1 2 1 2 2 1 2 2 2 1 2 2 2 2 1 2 ...... sum of the series upto 1230 terms

• 1 2 1 2 2 1 2 2 2 1 2 2 2 2 1 2 ......
  =(1+2),(1+2+2),(1+2+2+2),(1+2+2+2+2)......
  if N be no. of terms then
  N=(1+1)+(1+2)+(1+3)+(1+4)+...
  N= n+ n(n+1)/2 = (n^2+3n)/2
  now (n^2+3n)/2 = 1230 => n^2+3n=2460
  if n=48, (n^2+3n)=2448 & N=2448/2=1224
  so series will be
  sum of (1224 term + 6 term)
  [(1+2)+(1+2+2)+(1+2+2+2)+(1+2+2+2+2)+...+(1+2+2...48 times 2)]+(1+2+2+2+2+2)
  so no. of 1's in series=49
  no. of 2's in series=(1230-49)=1181
  
  sum of series= 49*1+1181*2 = 49+2362=2411
  
  ans 2411
• 10 years agoHelpfull: Yes(51) No(2)
• 1 2
  1 2 2
  1 2 2 2
  1 2 2 2 2
  so on, it means at first for 1st 1 we have one 2, for second 1 total num of twos are 1+2, for 3rd 1 total num of twos are 1+2+3.....so on. so if we take 48 one then total num of twos are 1+2+...+48 = 48*49/2 = 1176 twos. So total num of terms are 48 ones and 1176 tows = 1124 terms ...again after it one 1 comes it becomes 1125th term and after that 5 tows ...so total becomes 1130 terms. So sum is 48*1+1176*2+1+10 = 2411
• 10 years agoHelpfull: Yes(12) 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
• 9 years agoHelpfull: Yes(5) No(0)
• No of terms: Sums
  1 1 1
  2 21 3
  3 221 5
  4 2221 7
  5 22221 9
  =15 terms total total sums=25
  and so on..
  if we calculate the no.of terms column upto 49no. that will gives us the total no.of terms upto 49no. that will be equal to n(n+1)/2 = 49*25=1225.
  so, the total sum for 1225 terms will be equal to 49/2(2*1+(49-1)*2)=2401 (since the series in sum column is 1 3 5 7 9 ....)
  now the next 5 terms will be like this... 2 2 2 2 2 (can see the pattern)
  so the total sums will be (2*5+2401)=2411.
  
• 10 years agoHelpfull: Yes(4) No(0)
• Sum Cant be more than double the value of terms that is 1230. sum cnt be more than 2460 since none of the term is more than 2.
• 10 years agoHelpfull: Yes(1) No(0)
• 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
• 9 years agoHelpfull: Yes(1) No(0)
• Write the series as
  1
  2 1
  2 2 1
  2 2 2 1.......add each line of terms then...the series would be...
  1 3 5 7 ........upto 492 terms(first four numbers means sum of 10 numbers)
  last term=a+(n-1)d=1+(492-1)2=983
  sum of n terms =(n(a+l))/2=242664
• 10 years agoHelpfull: Yes(0) No(13)
• 1............(1)
  2 1..........(3)
  2 2 1.........(5)
  2 2 2 1.......(7)
  that means 10 terms sum is (1+3+5+7)..so next 26 terms(9+11+13+15)....
  so terms series would be...
  10 26 42 58 74 90 106 122 138 154 170 186(sum is 1176)...next one is 202..but if e add them it will be 1378..we need only 1230...still 54 terms are to be added.
  1+3+5+7+9+11...........48 terms(12*4)=sum is 2304
  in that 54 terms(48*2+1=97 i.e., total 49 terms still 5 2's are needed )
  so total would be 2304+97+(5*2)=2411
  ans 2411
  
• 10 years agoHelpfull: Yes(0) No(0)
• Rakesh bhai solution repeat n=48 kaise aaya
• 10 years agoHelpfull: Yes(0) No(0)
• can u pls tell me ..hw u taken n=48
• 9 years agoHelpfull: Yes(0) No(0)