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Example 6 - Sum of n terms of two APs are in ratio (3n + 8) - Examples

Example 6 - Chapter 9 Class 11 Sequences and Series - Part 2
Example 6 - Chapter 9 Class 11 Sequences and Series - Part 3 Example 6 - Chapter 9 Class 11 Sequences and Series - Part 4


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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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.