Slide18.JPG

Slide19.JPG

Slide20.JPG Slide21.JPG Slide22.JPG Slide23.JPG

 

 

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Misc 5 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 5 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(𝑟^𝑛− 1))/(𝑟 − 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(𝑟^𝑛− 1))/(𝑟 − 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(𝑟^2𝑛 −1))/(𝑟 − 1) = 5 (a(𝑟^2𝑛 −1))/(𝑟2 − 1) using a2 – b2 = (a – b)(a + b) (a(𝑟^2𝑛 −1))/(𝑟 − 1) = 5(a(𝑟^2𝑛 −1))/((𝑟 − 1)(𝑟 + 1)) (a(𝑟^2𝑛 −1))/(𝑟 − 1) (𝑟−1) = 5(a(𝑟^2𝑛 −1))/((𝑟 + 1)) a(r2n – 1) = 5(a(𝑟^2𝑛 −1))/(𝑟 + 1) (a(𝑟^2𝑛 −1))/(a(𝑟^2𝑛 −1)) = 5/(𝑟 + 1) 1 = (5 )/(𝑟 + 1) 1 = (5 )/(𝑟 + 1) (r + 1) × 1 = 5 r + 1 = 5 r = 5 – 1 r = 4 Hence the common ratio is 4

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.