what is the sum of these 1230 sequences such as 1,2,1,2,2,1,2,2,2,1,2,2,2,2.....

• Dont know wats wrong... couldnt able to post full solution...
  1,2,1,2,2,1,2,2,2,1,2,2,2,2....
  We have to find the no. of 1's nd no. of 2's in 1230 terms...
  Lets take 1st 2 terms ... 1, 2
  no. of 1's is 1 nd no. of 2's is 1 => (1+1)
  1st 5 terms ... 1,2,1,2,2
  => (1+1)+(1+2)
  1st 9 terms ... 1,2,1,2,2,1,2,2,2
  => (1+1)+(1+2)+(1+3)
  => (1+1+1)+(1+2+3)
  3 + 3(3+1)/2 ... [Its of the form n + n(n+1)/2]...
  So in 1st 9 terms, no. of 1's is "n", no. of 2's is n(n+1)/2...
  So sum of 1st 9 terms ll be n*1 + n(n+1)*2/2
  => 3*1 + 3(4)*2/2 => 15...
  Provided n + n(n+1)/2
  So for the 1230 terms,
  n + n(n+1)/2
  n=48
  48 + 48*49/2 = 1224 which is 1+2+2+2+2+2 = 11
  So total sum ll be = 48*1 + 1176*2 + 11 = 2411
  Ans : 2411
• 10 years agoHelpfull: Yes(23) No(1)
• The series goes like :
  1 2
  1 2 2
  1 2 2 2
  1 2 2 2 2
  1 2 2 2 2 2
  1 2 2 2 2 2 2...
  and so on
  
  we see that for nth 1 there are n number of 2s
  => given 'n' 1s, there are (n/2)[2a+(n-1)d]_____formula for sum of AP
  lets see
  for 50 1s, there are 1275 2s => total number of terms = 50+1275 which is >1230
  for 49 1s, there are 1200 2s => total number of terms = 49+1200=1249 which is >1230, but we can ignore the surplus 2s (since the last 49 terms are 2)
  extra terms = 19 (=1249-1230)
  Therefore numberof 2s = 1200-19=1181
  => (49*1) + (1181*2) = 2411
  
  
• 10 years agoHelpfull: Yes(18) No(0)
• This is my complete solution...
  
  1,2,1,2,2,1,2,2,2,1,2,2,2,2....
  
  We have to find the no. of 1's nd no. of 2's in 1230 terms...
  
  Lets take 1st 2 terms ... 1, 2
  no. of 1's is 1 nd no. of 2's is 1 => (1+1)
  1st 5 terms ... 1,2,1,2,2
  => (1+1)+(1+2)
  1st 9 terms ... 1,2,1,2,2,1,2,2,2
  => (1+1)+(1+2)+(1+3)
  => (1+1+1)+(1+2+3)
  3 + 3(3+1)/2 ... [Its of the form n + n(n+1)/2]...
  So in 1st 9 terms, no. of 1's is "n", no. of 2's is n(n+1)/2...
  So sum of 1st 9 terms ll be n*1 + n(n+1)*2/2
  => 3*1 + 3(4)*2/2 => 15...
  Provided n + n(n+1)/2
  
  So for the 1230 terms,
  n + n(n+1)/2
• 10 years agoHelpfull: Yes(7) No(1)
• So for the 1230 terms,
  n + n(n+1)/2
  n=48
  48 + 48*49/2 = 1224 which is close to 1230
  no. of 1's in 1224 terms = 48
  no. of 1's in 1224 terms = 48*49/2 = 1176
  So the next 6 terms after 1224 ll be 1,2,2,2,2,2 => sum = 1+2+2+2+2+2 = 11
  So total sum ll be = 48*1 + 1176*2 + 11 = 2411
  Ans : 2411
• 10 years agoHelpfull: Yes(5) No(0)
• 1,2,1,2,2,1,2,2,2,1,2,2,2,2....
  
  We have to find the no. of 1's nd no. of 2's in 1230 terms...
  
  Lets take 1st 2 terms ... 1, 2
  no. of 1's is 1 nd no. of 2's is 1 => (1+1)
  1st 5 terms ... 1,2,1,2,2
  => (1+1)+(1+2)
  1st 9 terms ... 1,2,1,2,2,1,2,2,2
  => (1+1)+(1+2)+(1+3)
  => (1+1+1)+(1+2+3)
  3 + 3(3+1)/2 ... [Its of the form n + n(n+1)/2]...
  So in 1st 9 terms, no. of 1's is "n", no. of 2's is n(n+1)/2...
  So sum of 1st 9 terms ll be n*1 + n(n+1)*2/2
  => 3*1 + 3(4)*2/2 => 15...
  Provided n + n(n+1)/2
• 10 years agoHelpfull: Yes(2) No(1)
• 1230 terms contain 49 1's and remaining 2's
  so 1181*2+49 = 2411
• 10 years agoHelpfull: Yes(2) No(1)
• can u plss explain hw n=48 came saraswathy???
• 10 years agoHelpfull: Yes(2) No(0)
• @ SARASWATHY
  i m not able to understand hw u r taking no of terms plz explain 1st 5 terms ... 1,2,1,2,2
  => (1+1)+(1+2) why r u taking 1+2???
  @ LOKI
  can u plz explain hw r u calculating no.of 1' s & 2's???
• 10 years agoHelpfull: Yes(0) No(0)
• I could able to post my complete solution...
  Ans ll be 2411
  1,2,1,2,2,1,2,2,2
  (1+1)+(1+2)+(1+3)
  here (1+1) represents no. of 1's is 1 nd no. of 2's is 1...
  (1+2) represents no. of 1's is 1 nd no. of 2's is 2...
  (1+3) represents no. of 1's is 1 nd no. of 2's is 3...
  (1+1)+(1+2)+(1+3)
  We can write as (1+1+1)+(1+2+3)
  3 + sum of 3 terms => 3+ 3(4)/2 => 9
  So in first 9 terms no. of 1's is 3 nd no. of 2's is 6...
  So sum ll be 3*1 + 6*2 = 15
  
  
• 10 years agoHelpfull: Yes(0) No(0)
• I guess 2412
  No.of 1's are 48 and remain 2's I.e., 1182
  1182*2+48*1=2412.
  Can anyone verify it....
• 10 years agoHelpfull: Yes(0) No(1)
• they said sequence not terms
• 10 years agoHelpfull: Yes(0) No(2)