Examples
Example 1 (ii)
Example 2
Example 3 Important
Example 4
Example 5
Example 6 Important You are here
Example 7
Example 8 Important
Example 9
Example 10 Important
Example 11
Example 12 Important
Example 13
Example 14 Important
Example 15 Important
Example 16
Example 17 Important
Example 18 Important
Example 19 Important Deleted for CBSE Board 2022 Exams
Example 20 Deleted for CBSE Board 2022 Exams
Example 21 Important
Example 22
Example 23
Example 24 Important
Examples
Last updated at Oct. 25, 2021 by Teachoo
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