Last updated at May 29, 2018 by Teachoo
Transcript
Ex 9.3, 21 Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18. Let a be the first term of G.P r be the common ratio of the G.P. We know that nth term of G.P = arn–1 i.e. an = arn–1 It is given that Third term is greater than the 1st term by 9 i.e. a3 = a1 + 9 ar3 – 1 = a + 9 ar2 = a + 9 Also given that 2nd term is greater than the 4th by 18 a2 = a4 + 18 ar2–1 = ar4–1 + 18 ar = ar3 + 18 We need to calculate a & r Multiplying (1) by r r(ar2) = r(a + 9) ar3 = ar + 9r Putting ar3 = ar + 9r in (2) ar = (ar + 9r) 18 ar – ar = 9r + 18 0 = 9r + 18 -9r = 18 r = 18/(−9) r = -2 put r = -2 in (1) ar2 = a + 9 a(–2)2 = a + 9 a × 4 = a + 9 4a = a + 9 4a – a = 9 4a – a = 9 3a = 9 a = 9/3 a = 3 Hence a = 3 & r = -2 We know that nth term of G.P = arn–1 1st term of G.P = a = 3 2nd term of G.P = ar2 – 1 a2 = 3 × (-2)2 – 1 = 3 × (-2) = -6 3rd term of G.P = ar3 – 1 a3 = 3(-2)3 – 1 = 3(-2)2 = 3 × 4 = 12 4th term of G.P = ar4 – 1 a4 = (3)(-2)4 – 1 = 3(-2)3 = 3(-2 × -2 × -2) = 3 × (-8) = -24 Hence G.P is 3, -6, 12, -24 …
Ex 9.3
Ex 9.3, 2
Ex 9.3, 3 Important
Ex 9.3, 4
Ex 9.3, 5
Ex 9.3, 6
Ex 9.3, 7
Ex 9.3, 8
Ex 9.3, 9
Ex 9.3, 10
Ex 9.3, 11 Important
Ex 9.3, 12
Ex 9.3, 13
Ex 9.3, 14
Ex 9.3, 15
Ex 9.3, 16
Ex 9.3, 17 Important
Ex 9.3, 18 Important
Ex 9.3, 19
Ex 9.3, 20
Ex 9.3, 21 You are here
Ex 9.3, 22 Important
Ex 9.3, 23
Ex 9.3, 24
Ex 9.3, 25
Ex 9.3, 26
Ex 9.3, 27 Important
Ex 9.3, 28 Important
Ex 9.3, 29 Important
Ex 9.3, 30
Ex 9.3, 31
Ex 9.3, 32
Ex 9.3
About the Author