Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 9.3, 8 Find the sum to n terms in the geometric progression √7 ,√21 ,3√7…… √7 ,√21 ,3√7…… Here, First term a = √7 Common ratio r = √21/√7 = √(7 × 3)/√7 = (√7 ×√3 )/√7 = √3 So r = √3 ≈ 1.73 Since, r > 1 ∴ Sn = (𝑎(𝑟^𝑛 − 1))/(𝑟 − 1) Sn = (𝑎(𝑟^𝑛 − 1))/(𝑟 − 1) where Sn = sum of n terms of GP n is the number of terms a is the first term r is the common ratio Now, Sum of n terms = (𝑎(𝑟^𝑛 − 1))/(𝑟 − 1) Putting values a = √7 , r =√3 Sn = (√7 ((√3)^𝑛−1))/(√3 − 1) Rationalizing the same = (√7 ((√3)n−1 )])/(√3 − 1) x (√3 + 1)/(√3 + 1) = (√7 (√3 𝑛 −1) (√3+ 1))/((√3 −1) (√3+ 1)) = (√7 (√3 𝑛 −1) (√3+ 1))/((√3 −1) (√3+ 1)) Using a2 – b2 = (a + b)(a – b) = (√7 (3^(1/2 𝑛) −1)(√3 +1))/((√3)2 − 1^2 ) =(√7 (3^(𝑛/2) − 1) (√3 + 1))/2 = √7/2(√3+1) (3^(𝑛/2)−1) Hence sum of n term is √7/2(√3+1) (3^(𝑛/2) −1)

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Chapter 9 Class 11 Sequences and Series

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.