a(๐^2๐ โ1))/(๐ โ 1) = 5 (a(๐^2๐ โ1))/(๐2 โ 1) using a2 โ b2 = (a โ b)(a + b) (a(๐^2๐ โ1))/(๐ โ 1) = 5(a(๐^2๐ โ1))/((๐ โ 1)(๐ + 1)) (a(๐^2๐ โ1))/(๐ โ 1) (๐โ1) = 5(a(๐^2๐ โ1))/((๐ + 1)) a(r2n โ 1) = 5(a(๐^2๐ โ1))/(๐ + 1) (a(๐^2๐ โ1))/(a(๐^2๐ โ1)) = 5/(๐ + 1) 1 = (5 )/(๐ + 1) 1 = (5 )/(๐ + 1) (r + 1) ร 1 = 5 r + 1 = 5 r = 5 โ 1 r = 4 Hence the common ratio is 4

Misc 1

Misc 2

Misc 3 Important

Misc 4

Misc 5

Misc 6

Misc 7 Important

Misc 8

Misc 9

Misc 10

Misc 11 You are here

Misc 12

Misc 13

Misc 14

Misc 15

Misc 16 Important

Misc 17

Misc 18

Misc 19 Important

Misc 20

Misc 21

Misc 22

Misc 23

Misc 24

Misc 25 Important

Misc 26

Misc 27

Misc 28 Important

Misc 29

Misc 30

Misc 31

Misc 32 Important

Chapter 9 Class 11 Sequences and Series

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.