IN THIS SEQUENCE 1 22 333 4444 11 2222 333333 44444444 111 222222 ........... what is 2926th term???

• 2926th term will be 22222....222(1464 times).
  considering first 4 terms at a time
  no. of 1 = 1
  no. of 2 = 2 X no. of 1
  no. of 3 = 3 X no. of 1
  no. of 4 = 4 X no. of 1
  also 1s repeat after every 4 terms and no. of 1s increases by 1.
  Now, 2926/4 = 731 plus remainder 2
  Hence 2925th term will be 111..11(732 times) and 2926th term is 2222...222(732 X 2 = 1464 times)
• 12 years agoHelpfull: Yes(22) No(4)
• 2926%4=2 so that term will in terms of 2.
  for 2,6,10,14,18.... series the no of 2's are increased in 2,4,6,8,10.....
  so by using the A.M.,T(n)=a+(n-1)d
  2926=2+(n-1)4 ==> n=732. in the 2's the 732 term contain 2+(732-1)2=1464.
• 12 years agoHelpfull: Yes(11) No(3)
• actually its 1,2,3,4,5..........3000 multiplication table.1 22 333 4444-1 table.11 2222 333333 44444444-2table,similarly 111,222222,333333333,444444444444-3rd table.so 3000th table wil have 24 1's and 42 2's and 72 3's and 96 4's so from 2904 to 3000 we hav 96 4's so we require 2926th term it is 4.so answer is 4
• 12 years agoHelpfull: Yes(11) No(1)
• no of 2 is 1464,that is the answer
• 12 years agoHelpfull: Yes(4) No(2)
• no of 2 is 1464,that is the ans
• 12 years agoHelpfull: Yes(2) No(1)
• 1s term arrives after every four terms i.e 1st,5th,9th term......so on are 1s terms with 1,2,3 no of 1s......so on. Now, 2926/4 =731 with a remainder of 2, so 2924th term(as 2924 is divisible by 4) is of 4s and 2921th term is of 731 no of 1s. 2925th term consists of 732 1s, so 2926th will be 732*2=1464 no of 2s.
• 9 years agoHelpfull: Yes(1) No(1)
• Another logic:
  4s term arrives after every four terms i.e 4st,8th,12th term......so on are 4s terms with 4,8,12 no of 4s......so on. Any 2s term which arrives immediately before the 4s term has 2s equal to half the no of 4s present in the 4s term. Now, 2926/4 =731 with a remainder of 2. So 2924th term consists of 2924 4s. 2928th term will have 2928 no of 4s. Thus, 2926th term has 2928/2=1464 no of 2s.
• 9 years agoHelpfull: Yes(1) No(1)
• pls explain clearly
• 12 years agoHelpfull: Yes(0) No(2)
• SRUJANA PONNA....plz calculate correctly....
• 12 years agoHelpfull: Yes(0) No(0)
• the sequence is
  1) 1 22 333 4444
  2) 11 2222 333333 44444444
  3) 111 222222 333333333 444444444444
  i.e. 1, 22, 333, 4444 are added to respective terms....
  i.e. the num of 2's in nth term are (2)(n)
  so in
  2926) 22222.........22222(5852 times)
• 12 years agoHelpfull: Yes(0) No(3)