1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4......
In the above sequence what is the number of the position 2888 of the sequence.
option
a) 1
b) 4
c) 3
d) 2

• sum of n repeated numbers
  c)3
• 10 years agoHelpfull: Yes(23) No(5)
• First if we add 1223334444 they are 10..
  Next they are 20..
  Next they are 30... and so on...
  we will add all of them and fnally the 2860th term will be 4...from 2861 to 2884th term it will be 1 and thus the next 48terms will be2 so the answer is 2..
• 10 years agoHelpfull: Yes(12) No(17)
• answer is 3
• 10 years agoHelpfull: Yes(8) No(2)
• answer is 3
• 10 years agoHelpfull: Yes(5) No(2)
• its a GP having:10,20,40.......having first term=10 nd r=2
  so, (2*r^n)/2-1
• 10 years agoHelpfull: Yes(2) No(2)
• answer will be 3 only
• 10 years agoHelpfull: Yes(2) No(0)
• answer wil b 2.
• 10 years agoHelpfull: Yes(1) No(3)
• Answer:
  
  3
  
  Step-by-step explanation:
  
  1st series: 1,22,333,4444 no.of terms(n.o.t.)=10
  
  2nd: 1,1,2222,333333,44444444 n.o.t=20
  
  3rd: 1(3),2(6),3(12),4(15) n.o.t=30
  
  sum of number of terms till nth series = 10+20+30+...10*n
  
  = 10(1+2+3+4...+n) = 10*n(n+1)/2
  
  Required position = 2888
  
  10*n(n+1)/2
• 3 years agoHelpfull: Yes(0) No(0)