Misc 14 - Let S be sum, P product, R sum of reciprocals in GP  - Miscellaneous

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise

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‘Ÿ^๐‘›โˆ’ 1))/(r โˆ’ 1) โˆด S = (a(๐‘Ÿ^๐‘›โˆ’ 1))/(r โˆ’ 1) Now, finding P Now P is the product of n terms P = a1 ร— a2 ร— a3 ร— โ€ฆ an = a ร— ar ร— ar2 ร— ar3 โ€ฆ ร— arn โ€“ 1 = (a ร— a ร— โ€ฆ a) ร— (r ร— r2 ร— โ€ฆrnโ€“1) = an ๐‘Ÿ^(1+2+โ€ฆ+(๐‘›โˆ’1)) = an ๐‘Ÿ^((๐‘›(๐‘› โˆ’ 1))/2) Thus, P = an ๐‘Ÿ^((๐‘›(๐‘› โˆ’ 1))/2) & R be the sum of reciprocals of n terms in GP i.e. R = 1/a + 1/ar + 1/ar^2 + โ€ฆ.+ to n terms 1/a + 1/ar + 1/ar^2 โ€ฆ.+ to n terms is a GP where first term = 1/๐‘Ž & common ratio = 1/๐‘Ÿ & sum of n terms = R We know that sum of n terms of a GP is Sn = (a(๐‘Ÿ^๐‘›โˆ’1))/(rโˆ’1) Putting r = 1/๐‘Ÿ & a = 1/๐‘Ž & S = R R = (1/๐‘Ž ((1/๐‘Ÿ)^๐‘› โˆ’ 1))/(1/๐‘Ÿ โˆ’ 1) R = (1/๐‘Ž ((1/๐‘Ÿ)^๐‘› โˆ’ 1))/((1 โˆ’ ๐‘Ÿ)/๐‘Ÿ ) = 1/๐‘Ž ((1/๐‘Ÿ)^๐‘›โˆ’1)1ร— (๐‘Ÿ/(1 โˆ’ ๐‘Ÿ)) = 1/๐‘Ž (๐‘Ÿ/(1 โˆ’ ๐‘Ÿ))"ร—" (1/๐‘Ÿ^๐‘› โˆ’1) = 1/๐‘Ž (๐‘Ÿ/(1 โˆ’ ๐‘Ÿ))"ร—" ((1 โˆ’ ๐‘Ÿ^๐‘›)/๐‘Ÿ^๐‘› ) = (๐‘Ÿ(1 โˆ’๐‘Ÿ^๐‘›))/(๐‘Ž(1 โˆ’ ๐‘Ÿ)๐‘Ÿ^๐‘› ) We need to prove P2Rn = Sn Taking L.H.S P2 Rn = ["an . " ๐‘Ÿ^((๐‘›(๐‘› โˆ’ 1))/2 ) ]^2 [((1 โˆ’๐‘Ÿ^๐‘›))/(๐‘Ž(1 โˆ’ ๐‘Ÿ) ๐‘Ÿ^(๐‘› โˆ’ 1) )]^๐‘› = ["an . " ๐‘Ÿ^((๐‘›(๐‘› โˆ’ 1))/2 ) ]^2 [((1 โˆ’๐‘Ÿ^๐‘›))/(๐‘Ž(1 โˆ’ ๐‘Ÿ) ๐‘Ÿ^(๐‘› โˆ’ 1) )]^๐‘› = ["a2n . " ๐‘Ÿ^((๐‘›(๐‘› โˆ’ 1))/2 ร— 2) ] [(1 โˆ’๐‘Ÿ^๐‘› )๐‘›/((๐‘Ž๐‘› . (1 โˆ’ ๐‘Ÿ)๐‘›ใ€–(๐‘Ÿใ€—^(๐‘› โˆ’ 1) )๐‘›))] = (๐‘Ž2๐‘› . ๐‘Ÿ^(๐‘›(๐‘› โˆ’ 1)). (1 โˆ’๐‘Ÿ^๐‘› )๐‘›)/(๐‘Ž๐‘› .(1 โˆ’ ๐‘Ÿ)๐‘›. (๐‘Ÿ^(๐‘› โˆ’ 1)๐‘›)) = (๐‘Ž^(2๐‘› โˆ’ ๐‘›) . ๐‘Ÿ^(๐‘›(๐‘› โˆ’ 1) โˆ’ (๐‘› โˆ’ 1)๐‘›). (1 โˆ’๐‘Ÿ^๐‘› )๐‘›)/(1 โˆ’ ๐‘Ÿ)๐‘› = (๐‘Ž๐‘› . ๐‘Ÿ0 . (1 โˆ’๐‘Ÿ^๐‘› )๐‘›)/(1 โˆ’ ๐‘Ÿ)๐‘› = (an . 1 .(1 โˆ’๐‘Ÿ^๐‘› )n)/(1โˆ’r)n = an(๐‘Ÿ^๐‘›โˆ’1)n/(rโˆ’1)n = [a(1 โˆ’๐‘Ÿ^๐‘› )/((rโˆ’1) )]^๐‘› Taking R.H.S Sn Putting values from (1) = [a(1 โˆ’๐‘Ÿ^๐‘› )/((rโˆ’1) )]^๐‘› = L.H.S Hence, L.H.S = R.H.S Thus, P2 Rn = Sn Hence proved

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.