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

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Question 8 The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P. We know that an = a + (n – 1) d Now, let’s find the 4th, 8th, 6th and 10th term of AP Now, 4th term = a4 = a + (4 – 1) d = a + 3d 8th term = a8 = a + (8 – 1) d = a + 7d And, 6th term = a6 = a + (6 – 1) d = a + 6d Given that Sum of 4th and 8th term of A.P. is 24 4th term + 8th term = 24 a4 + a8 = 24 (a + 3d) + (a + 7d) = 24 2a + 10d = 24 10th term = a8 = a + (10 – 1) d = a + 9d Similarly, Sum of 6th and 10th term of A.P. is 44 6th term + 10th term = 44 a6 + a10 = 44 (a + 5d) + (a + 9d) = 44 2a + 14d = 44 Now, our equations are 2a + 10d = 24 ...(1) 2a + 14d = 44 …(2) Doing (2) – (1) (2a + 14d) – (2a + 10d) = 44 – 24 2a + 14d – 2a – 10d = 20 2a – 2a + 14d – 10d = 20 4d = 20 d = 20/4 d = 5 Putting d = 5 in (1) 2a + 10d = 24 2a + 10 × 5 = 24 2a + 50 = 24 2a = 24 – 50 2a = –26 a = (−26)/2 a = –13 We need to find first 3 terms of AP Finding first three terms of the AP First term = a = –13 Second term = First term + common difference = –13 + 5 = –8 Third term = Second term + common difference = –8 + 5 = –3 So, first these terms are –13, – 8, –3