Question 5 - End-of-Chapter Exercises - Chapter 8 - Predicting What Comes Next: Exploring Sequences & Progress

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Question 5 Find a GP for which the 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 5th term of G.P. = a × r = ar4 Hence terms of G.P are a, ar, ar2, ar3, ar4,… Given that Sum of first two term is = 4 i.e. a + ar = −4 a(1 + r) = −4 Also, given that Fifth term is 4 times the 3rd term i.e. a5 = 4a3 (ar4) = 4ar2 𝑎𝑟4/𝑎𝑟2 = 4 r2 = 4 r = ±√4 r = ±2 Thus, r = 2, r = −2 Let’s put both values and find a 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) a = 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…