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Ex 5.2, 18 - The sum of 4th and 8th terms of an AP is 24 - Finding AP

  1. Chapter 5 Class 10 Arithmetic Progressions
  2. Serial order wise
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Ex 5.2 ,18 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 So, a4 = a + (4 – 1) d a4 = a + 3d Also, a8 = a + (8 – 1)d a8 = a + 7d Similarly, a6 = a + (6 – 1)d a6 = a + 5d And a10 = a + (10 – 1)d a10 = a + 9d 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 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 2a = 44 – 14d a = (44 −14𝑑)/2 Putting (6) in (5) 2a + 10 d = 24 2 × [(44 − 14𝑑)/2] + 10 d = 24 44 – 14 d + 10 d = 24 44 – 4d = 24 44 – 24 = 4d 20 = 4d 20/4 = d 5 = d d = 5 From (6) a = (44 − 14𝑑)/2 = (44 − 14 × 5)/2 = (44 − 70)/2 = (− 26)/2 = – 13 Finding first three terms of the AP (Method 1) 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 Finding first three terms of AP ( Method 2) We know that an = a + (n – 1) d 2nd term = a2 = a + (2 – 1) d = a + d = – 13 + 1 × 5 = – 13 + 5 = – 8 3rd term = a3 = a + (3 – 1)d = a + 2d = – 13 + 2 × 5 = – 13 + 10 = – 3

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