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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.