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  1. Chapter 5 Class 10 Arithmetic Progressions
  2. Serial order wise

Transcript

Ex 5.2, 16 Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12 We know that an = a + (n – 1) d Let’s find the 3rd, 5th and 7th term a3 a3 = a + (3 – 1) d 16 = a + 2d a + 2d = 16 a5 a3 = a + (5 – 1)d = a + 4d a7 a7 = a + (7 – 1)d = a + 6d Given that 7th term exceeds the 5th term by 12 7th term – 5th term = 12 a7 – a5 = 12 (a + 6d) – (a + 4d) = 12 a + 6d – a – 4d = 12 a – a + 6d – 4d = 12 0 + 6d – 4d = 12 2d = 12 d = 12/2 d = 6 Putting the value of d in (1) a = 16 – 2d a = 16 – 2 × 6 a = 16 – 12 a = 4 Hence, First term of AP = a = 4 Second term of AP = First term + Common difference = 4 + 6 = 10 Third term of AP = Second term + Common Difference = 10 + 6 = 16 So, the AP is 4, 10, 16, …

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.