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

Misc 14 - Chapter 9 Class 11 Sequences and Series - Part 2
Misc 14 - Chapter 9 Class 11 Sequences and Series - Part 3 Misc 14 - Chapter 9 Class 11 Sequences and Series - Part 4 Misc 14 - Chapter 9 Class 11 Sequences and Series - Part 5 Misc 14 - Chapter 9 Class 11 Sequences and Series - Part 6

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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‘Ÿ^π‘›βˆ’ 1))/(r βˆ’ 1) ∴ S = (a(π‘Ÿ^π‘›βˆ’ 1))/(r βˆ’ 1) Now, finding P Now P is the product of n terms P = a1 Γ— a2 Γ— a3 Γ— … an = a Γ— ar Γ— ar2 Γ— ar3 … Γ— arn – 1 = (a Γ— a Γ— … a) Γ— (r Γ— r2 Γ— …rn–1) = an π‘Ÿ^(1+2+…+(π‘›βˆ’1)) = an π‘Ÿ^((𝑛(𝑛 βˆ’ 1))/2) Thus, P = an π‘Ÿ^((𝑛(𝑛 βˆ’ 1))/2) & R be the sum of reciprocals of n terms in GP i.e. R = 1/a + 1/ar + 1/ar^2 + ….+ to n terms 1/a + 1/ar + 1/ar^2 ….+ to n terms is a GP where first term = 1/π‘Ž & common ratio = 1/π‘Ÿ & sum of n terms = R We know that sum of n terms of a GP is Sn = (a(π‘Ÿ^π‘›βˆ’1))/(rβˆ’1) Putting r = 1/π‘Ÿ & a = 1/π‘Ž & S = R R = (1/π‘Ž ((1/π‘Ÿ)^𝑛 βˆ’ 1))/(1/π‘Ÿ βˆ’ 1) R = (1/π‘Ž ((1/π‘Ÿ)^𝑛 βˆ’ 1))/((1 βˆ’ π‘Ÿ)/π‘Ÿ ) = 1/π‘Ž ((1/π‘Ÿ)^π‘›βˆ’1)1Γ— (π‘Ÿ/(1 βˆ’ π‘Ÿ)) = 1/π‘Ž (π‘Ÿ/(1 βˆ’ π‘Ÿ))"Γ—" (1/π‘Ÿ^𝑛 βˆ’1) = 1/π‘Ž (π‘Ÿ/(1 βˆ’ π‘Ÿ))"Γ—" ((1 βˆ’ π‘Ÿ^𝑛)/π‘Ÿ^𝑛 ) = (π‘Ÿ(1 βˆ’π‘Ÿ^𝑛))/(π‘Ž(1 βˆ’ π‘Ÿ)π‘Ÿ^𝑛 ) We need to prove P2Rn = Sn Taking L.H.S P2 Rn = ["an . " π‘Ÿ^((𝑛(𝑛 βˆ’ 1))/2 ) ]^2 [((1 βˆ’π‘Ÿ^𝑛))/(π‘Ž(1 βˆ’ π‘Ÿ) π‘Ÿ^(𝑛 βˆ’ 1) )]^𝑛 = ["an . " π‘Ÿ^((𝑛(𝑛 βˆ’ 1))/2 ) ]^2 [((1 βˆ’π‘Ÿ^𝑛))/(π‘Ž(1 βˆ’ π‘Ÿ) π‘Ÿ^(𝑛 βˆ’ 1) )]^𝑛 = ["a2n . " π‘Ÿ^((𝑛(𝑛 βˆ’ 1))/2 Γ— 2) ] [(1 βˆ’π‘Ÿ^𝑛 )𝑛/((π‘Žπ‘› . (1 βˆ’ π‘Ÿ)𝑛〖(π‘Ÿγ€—^(𝑛 βˆ’ 1) )𝑛))] = (π‘Ž2𝑛 . π‘Ÿ^(𝑛(𝑛 βˆ’ 1)). (1 βˆ’π‘Ÿ^𝑛 )𝑛)/(π‘Žπ‘› .(1 βˆ’ π‘Ÿ)𝑛. (π‘Ÿ^(𝑛 βˆ’ 1)𝑛)) = (π‘Ž^(2𝑛 βˆ’ 𝑛) . π‘Ÿ^(𝑛(𝑛 βˆ’ 1) βˆ’ (𝑛 βˆ’ 1)𝑛). (1 βˆ’π‘Ÿ^𝑛 )𝑛)/(1 βˆ’ π‘Ÿ)𝑛 = (π‘Žπ‘› . π‘Ÿ0 . (1 βˆ’π‘Ÿ^𝑛 )𝑛)/(1 βˆ’ π‘Ÿ)𝑛 = (an . 1 .(1 βˆ’π‘Ÿ^𝑛 )n)/(1βˆ’r)n = an(π‘Ÿ^π‘›βˆ’1)n/(rβˆ’1)n = [a(1 βˆ’π‘Ÿ^𝑛 )/((rβˆ’1) )]^𝑛 Taking R.H.S Sn Putting values from (1) = [a(1 βˆ’π‘Ÿ^𝑛 )/((rβˆ’1) )]^𝑛 = L.H.S Hence, L.H.S = R.H.S Thus, P2 Rn = Sn Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.