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

Ex 9.3, 14 - Chapter 9 Class 11 Sequences and Series - Part 2
Ex 9.3, 14 - Chapter 9 Class 11 Sequences and Series - Part 3 Ex 9.3, 14 - Chapter 9 Class 11 Sequences and Series - Part 4 Ex 9.3, 14 - Chapter 9 Class 11 Sequences and Series - Part 5 Ex 9.3, 14 - Chapter 9 Class 11 Sequences and Series - Part 6


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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 = (𝑎(1 − 𝑟^𝑛))/(1 − 𝑟 ) 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 (𝑎(1 − 𝑟3 ))/(1 − 𝑟 ) = 16 Also, S6 = 144 Putting n = 6 in Sn (a(1 − r6))/(1− r) = 144 We need to find first term (a) & common ratio (r) Dividing (2) from (1) ((𝑎(1 − 𝑟6))/(1− 𝑟))/((𝑎(1 − 𝑟3))/(1 − 𝑟)) = 144/16 (𝑎(1 − 𝑟6))/(1− 𝑟) × (1− 𝑟)/(𝑎(1 − 𝑟3)) = 144/16 (𝑎(1 − 𝑟6)(1− 𝑟))/(𝑎(1 − 𝑟3)(1− 𝑟)) = 9/1 (1 −𝑟6)/(1 −𝑟3) = 9/1 (12 −(𝑟3)2)/(1 − 𝑟3) = 9/1 Using a2 – b2 = (a – b)(a + b) ((1 − 𝑟3)(1 + 𝑟3))/((1 − 𝑟3)) = 9/1 ((1+ 𝑟3))/1 = 9/1 ((1+ 𝑟3))/1 = 9/1 1 + r3 = 9 r3 = 9 – 1 r3 = 8 r =∛8 r = ∛(2 ×2 ×2) r = 2 Hence common ratio r = 2 Putting r = 2 in (1) (𝑎(1 −𝑟3))/(1 −𝑟) = 16 (𝑎(1 −23))/(1 −2) = 16 (𝑎(1 −8))/(−1) = 16 (𝑎(1 −8))/(−1) = 16 (𝑎(−7))/(−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 = (𝑎(𝑟^𝑛 − 1))/(𝑟 − 1) Sum of n terms of G.P = (𝑎(𝑟^𝑛 − 1))/(𝑟 − 1) Putting a = 16/7 & r = 2 Sn = (16/7(2𝑛 − 1 ))/(2 − 1 ) = (16/7(2𝑛 − 1 ))/1 = 16/7 (2n – 1) Hence sum of n terms is 16/7 (2n – 1)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.