If the sum of n terms of two series of A.P are in the ratio 5n+4:9n+6 .find the ratio of their 13th
terms

• Ratio of sum of n terms 2 A.P = 5n+4 : 9n+6
  
  So, Ratio Sum of 13 terms = 5*13+4 : 9*13+4 (put n=13)
  = 69 : 123
  Now, Ratio Sum of 12 terms = 5*12+4 : 9*12+4 (put n=12)
  = 64 : 114
  
  So to claculate 13th term we will simple subtract sum of first 12 terms from sum of first 13 terms
  
  i.e, Sum(13) - Sum(12) = = 69 : 123 - 64 : 114
  
  (The important thing is tht we will not subtract it by taking LCM as these are ratios, We will subtract Numerator to Numerator & Denomenator to Denomenator)
  So,
  Ratio of 13th term of 2 APs will be = 5 : 9
  
  
• 12 years agoHelpfull: Yes(47) No(21)
• given, 2A+(n-1)D:2a+(n-1)d= 5n+4:9n+6
  now, put n-1=24...i.e. n=25
  we get, 2A+24D:2a+24d= 2*(13th term of 1st ap):2*(13th term of 2nd ap)
  hence ratio= (5*25+4):(9*25+6)=129:231
• 12 years agoHelpfull: Yes(13) No(2)
• Ans-5:9
  Ratio of sum of n terms of 2 A.P = 5n+4:9n+6 (given)
  now, ratio of sum of 13terms S(13)= 69:123 (put n=13)
  and ratio of sum of 12 terms S(12)= 64:114 (put n=12)
  So, 13th term [S(13)-S(12)]= 5:9
• 12 years agoHelpfull: Yes(7) No(5)
• 129/231
  Sn=n*middle term
  let 13th term is middle term then total no. of terms =25.put n=25
• 12 years agoHelpfull: Yes(6) No(10)
• Ajit Ankit, pls explain
  how it came n-1 = 24
• 12 years agoHelpfull: Yes(4) No(2)
• Let the 2 series have the sum S(1,n) and S(2,n).
  now, S(1,n):S(2,n) = (5n+4):(9n+6)
  = {5(n-1)+5+4}:{9(n-1)+9+6}
  = {2*(9/2)+5(n-1)}:{2*(15/2)+9(n-1)}
  
  = [(n/2){2*(9/2)+5(n-1)}]:[(n/2){2*(15/2)+9(n-1)}]
  hence nth terms will be in the ratio
  T(1,n):T(2,n)= [(9/2)+5(n-1)]:[(15/2)+9(n-1)]
  
  =>T(1,13):T(2,13)=[(9/2)+5(13-1)]:[(15/2)+9(13-1)]
  =(129/2):(231/2)
  =43/77
• 12 years agoHelpfull: Yes(4) No(3)
• We should bring a solution that will be
  
  13th term of 1st AP : 13th term of 2nd AP
  
  => a+(13-1)d of AP 1 / a+(13-1)d of AP 2
  
  => 2[ a+(13-1)d] of AP 1 / 2[a+(13-1)d] of AP 2
  => 2a+(12*2)d of AP 1 / 2a+(12*2)d of AP 2
  => 2a + 24d of AP 1 / 2a+ 24d of AP 2
  => n=25 because for n=25, sum of n terms is 2a+ 24d
  
  thus substitute n=25
  then,
  5*25+4 / 9*25+6
  =>129/231.
• 9 years agoHelpfull: Yes(2) No(0)
• Ans-5:9
  Ratio of sum of n terms of 2 A.P = 5n+4:9n+6 (given)
  now, ratio of sum of 13terms S(13)= 69:123 (put n=13)
  and ratio of sum of 12 terms S(12)= 64:114 (put n=12)
  So, 13th term [S(13)-S(12)]= 5:9
  
• 9 years agoHelpfull: Yes(1) No(1)