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Misc 11 - A GP consists of even number of terms. If sum - Geometric Progression(GP): Formulae based

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

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Misc 11 A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio. Introduction Let G.P. 10,100,1000,10000,100000,1000000 Here, number of terms = 6 which is even Common ratio = 100/10 = 10 Sum of all terms = 10 + 100 + 1000 + 10000 + 100000 + 1000000 Sum of terms occupying odd places = 10 + 1000 + 100000 This is also a GP with common ratio = 1000/10 = 100 = 102 Misc 11 A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio. We know that Sn = (a(π‘Ÿ^π‘›βˆ’ 1))/(π‘Ÿ βˆ’ 1) Since GP has even number of terms, We take number of terms as 2n (so that it is always even) Let G.P. be a, ar, ar2, ar3, … to 2n terms Now, finding sum of all terms & sum of terms occupying odd places G.P. be a, ar, ar2, ar3, … to 2n terms & Sn = (a(π‘Ÿ^π‘›βˆ’ 1))/(π‘Ÿ βˆ’ 1) It is given that Sum of all the terms is 5 times the sum of terms occupying odd places i.e. S = 5S1 Putting values from (1) & (2) (a(π‘Ÿ^2𝑛 βˆ’1))/(π‘Ÿ βˆ’ 1) = 5 (a(π‘Ÿ^2𝑛 βˆ’1))/(π‘Ÿ2 βˆ’ 1) using a2 – b2 = (a – b)(a + b) (a(π‘Ÿ^2𝑛 βˆ’1))/(π‘Ÿ βˆ’ 1) = 5(a(π‘Ÿ^2𝑛 βˆ’1))/((π‘Ÿ βˆ’ 1)(π‘Ÿ + 1)) (a(π‘Ÿ^2𝑛 βˆ’1))/(π‘Ÿ βˆ’ 1) (π‘Ÿβˆ’1) = 5(a(π‘Ÿ^2𝑛 βˆ’1))/((π‘Ÿ + 1)) a(r2n – 1) = 5(a(π‘Ÿ^2𝑛 βˆ’1))/(π‘Ÿ + 1) (a(π‘Ÿ^2𝑛 βˆ’1))/(a(π‘Ÿ^2𝑛 βˆ’1)) = 5/(π‘Ÿ + 1) 1 = (5 )/(π‘Ÿ + 1) 1 = (5 )/(π‘Ÿ + 1) (r + 1) Γ— 1 = 5 r + 1 = 5 r = 5 – 1 r = 4 Hence the common ratio is 4

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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.