1,2,2,3,3,3,4,4,4,4,1,1,2,2,2,3,3,3,3,4,4,4,4,4..........................what would be 2320th element of this series??

• Answer=> 2320th element is 1.
  Solution:
  set 1= 1,2,2,3,3,3,4,4,4,4 =10 elements
  set 2= 1,1,2,2,2,3,3,3,3,4,4,4,4,4 = 14 elements
  therefore
  10+14+.......+n-1+n+M=2320
  set1+set2+....setn+M=2320
  [M included because we dont know that 2320 is an exact sum of this A.P series]
  set1+set2+...set16=2304
  so M=16
  After set 16, set 17 has 17 ones. So M=16 so 2320th element is "1".
• 12 years agoHelpfull: Yes(13) No(3)
• 1 will be ans
• 12 years agoHelpfull: Yes(8) No(5)
• can anyone explain properly???? in exam they are changing the value of element.. so ometime ans is one, and sometime it is four,two ,three... we cant just learn...
• 12 years agoHelpfull: Yes(7) No(0)
• we can see in first 10 terms there is one 1,then there will be 14 terms where two 1s next in 18 there are three 1s....
  so i have to do first, n/2((2*10)+(n-1)*4))2n^2 + 8n2(32)^2 + 8*32=2304 n=32 where this is satisfying and the 32nd term of(10,14,18..)will be 134 and in 134 there are 32 ones and in the next term there will be 33 ones 34 two but after 134 the next (2320-2304)=16 th term is 2320th term and there will be a one becoz in 138 there is 33 ones....... This is the one thing i am trying to explain this is the correct way according to me.....
• 12 years agoHelpfull: Yes(6) No(3)
• please some1 explain properly how to solve such problems..
  
• 12 years agoHelpfull: Yes(3) No(0)
• ans is 4..
• 12 years agoHelpfull: Yes(2) No(6)
• after the 2304th element there will be 32 1s. so the 2320th element must be 1
  
• 12 years agoHelpfull: Yes(1) No(0)
• the terms are increasing by 3 like 10,13,16 like this...
  10+13+16+....+115=2250,in ten there is one 1,in 13,two 1 like this in 118 there is 36 1,37 2...so as 2320-2250=70
• 12 years agoHelpfull: Yes(1) No(3)
• sorry in 118 there will be 37 ones nd 38 twos, it is also(37+38)=75>70,so ans is 2...
• 12 years agoHelpfull: Yes(1) No(2)
• as we all know the international formula of A.P ...
  Sn=n/2[2a+(n-1)d]
  so here on first set 10,second set 14 and must be the third 18 so d=4.
  a=10,we just need to find out the n ;
  as here
  n/2[2a+(n-1)d]
• 9 years agoHelpfull: Yes(1) No(0)
• 4 is the correct answer
• 12 years agoHelpfull: Yes(0) No(6)
• send me recent tcs paper at kumar.gopal308@gmail.com pleae
• 12 years agoHelpfull: Yes(0) No(3)
• 1 sure........
• 12 years agoHelpfull: Yes(0) No(1)
• sorry frnds the common diff is 4 i hae mistaken with 3,so it will be 10,14,18,........134 and sum of them is 2276 and 2320-2304=16 in 10 there is one 1 in 14 there is two 1.... so in 138 there will be 33 ones which is greater than 16 so ans is one... plz forgive for the previous two wrong ans......
• 12 years agoHelpfull: Yes(0) No(1)
• the 2320th term of the series is 2 which will be 732 times
• 12 years agoHelpfull: Yes(0) No(1)
• yes its 1..my previous ans ws wrong..
• 12 years agoHelpfull: Yes(0) No(0)
• surely ans. is 1....
• 12 years agoHelpfull: Yes(0) No(0)
• as we all know the international formula of A.P ...
  Sn=n/2[2a+(n-1)d]
  so here on first set 10,second set 14 and must be the third 18 so d=4.
  a=10,we just need to find out the n
  as here
  n/2[2a+(n-1)d]
• 9 years agoHelpfull: Yes(0) No(0)
• In first term 1,2,2,3,3,3,4,4,4,4 then count=10
  In second term count is 20
  next is 30 and so on.
  so using n(n+1)/2 *10
• 9 years agoHelpfull: Yes(0) No(0)
• in first set: 1,2,2,3,3,3,4,4,4,4 total terms=10 terms and next set will start with value '1'
  2 set: 1,1,2,2,,2,2,3,3,3,3,3,3,4,4,4,,4,4,4,4,,4=20 terms and next set will start with value '1'
  3 set: each term is 3 times of its value ,e.g= 2*3 ie 2 will be 6 times ,therefore in all =30 terms
  
  similarly
  231^th set will have 2310 terms and next set will start with '1'
  now in next set which is 232 will have value '1' 232 times i.e from 2311 to (2310+232)i.e upto 2543 term value will remain 1 ,and we need 2320 term which will also be '1'
• 3 years agoHelpfull: Yes(0) No(0)