Question 11 - End-of-Chapter Exercises - Chapter 8 - Predicting What Comes Next: Exploring Sequences & Progress

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Question 11 The sum of the first three terms of a GP is 13/12 and their product is -1. Find the common ratio and the terms. Let the three terms in G.P. be 𝒂/𝒓, a, ar Given that Sum of first three terms of G.P. = 13/12 𝒂/𝒓 + a + ar = 𝟏𝟑/𝟏𝟐 Also, Product of first three term = –1 (𝒂/𝒓) × (a) × (ar) = –1 a3 = –1 a3 = (–1)3 a = – 1 Putting value of a in (1) a/r + a + ar = 13/12 (−𝟏)/𝒓 + (–1) + (–1)r = 𝟏𝟑/𝟏𝟐 (−1)/𝑟 – 1 – r = 13/12 (−1)/r – 1 – r = 13/12 (−1 −1 × 𝑟 − 𝑟 × 𝑟)/𝑟 = 13/12 (−𝟏 − 𝒓 − 𝒓𝟐)/𝒓 = 𝟏𝟑/𝟏𝟐 12(–1 – r – r2) = 13r –12 – 12r – 12r2 = 13r 0 = 13r + 12 + 12r + 12r2 0 = 12r2 + 25r + 12 12r2 + 25r + 12 = 0 We need to factorise this, Let’s use splitting the middle term 12r2 + 25r + 12 = 0 12r2 + 16r + 9r + 12 = 0 4r(3r + 4) + 3(3r + 4) = 0 (4r + 3) (3r + 4) = 0 Solving this Now we have a = –1 r = (−𝟑)/𝟒 & r = (−𝟒)/𝟑 4r + 3 = 0 4r = – 3 r = (−𝟑)/𝟒 3r + 4 = 0 3r = –4 r = (−𝟒)/𝟑 Splitting the middle term method We need to find two numbers whose Sum = 25 Product = 12 × 12 = 144 Since sum and product both are positive, both numbers are positive We need to find common ratio and the terms. For a = –1 and r = (−𝟑)/𝟒 Since our terms are𝒂/𝒓, a, ar a/r = (−1)/((−3)/4) = 𝟒/𝟑 a = −1 ar = (–1) ((−3)/4) = 𝟑/𝟒 Hence, the three term of G.P. are 𝟒/𝟑, −1, 𝟑/𝟒 Hence, the three term of G.P. are 𝟒/𝟑, −1, 𝟑/𝟒 For a = –1 and r = (−𝟒)/𝟑 Since our terms are𝒂/𝒓, a, ar a/r = (−1)/((−4)/3) = 𝟑/𝟒 a = −1 ar = (–1) ((−4)/3) = 𝟒/𝟑 Hence the three term of G.P. are 𝟑/𝟒, –1, 𝟒/𝟑 Hence first three terms of G.P. are For r = (−𝟑)/𝟒 : 𝟒/𝟑, −1, 𝟑/𝟒 For r = (−𝟒)/𝟑 : 𝟑/𝟒, –1, 𝟒/𝟑