Given series 12122122212222122222... .what is the sum of first 1234 terms

• series is 12 , 122 , 1222... hence after n no. of 1s.. n no. of 2s are added to series. so.. x + x*(x+1)/2=m , where x is no of 1s and m is total no. of terms.
  now, if we put x=48 we get m=1224. so to get 1234 10 terms are remaining. so after 1224th term ie 1225th terms should b 1 and reamining 9 terms should be 2.
  so.-> 48+ 2*(48*49/2)+1+(2*9)=2419.
  
• 11 years agoHelpfull: Yes(16) No(3)
• 12 122 1222 12222......
  =>2+3+4+5+....=1234
  =>(n/2)(2*a + (n-1)*d) = 1234
  =>(n/2)(2*2 + (n-1)*1) = 1234
  =>(n/2)(3+n) = 1234
  =>n^2 + 3n = 1234*2
  =>n^2 + 3n = 2468
  =>n^2 + 3n -2468 = 0
  =>n = [-3 +- (3^2 - 4*1*-2468)^1/2]
  =>n = (-3+-99)/2
  =>n = 48
  
  now, sum of 1st 48 number of terms,
  2+3+4+5....+48 = 1224th term
  So, our 1225th term will contain 1 followed by 48 2s
  so, 1234th term will be 1224th + 10th term(1222222222)
  
  now , 12 122 1222....
  sum of terms, 3,5,9,....48th term + (1+2+2+2+2+2+2+2+2+2)
  =>(48/2)*(2*3 + (48-1)*2) + 19 = 2400+19 =2419
• 11 years agoHelpfull: Yes(3) No(0)
• question not clear ,please provide comma separated value
• 11 years agoHelpfull: Yes(2) No(0)
• if n=34, we get n*(n+1)/2=1190.
  also taking 34 1's we get 34+1190=1224
  so, the answer will contain..
  35 1s and 1190+9=1199 2s
  so answer= 35+2*1199=2433
• 11 years agoHelpfull: Yes(1) No(9)
• Ans: 2420
  if you se in first 1234 terms: 12(1st)122(2nd)1222(3rd)........12222...(49th)12222..(50th). In the 50th term there are 10 elements(one 1 nine 2's).
  so total sum:50+2352(sum 2's in first 49 terms)+18(sum of nine 2's in the 50th term)=2420.
• 11 years agoHelpfull: Yes(0) No(1)
• Series is 1,2,1,2,2,1,2,2,2,1,2,2,2,2,1....find sum of first 1234 terms?
• 11 years agoHelpfull: Yes(0) No(1)