find the sum of the series
1,2,1,2,2,1,2,2,2,1,2,2,2,2,.... till 1234th digit.

• 1--2,1--2,2,1--2,2,2,1--2,2,2,2,1 and so on........
  no. of terms=1,2,3,4,5------49=1225 terms+9 terms gives 1234 terms
  so 1234th digit is 2
  In 49 terms 1 will come 49 times=1*49=49
  In 49 terms 2 will come 1176 times=2*[1+2+3+----48]=2352
  (becz in 2nd terms 2 comes once and in 3rd terms 2 comes 2times and so on..
  and adding we get 2+2*2+2*3+2*4+2*5--2*48=2352
  In next 9 terms 2 only comes 9 times=2*9=18
  so sum of series=49+2352+18=2419
  Answer is 2419
  
• 10 years agoHelpfull: Yes(1) No(0)
• the position of occurance of 1 are
  1,3,6,10,15,etc
  so it is of the form n(n+1)/2
  eg, the 10th position = 4(4+1)/2
  here upto the 10th position the total no. of 1 occured = 4
  ie, at position 1,3,6 and 10,
  total = 4, it is the value of n when calculated by solving the above equation.
  so the total no. of 1 that will be in the series is given by
  n(n+1)/2 = 1234
  on solving n = 50 or 49 , taking only integers.
  when 50 is substituted in n
  then n(n+1) /2 = 50*51/2= 2550
  which is not possible
  so the total no. of 1 in the series is 49
  rest numbers will be 2
  no. of 2 = 1234-49= 1185
  total sum = 1185*2+49*1 = 2419
  2419 is the answer
• 10 years agoHelpfull: Yes(1) No(0)