Arithmetic Progression

Chapter 8 Class 11 Sequences and Series
Serial order wise

### Transcript

Question16 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 (7^๐กโ ๐๐ข๐๐๐๐)/((๐ โ 1)^๐กโ ๐๐ข๐๐๐๐) = 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