Ex 9.3, 14 - Sum of first three terms of GP is 16, sum of - Geometric Progression(GP): Formulae based

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise

Transcript

Ex 9 .3, 14 The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P. Given Now, Sn = (๐‘Ž(1 โˆ’ ๐‘Ÿ^๐‘›))/(1 โˆ’ ๐‘Ÿ ) where Sn = sum of n terms of GP n is the number of terms a is the first term r is the common ratio We know that S3 = 16 Putting n = 3 in Sn (๐‘Ž(1 โˆ’ ๐‘Ÿ3 ))/(1 โˆ’ ๐‘Ÿ ) = 16 Also, S6 = 144 Putting n = 6 in Sn (a(1 โˆ’ r6))/(1โˆ’ r) = 144 We need to find first term (a) & common ratio (r) Dividing (2) from (1) ((๐‘Ž(1 โˆ’ ๐‘Ÿ6))/(1โˆ’ ๐‘Ÿ))/((๐‘Ž(1 โˆ’ ๐‘Ÿ3))/(1 โˆ’ ๐‘Ÿ)) = 144/16 (๐‘Ž(1 โˆ’ ๐‘Ÿ6))/(1โˆ’ ๐‘Ÿ) ร— (1โˆ’ ๐‘Ÿ)/(๐‘Ž(1 โˆ’ ๐‘Ÿ3)) = 144/16 (๐‘Ž(1 โˆ’ ๐‘Ÿ6)(1โˆ’ ๐‘Ÿ))/(๐‘Ž(1 โˆ’ ๐‘Ÿ3)(1โˆ’ ๐‘Ÿ)) = 9/1 (1 โˆ’๐‘Ÿ6)/(1 โˆ’๐‘Ÿ3) = 9/1 (12 โˆ’(๐‘Ÿ3)2)/(1 โˆ’ ๐‘Ÿ3) = 9/1 Using a2 โ€“ b2 = (a โ€“ b)(a + b) ((1 โˆ’ ๐‘Ÿ3)(1 + ๐‘Ÿ3))/((1 โˆ’ ๐‘Ÿ3)) = 9/1 ((1+ ๐‘Ÿ3))/1 = 9/1 ((1+ ๐‘Ÿ3))/1 = 9/1 1 + r3 = 9 r3 = 9 โ€“ 1 r3 = 8 r =โˆ›8 r = โˆ›(2 ร—2 ร—2) r = 2 Hence common ratio r = 2 Putting r = 2 in (1) (๐‘Ž(1 โˆ’๐‘Ÿ3))/(1 โˆ’๐‘Ÿ) = 16 (๐‘Ž(1 โˆ’23))/(1 โˆ’2) = 16 (๐‘Ž(1 โˆ’8))/(โˆ’1) = 16 (๐‘Ž(1 โˆ’8))/(โˆ’1) = 16 (๐‘Ž(โˆ’7))/(โˆ’1) = 16 a(7) = 16 a = 16/7 Hence first term (a) =16/7 Now, a = 16/7 , r = 16/7 Now we need to find sum of n terms As r > 1 Sum of n terms of G.P = (๐‘Ž(๐‘Ÿ^๐‘› โˆ’ 1))/(๐‘Ÿ โˆ’ 1) Sum of n terms of G.P = (๐‘Ž(๐‘Ÿ^๐‘› โˆ’ 1))/(๐‘Ÿ โˆ’ 1) Putting a = 16/7 & r = 2 Sn = (16/7(2๐‘› โˆ’ 1 ))/(2 โˆ’ 1 ) = (16/7(2๐‘› โˆ’ 1 ))/1 = 16/7 (2n โ€“ 1) Hence sum of n terms is 16/7 (2n โ€“ 1)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.