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Last updated at Sept. 3, 2021 by Teachoo

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(π^πβ 1))/(r β 1) β΄ S = (a(π^πβ 1))/(r β 1) Now, finding P Now P is the product of n terms P = a1 Γ a2 Γ a3 Γ β¦ an = a Γ ar Γ ar2 Γ ar3 β¦ Γ arn β 1 = (a Γ a Γ β¦ a) Γ (r Γ r2 Γ β¦rnβ1) = an π^(1+2+β¦+(πβ1)) = an π^((π(π β 1))/2) Thus, P = an π^((π(π β 1))/2) & R be the sum of reciprocals of n terms in GP i.e. R = 1/a + 1/ar + 1/ar^2 + β¦.+ to n terms 1/a + 1/ar + 1/ar^2 β¦.+ to n terms is a GP where first term = 1/π & common ratio = 1/π & sum of n terms = R We know that sum of n terms of a GP is Sn = (a(π^πβ1))/(rβ1) Putting r = 1/π & a = 1/π & S = R R = (1/π ((1/π)^π β 1))/(1/π β 1) R = (1/π ((1/π)^π β 1))/((1 β π)/π ) = 1/π ((1/π)^πβ1)1Γ (π/(1 β π)) = 1/π (π/(1 β π))"Γ" (1/π^π β1) = 1/π (π/(1 β π))"Γ" ((1 β π^π)/π^π ) = (π(1 βπ^π))/(π(1 β π)π^π ) We need to prove P2Rn = Sn Taking L.H.S P2 Rn = ["an . " π^((π(π β 1))/2 ) ]^2 [((1 βπ^π))/(π(1 β π) π^(π β 1) )]^π = ["an . " π^((π(π β 1))/2 ) ]^2 [((1 βπ^π))/(π(1 β π) π^(π β 1) )]^π = ["a2n . " π^((π(π β 1))/2 Γ 2) ] [(1 βπ^π )π/((ππ . (1 β π)πγ(πγ^(π β 1) )π))] = (π2π . π^(π(π β 1)). (1 βπ^π )π)/(ππ .(1 β π)π. (π^(π β 1)π)) = (π^(2π β π) . π^(π(π β 1) β (π β 1)π). (1 βπ^π )π)/(1 β π)π = (ππ . π0 . (1 βπ^π )π)/(1 β π)π = (an . 1 .(1 βπ^π )n)/(1βr)n = an(π^πβ1)n/(rβ1)n = [a(1 βπ^π )/((rβ1) )]^π Taking R.H.S Sn Putting values from (1) = [a(1 βπ^π )/((rβ1) )]^π = L.H.S Hence, L.H.S = R.H.S Thus, P2 Rn = Sn Hence proved