The sum of 3 consecutive no of 4 no A,B,C,D are 4613,4961,5010,5099 then what is the largest no among A,B,C,d?

• Let, S1,S2,S3,S4 be the sums of A,B,C,D taking 3 of them at a time.
  
  S1+S2+S3+S4 = 4613+4961+5010+5099
  =>S1+S2+S3+S4 = 19683....[eqn 1]
  
  S1= A+B+C
  S2= B+C+D
  S3= C+D+A
  S4= D+A+B
  
  Hence
  S1+S2+S3+S4 = 3(A+B+C+D)
  3(A+B+C+D)= 19683
  
  (A+B+C+D)=6561
  
  So,the greatest term among them =sum of all four - sum of lowest three terms
  So,the greatest term among them=6561-4613 = 1948.
  
• 12 years agoHelpfull: Yes(10) No(0)