What is the sum of 2320th term of the series 1,2,3,4,1,1,2,2,3,3,4,4,.........??

• sum of 1st 4 terms=10
  sum of 2nd 8 terms=20
  sum of 3rd 12 terms=30
  so,last term can be written as 2320/4=580
  
  sum of 580th term=n/2(2a+(n-1)d)
  =580/2(2(10)+(579*10))
  =290*5810
  =1684900 answer
• 11 years agoHelpfull: Yes(2) No(0)
• 2320th term of the series=
  1+1+.....up to 2320times+(2+2+....upto 2320times)+(3+3+....2320times)+(4+4+....2320times)
  this is because in series there are in first term every digit comes 1 times and in 2nd term every digit comes 2 times and so on.
  hence sum=1*2320+2*2320+3*2320+4*2320=23200
• 11 years agoHelpfull: Yes(1) No(7)