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Last updated at May 29, 2018 by Teachoo
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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
Chapter 9 Class 11 Sequences and Series
Ex 9.2, 5 Important
Ex 9.2, 9 Important
Ex 9.2, 15 Important
Ex 9.2, 17
Example 14 Important
Example 15 Important
Ex 9.3, 3 Important
Ex 9.3, 11 Important
Ex 9.3, 17 Important
Ex 9.3, 18 Important
Ex 9.3, 22 Important
Ex 9.3, 28
Ex 9.3, 29 Important
Ex 9.4.4 Important Deleted for CBSE Board 2022 Exams
Ex 9.4, 7 Important Deleted for CBSE Board 2022 Exams
Ex 9.4, 9 Important Deleted for CBSE Board 2022 Exams
Ex 9.4, 10 Deleted for CBSE Board 2022 Exams
Example 23
Misc 16 Important
Misc 19 Important
Misc 25 Important Deleted for CBSE Board 2022 Exams
Misc 28 Important
Misc 32 Important
Chapter 9 Class 11 Sequences and Series
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