# Ex 5.2, 16 - Chapter 5 Class 10 Arithmetic Progressions

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

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 So, a3 = a + (3 1) d a3 = a + 2d 16 = a + 2d a + 2d = 16 Also, a7 = a + (8 1)d a7 = a + 6d Similarly, a5 = a + (4 1)d a5 = a + 4d 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 = 16 2 6 = 16 12 = 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,

Chapter 5 Class 10 Arithmetic Progressions

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.