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

 

  1. Chapter 9 Class 11 Sequences and Series
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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(๐‘Ÿ^๐‘›โˆ’ 1))/(๐‘Ÿ โˆ’ 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, โ€ฆ to 2n terms Now, finding sum of all terms & sum of terms occupying odd places G.P. be a, ar, ar2, ar3, โ€ฆ to 2n terms & Sn = (a(๐‘Ÿ^๐‘›โˆ’ 1))/(๐‘Ÿ โˆ’ 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(๐‘Ÿ^2๐‘› โˆ’1))/(๐‘Ÿ โˆ’ 1) = 5 (a(๐‘Ÿ^2๐‘› โˆ’1))/(๐‘Ÿ2 โˆ’ 1) using a2 โ€“ b2 = (a โ€“ b)(a + b) (a(๐‘Ÿ^2๐‘› โˆ’1))/(๐‘Ÿ โˆ’ 1) = 5(a(๐‘Ÿ^2๐‘› โˆ’1))/((๐‘Ÿ โˆ’ 1)(๐‘Ÿ + 1)) (a(๐‘Ÿ^2๐‘› โˆ’1))/(๐‘Ÿ โˆ’ 1) (๐‘Ÿโˆ’1) = 5(a(๐‘Ÿ^2๐‘› โˆ’1))/((๐‘Ÿ + 1)) a(r2n โ€“ 1) = 5(a(๐‘Ÿ^2๐‘› โˆ’1))/(๐‘Ÿ + 1) (a(๐‘Ÿ^2๐‘› โˆ’1))/(a(๐‘Ÿ^2๐‘› โˆ’1)) = 5/(๐‘Ÿ + 1) 1 = (5 )/(๐‘Ÿ + 1) 1 = (5 )/(๐‘Ÿ + 1) (r + 1) ร— 1 = 5 r + 1 = 5 r = 5 โ€“ 1 r = 4 Hence the common ratio is 4

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