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,…………………………………..
Then what is the 2320 position of the number in the sequence?
b) 2 b) 1c) 3 d) 4

• ans will be 4
  10+20+40+80+160+320+640+1280=2550
  we need to find out 2320 position..
  number of 4's in last 1280 digit=512
  so, ans=4
• 12 years agoHelpfull: Yes(41) No(33)
• answer is b)1
  1,2,3,4(1-1time 2- 2times 3-3 4-4)=10 terms /completes cycle and starts from 1
  1,2,3,4(1-2 2-4 3-6 4-8)= 20 terms /completes cycle and starts from 1
  (1,2,3,4)each digit 3 time to its value =30 terms/completes cycle and starts from 1
  
  10+20+30+40+50+.......=x
  x is nearer value to 2320 solving n(n+1)/2
  10.(21.22)/2= 2310
  analysing it 2310 completes cycle and starts from 1 again
  now it 22 times
  (1-22 times 2-44 times ......)
  2320 position will occupied by 1
• 12 years agoHelpfull: Yes(40) No(27)
• as nagarjuna said a series will build up after cycles being formed
  10+20+30+40+50+60+...=x
  so 10(1+2+3+4+5+6+....)=x
  10*n(n+1)/2=x
  hence we will find the closest term to the desired position by hit and trail.
  so n=21 and x=2310 therefore 21st cycle is completed.
  so answer is b)1
  
  
• 12 years agoHelpfull: Yes(24) No(10)
• total grandtotal
  1 2 3 4 10 10
  2 4 6 8 20 30
  4 8 12 16 40 70
  8 16 24 32 80 150
  16 32 48 64 160 310
  32 64 96 128 320 630
  64 128 192 256 640 1270
  128 256 384 512 1280 2550
  
  the needed value is a grandtotal of 2320 so it is in the 1270+128+384+512 position therefore it is 4
• 10 years agoHelpfull: Yes(6) No(8)
• plz explain..can't understand
• 12 years agoHelpfull: Yes(5) No(0)
• ANS IS 1
  
  Number of terms=(1+2+3+4)+2^1(1+2+3+4)+2^2(1+2+3+4)+2^3(1+2+3+4)+.......
  where 1,2,3,4 represents number of 1's,2's,3's and 4's respectively for each repetition.
  sum=> 1+2+3+4(1+2^1+2^2..........n terms)=2888
  => 10*((2^n - 1)/(2-1))=2888
  => 2^n * 10 = 2898
  we have to find a number less than 2898 which can be represented in LHS form
  so 2560 can be written as 256*10=> 2^8 * 10
  thus, n = 8
  so 2560th term is last term of the repetition,thus is 4...2561th term represents new repetition starting with 1's...
  so this new repetition will have 2^9 1's, 2*(2^9) 2's and so on....
  so remaining terms 2888-2560=338
  1 will be repeated 2^9 times which is 512, thus 338th repetition after 2560th term is 1....Thus 2888th term is 1....
  
  I think this explanation will help you....enjoy :)
• 11 years agoHelpfull: Yes(4) No(5)
• plz explain clearly
  
• 12 years agoHelpfull: Yes(3) No(1)
• m4maths.com is a good website. I got 15 questions exactly from this in TCS online test.
  I got placed in TCS campus drive in Aligarh Muslim University.......
• 8 years agoHelpfull: Yes(3) No(1)
• pls explain clearly i am preparing for tcs pls help pls pls pls pls i w
• 7 years agoHelpfull: Yes(2) No(1)
• Ans is 1...
  
• 9 years agoHelpfull: Yes(1) No(1)
• 1 2 3 4
  1 + 2 + 3 + 4
  2 + 4 + 6 + 8
  4 + 8 + 12 + 16
  8 + 16 + 24 + 32
  16 + 32 + 48 + 64
  32 + 64 + 96 + 128
  64 + 128 + 192 + 256
  128 + 256 + 384 + 512
  256 + 512 + 768 + 1024=2560
  
  2560-200=2360 (aprox) 4 will be in that margin
  
  
  Ans will be 4
• 7 years agoHelpfull: Yes(1) No(0)
• First if we count 1223334444. they are 10
  In the next term they are 20
  Next they are 30 and so on
  So Using n(n+1)2×10≤2888
  
  For n = 23 we get LHS as 2760. Remaining terms 128.
  
  Now in the 24th term, we have 24 1's, and next 48 terms are 2's. So next 72 terms are 3's.
  The 2888 term will be “3”.
• 9 years agoHelpfull: Yes(0) No(6)
• In first 10terms 1s-1, 2s-2, 3s-3, 4s-4
  In next 20terms 1s-2, 2s-4, 3s-6, 4s-8
  similarly in next will be of 40terms
  sum of the series is 10+20+40+80+160+...1280 = 2550
  now 4s are in the series as 4,8,16,32,64,128,256,512
  that means in last 1280terms there are 512 4s, that comes within 2320
  so ans is 4
• 9 years agoHelpfull: Yes(0) No(0)
• First if we count 1223334444. they are 10
  In the next term they are 20
  Next they are 30 and so on
  So Using n(n+1)/2×10≤2320
  
  For n = 21 we get LHS as 2310. Remaining terms 10.
  
  Now in the 21st term, we have 21 1's, and next 42 terms are 2's. So next 126 terms are 3's.
  The 2320 term will be “1”.
  
• 9 years agoHelpfull: Yes(0) No(1)
• First if we count 1223334444. they are 10
  In the next term they are 20
  Next they are 30 and so on
  So Using n(n+1)/2×10≤2320
  
  For n = 21 we get LHS as 2310. Remaining terms 10.
  
  Now in the 21st term, we have 21 1's, and next 42 terms are 2's. So next 126 terms are 3's.
  The 2320 term will be 1 . hence option 2
• 9 years agoHelpfull: Yes(0) No(1)