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Ex 5.2, 7 - Find 31st term of an A.P. whose 11th term - Ex 5.2

Ex 5.2, 7 - Chapter 5 Class 10 Arithmetic Progressions - Part 2
Ex 5.2, 7 - Chapter 5 Class 10 Arithmetic Progressions - Part 3

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Ex 5.2, 7 Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73 We know that an = a + (n – 1) d Given 11th term is 38 a11 = a + (11 – 1) d 38 = a + (11 – 1)d 38 = a + 10 d 38 – 10 d = a a = 38 – 10d Given 16th term is 73 a16 = a + (16 – 1)d a16 = a + 15d 73 = a + 15d 73 – 15d = a a = 73 – 15d From (1) & (2) 38 – 10d = 73 – 15d 38 – 73 = –15d – 10d –35 = –5d (−35)/(−5) = d 7 = d d = 7 Putting the value of d in (1) a = 38 – 10 d a = 38 – 10× 7 a = 38 – 70 a = –32 We need to find the 31st term So, n = 31, a = – 32, d = 7 We need to find an an = a + (n – 1) d Putting values a31 = –32 + (31 – 1) 7 = –32 + 30 × 7 = –32 + 210 = 178 Therefore, the 31st term of AP is 178

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.