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

  1. Class 11
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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 APโ€™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 = ๐‘›/2 (2a + (n โ€“ 1)d) & nth term = an = a + (n โ€“ 1)d Similarly for second AP Let first term = A common difference = D Sn = ๐‘›/2 (2A + (n โ€“ 1)D) & nth term = An = A + (n โ€“ 1)D We need to find ratio of 12th term i.e. (๐‘Ž12 ๐‘œ๐‘“ ๐‘“๐‘–๐‘Ÿ๐‘ ๐‘ก ๐ด๐‘ƒ )/(๐ด12 ๐‘œ๐‘“ ๐‘ ๐‘’๐‘๐‘œ๐‘›๐‘‘ ๐ด๐‘ƒ) = (a +(12 โˆ’ 1)d)/(A +(12 โˆ’ 1)D) = (a + 11d)/(A + 11๐ท) It is given that (๐‘†๐‘ข๐‘š ๐‘œ๐‘“ ๐‘› ๐‘ก๐‘’๐‘Ÿ๐‘š๐‘  ๐‘œ๐‘“ 1๐‘ ๐‘ก ๐ด๐‘ƒ )/(๐‘†๐‘ข๐‘š ๐‘œ๐‘“ ๐‘› ๐‘ก๐‘’๐‘Ÿ๐‘š๐‘  ๐‘œ๐‘“ 2๐‘›๐‘‘ ๐ด๐‘ƒ) = (3๐‘› + 8)/(7๐‘› + 15 ) (๐‘›/2[2๐‘Ž+(๐‘› โˆ’ 1)๐‘‘])/((๐‘› )/2[2๐ด+(๐‘› โˆ’ 1)๐ท]) = (3n+8)/(7n+15) ( [2a + (n โˆ’ 1)d])/( [2A + (n โˆ’ 1)D]) = (3n+8)/(7n+15) ( 2(a +((๐‘› โˆ’1)/2)d))/( 2(A +((๐‘› โˆ’1)/2)D) ) = (3n+8)/(7n+15) ( (a +((๐‘› โˆ’1)/2)d))/( (A +((๐‘› โˆ’1)/2)D) ) = (3n+8)/(7n+15) We need to find (a + 11d)/(A + 11D) Hence (๐‘› โˆ’ 1)/2 = 11 n โ€“ 1 = 22 n = 23 Putting n = 23 in (1) ("a + (" (23 โˆ’ 1)/2 ")d" )/(๐ด+"(" (23 โˆ’ 1)/2 ")" ๐ท) = (3 ร— 23 + 8)/(7 ร— 23 +15) ("a + (" 22/2 ")d" )/(A+"(" 22/2 ")" D) = (69 + 8)/(161 +15) (a+11d)/(A + 11๐ท) = 77/176 (a+11d)/(A + 11๐ท) = 7/16 Hence ratio of their 12th term is 7/16 i.e. 7 : 16

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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.