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Ex 9.2, 16 - Between 1 and 31, m numbers are inserted - Ex 9.2

Ex 9.2, 16 - Chapter 9 Class 11 Sequences and Series - Part 2
Ex 9.2, 16 - Chapter 9 Class 11 Sequences and Series - Part 3 Ex 9.2, 16 - Chapter 9 Class 11 Sequences and Series - Part 4

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Ex 9.2,16 Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m โ€“ 1)th numbers is 5 : 9. Find the value of m. We know that to insert n numbers between a & b common difference (d) = (๐‘ โˆ’ ๐‘Ž)/(๐‘› + 1) Here, We have to insert m numbers between 1 and 31 So , b = 31 , a = 1 & number of terms to be inserted = n = m Therefore, d = (31 โˆ’ 1)/(๐‘š + 1) = 30/(๐‘š + 1 ) Now, a = 1 , d = 30/(๐‘š + 1 ), b = 31 We need to find 7th and (m โ€“ 1)th numbers inserted Now it is given that ratio of (7^๐‘กโ„Ž ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ)/((๐‘š โˆ’ 1)^๐‘กโ„Ž ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ) = 5/9 (1 + 7๐‘‘)/(1 + (๐‘š โˆ’ 1)๐‘‘) = 5/9 (1 + 7๐‘‘)/(1 + (๐‘š โˆ’ 1)๐‘‘) = 5/9 (1 + 7d)9 = 5[1 + (m โ€“ 1)d] 9 + 63d = 5 + 5d(m โ€“ 1) 9 + 63d = 5 + 5dm โ€“ 5d 9 โ€“ 5 + 63d + 5d = 5dm 4 + 63d = 5dm Putting d = 30/(๐‘š + 1) 4 + 63(30/(๐‘š + 1)) = 5(30/(๐‘š + 1))m (4(๐‘š + 1) + 63 ร— 30)/(๐‘š + 1 ) = (5 ร— 30 ร— ๐‘š)/(๐‘š + 1) 4(m + 1) + 2040 = 150m 4m + 4 + 2040 = 150m 2044 = 150m โ€“ 4m 2044 = 150m โ€“ 4m 2044 = 146m m = 2044/146 m = 14 Hence m = 14

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.