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Miscellaneous

Misc 1
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Misc 11 You are here

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Last updated at Dec. 8, 2016 by Teachoo

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^πβ 1))/(π β 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, β¦ to 2n terms Now, finding sum of all terms & sum of terms occupying odd places G.P. be a, ar, ar2, ar3, β¦ to 2n terms & Sn = (a(π^πβ 1))/(π β 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^2π β1))/(π β 1) = 5 (a(π^2π β1))/(π2 β 1) using a2 β b2 = (a β b)(a + b) (a(π^2π β1))/(π β 1) = 5(a(π^2π β1))/((π β 1)(π + 1)) (a(π^2π β1))/(π β 1) (πβ1) = 5(a(π^2π β1))/((π + 1)) a(r2n β 1) = 5(a(π^2π β1))/(π + 1) (a(π^2π β1))/(a(π^2π β1)) = 5/(π + 1) 1 = (5 )/(π + 1) 1 = (5 )/(π + 1) (r + 1) Γ 1 = 5 r + 1 = 5 r = 5 β 1 r = 4 Hence the common ratio is 4