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Ex 9.3, 16 - Find a GP for which sum of first two terms is -4 - Geometric Progression(GP): Formulae based

Ex 9.3, 16 - Chapter 9 Class 11 Sequences and Series - Part 2
Ex 9.3, 16 - Chapter 9 Class 11 Sequences and Series - Part 3 Ex 9.3, 16 - Chapter 9 Class 11 Sequences and Series - Part 4 Ex 9.3, 16 - Chapter 9 Class 11 Sequences and Series - Part 5

Ex 9.3, 16 - Chapter 9 Class 11 Sequences and Series - Part 6 Ex 9.3, 16 - Chapter 9 Class 11 Sequences and Series - Part 7 Ex 9.3, 16 - Chapter 9 Class 11 Sequences and Series - Part 8 Ex 9.3, 16 - Chapter 9 Class 11 Sequences and Series - Part 9

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Ex9.3, 16 (Method 1) Find a G.P. for which sum of the first two terms is –4 and the fifth term is 4 times the third term. Let a be the first term & r be the common ratio of G.P. It is given that Sum of first two term is = -4 i.e. S2 = -4 Also fifth term is 4 times of third term i.e. a5 = 4 × a3 We know that nth term of G.P. an = arn-1 a5 = ar5-1 a5 = ar4 & a3 = ar3-1 a3 = ar2 Putting a5 & a3 in (1) a5 = 4a3 ar4 = 4(ar2) "ar4" /"ar2" = 4 r2 = 4 r = ±√4 r = ± 2 ∴ r = 2 and r = -2 Taking r = 2 if r = 2 (>1) We know that sum of n terms = (𝑎(𝑟^𝑛− 1))/(𝑟 − 1) S2 = (𝑎(22 − 1))/(2 − 1) -4 = (𝑎(4 − 1))/1 -4 = (𝑎(4 − 1))/1 -4 = a(3) (−4 )/3 = a a = (−4 )/3 Now we have r = 2 & a = (−4)/3 So, Terms of G.P. are (−4)/3, (−4)/3 × 2, (−4)/3 × 2 × 2… = (−4)/3, (−8)/3, (−16)/3 … Taking r = -2 when r = -2 (r < 1) Sum of n terms = (a[1 − 𝑟^𝑛])/(1−r), Sn = (a[1 − (−2)^𝑛])/(1−(−2)) Sn = (a [1 −(−2)^𝑛])/(1+ 2) Given that S2 = -4 S2 = (a [1 −(−2)^2])/(1+ 2) -4 = (a[1 − 4])/(1 + 2) -4 = (−3𝑎)/3 -4 = -a 4 = a a = 4 When a = 4 & r = -2 Terms of G.P. are 4, 4 × (-2), 4 × (-2) (-2),… 4, -8, 16, -32,… Ex 8.2, 16 (Method 2) Find a G.P. for which sum of the first two terms is –4 and the fifth term is 4 times the third term. Let a be the first term & r be the common difference 1st term of G.P. is = a 2nd term of G.P. is = a × r = ar 3rd term of G.P. = a × r = ar2 4th term of G.P. = a × r = ar3 5rd term of G.P. = a × r = ar4 Hence terms of G.P are a, ar, ar2, ar3, ar4,… It is given that sum of first two term is = 4 i.e. a + ar = -4 a(1 + r) = -4 Slso, given that fifth term is 4 time of 3rd term i.e. a5 = 4a3 (ar4) = 4ar2 𝑎𝑟4/𝑎𝑟2 = 4 r2 = 4 r = ±√4 r = ±2 r = 2, r = -2 Putting r = 2 in (1) a(1 + 2) = -4 a(3) = -4 a = (−4)/3 When a = (−4)/3 & r = 2 Terms of G.P. are a, ar, ar2, ar3 = (−4)/3, (−4)/3 × 2, (−4)/3 × 22, (−4)/3 × 23, … = (−4)/3, (−8)/3, (−16)/3, (−32)/3, … Putting r = -2 in (1) a(1 + (-2)) = -4 a(1 – 2) = -4 a(-1) = -4 a = (−4)/(−1) = 4 When a = 4 & r = -2 Terms of G.P are a, ar, ar2, ar3 … = 4, 4(-2) , 4(-2)2, 4(-2)3 = 4, -8, 16, -32, 64…

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.