Solve all your doubts with Teachoo Black (new monthly pack available now!)
Ex 9.2
Ex 9.2, 2 Deleted for CBSE Board 2023 Exams
Ex 9.2, 3 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 4 Deleted for CBSE Board 2023 Exams
Ex 9.2, 5 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 6 Deleted for CBSE Board 2023 Exams
Ex 9.2, 7 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 8 Deleted for CBSE Board 2023 Exams
Ex 9.2, 9 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 10 Deleted for CBSE Board 2023 Exams
Ex 9.2, 11 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 12 Deleted for CBSE Board 2023 Exams
Ex 9.2, 13 Deleted for CBSE Board 2023 Exams
Ex 9.2, 14 Important
Ex 9.2, 15 Important
Ex 9.2, 16 Important Deleted for CBSE Board 2023 Exams You are here
Ex 9.2, 17 Deleted for CBSE Board 2023 Exams
Ex 9.2, 18 Important Deleted for CBSE Board 2023 Exams
Ex 9.2
Last updated at Nov. 15, 2018 by Teachoo
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 โ ๐)/(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 โ 1)/(๐ + 1) = 30/(๐ + 1 ) Now, a = 1 , d = 30/(๐ + 1 ), b = 31 We need to find 7th and (m โ 1)th numbers inserted Now it is given that ratio of (7^๐กโ ๐๐ข๐๐๐๐)/((๐ โ 1)^๐กโ ๐๐ข๐๐๐๐) = 5/9 (1 + 7๐)/(1 + (๐ โ 1)๐) = 5/9 (1 + 7๐)/(1 + (๐ โ 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 + 1) 4 + 63(30/(๐ + 1)) = 5(30/(๐ + 1))m (4(๐ + 1) + 63 ร 30)/(๐ + 1 ) = (5 ร 30 ร ๐)/(๐ + 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