Last updated at May 29, 2018 by Teachoo

Transcript

Example 6 The sum of n terms of two arithmetic progressions are in the ratio (3n + 8) : (7n + 15). Find the ratio of their 12th terms. There are 2 AP s with different first term and common difference For the first AP Let first term be a common difference be d Sum of n term = Sn = /2 (2a + (n 1)d) & nth term = an = a + (n 1)d Similarly for second AP Let first term = A common difference = D Sn = /2 (2A + (n 1)D) & nth term = An = A + (n 1)D We need to find ratio of 12th term i.e. ( 12 )/( 12 ) = (a +(12 1)d)/(A +(12 1)D) = (a + 11d)/(A + 11 ) It is given that ( 1 )/( 2 ) = (3 + 8)/(7 + 15 ) ( /2[2 +( 1) ])/(( )/2[2 +( 1) ]) = (3n+8)/(7n+15) ( [2a + (n 1)d])/( [2A + (n 1)D]) = (3n+8)/(7n+15) ( 2(a +(( 1)/2)d))/( 2(A +(( 1)/2)D) ) = (3n+8)/(7n+15) ( (a +(( 1)/2)d))/( (A +(( 1)/2)D) ) = (3n+8)/(7n+15) We need to find (a + 11d)/(A + 11D) Hence ( 1)/2 = 11 n 1 = 22 n = 23 Putting n = 23 in (1) ("a + (" (23 1)/2 ")d" )/( +"(" (23 1)/2 ")" ) = (3 23 + 8)/(7 23 +15) ("a + (" 22/2 ")d" )/(A+"(" 22/2 ")" D) = (69 + 8)/(161 +15) (a+11d)/(A + 11 ) = 77/176 (a+11d)/(A + 11 ) = 7/16 Hence ratio of their 12th term is 7/16 i.e. 7 : 16

Examples

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6 Important You are here

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13

Example 14 Important

Example 15 Important

Example 16

Example 17 Important

Example 18 Important

Example 19 Important

Example 20

Example 21 Important

Example 22

Example 23 Important

Example 24 Important

Chapter 9 Class 11 Sequences and Series

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.