Last updated at May 29, 2018 by Teachoo

Transcript

Misc 14 Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn Let a be the first term of GP & r be the common ratio of GP We know that Sum of n term of GP = (a(๐^๐โ 1))/(r โ 1) โด S = (a(๐^๐โ 1))/(r โ 1) Now, finding P Now P is the product of n terms P = a1 ร a2 ร a3 ร โฆ an = a ร ar ร ar2 ร ar3 โฆ ร arn โ 1 = (a ร a ร โฆ a) ร (r ร r2 ร โฆrnโ1) = an ๐^(1+2+โฆ+(๐โ1)) = an ๐^((๐(๐ โ 1))/2) Thus, P = an ๐^((๐(๐ โ 1))/2) & R be the sum of reciprocals of n terms in GP i.e. R = 1/a + 1/ar + 1/ar^2 + โฆ.+ to n terms 1/a + 1/ar + 1/ar^2 โฆ.+ to n terms is a GP where first term = 1/๐ & common ratio = 1/๐ & sum of n terms = R We know that sum of n terms of a GP is Sn = (a(๐^๐โ1))/(rโ1) Putting r = 1/๐ & a = 1/๐ & S = R R = (1/๐ ((1/๐)^๐ โ 1))/(1/๐ โ 1) R = (1/๐ ((1/๐)^๐ โ 1))/((1 โ ๐)/๐ ) = 1/๐ ((1/๐)^๐โ1)1ร (๐/(1 โ ๐)) = 1/๐ (๐/(1 โ ๐))"ร" (1/๐^๐ โ1) = 1/๐ (๐/(1 โ ๐))"ร" ((1 โ ๐^๐)/๐^๐ ) = (๐(1 โ๐^๐))/(๐(1 โ ๐)๐^๐ ) We need to prove P2Rn = Sn Taking L.H.S P2 Rn = ["an . " ๐^((๐(๐ โ 1))/2 ) ]^2 [((1 โ๐^๐))/(๐(1 โ ๐) ๐^(๐ โ 1) )]^๐ = ["an . " ๐^((๐(๐ โ 1))/2 ) ]^2 [((1 โ๐^๐))/(๐(1 โ ๐) ๐^(๐ โ 1) )]^๐ = ["a2n . " ๐^((๐(๐ โ 1))/2 ร 2) ] [(1 โ๐^๐ )๐/((๐๐ . (1 โ ๐)๐ใ(๐ใ^(๐ โ 1) )๐))] = (๐2๐ . ๐^(๐(๐ โ 1)). (1 โ๐^๐ )๐)/(๐๐ .(1 โ ๐)๐. (๐^(๐ โ 1)๐)) = (๐^(2๐ โ ๐) . ๐^(๐(๐ โ 1) โ (๐ โ 1)๐). (1 โ๐^๐ )๐)/(1 โ ๐)๐ = (๐๐ . ๐0 . (1 โ๐^๐ )๐)/(1 โ ๐)๐ = (an . 1 .(1 โ๐^๐ )n)/(1โr)n = an(๐^๐โ1)n/(rโ1)n = [a(1 โ๐^๐ )/((rโ1) )]^๐ Taking R.H.S Sn Putting values from (1) = [a(1 โ๐^๐ )/((rโ1) )]^๐ = L.H.S Hence, L.H.S = R.H.S Thus, P2 Rn = Sn Hence proved

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

Misc 13

Misc 14 You are here

Misc 15

Misc 16 Important

Misc 17

Misc 18

Misc 19 Important

Misc 20

Misc 21

Misc 22

Misc 23

Misc 24 Deleted for CBSE Board 2022 Exams

Misc 25 Important Deleted for CBSE Board 2022 Exams

Misc 26 Deleted for CBSE Board 2022 Exams

Misc 27

Misc 28 Important

Misc 29

Misc 30

Misc 31

Misc 32 Important

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.