Q. 1,2,1,2,2,1,2,2,2,1,2,2,2,2... then sum of 1230 terms...?

• (1,2),(1,2,2),(1,2,2,2),(1,2,2,2,2)........
  (2+1),(1+2+2),(1+2+2+2)..........
  3,5,7,9........
  Here a=3, d=2, n=1230
  last term=a+(n-1)d=3+(1230-1)2=2461
  sum=n/2[2a+(n-1)d)=1230/2[(2*3)+(1230-1)*2]=1515360 ans
• 10 years agoHelpfull: Yes(6) No(5)
• (12) (122)(1222)..............
  let there be n series of(12222.....)have the 1230 terms
  2+3+4+5+...........n=1230
  n(n+1)=2462
  we have..49*50+12=2462(so n=49)
  so series will be
  (12)(122).........................(1222...49 times)(12222....11 times)
  now summing up these terms
  2(1+2+3+.......+48+11)+50(1)
  =2444
  
• 10 years agoHelpfull: Yes(1) No(6)
• (1,2),(1,2,2),(1,2,2,2),(1,2,2,2,2)........
  (3) , (5) , (7) , (9) .........
  (2) , (3) , (4) , (5) ........ now these series sum equals 1230 according to question
  sum=n/2[2a+(n-1)d)=>n=48.12 (sum=1230.a=2,d=1)
  find sum upto 48 term by formula ( sum=n/2[2a+(n-1)d])=>1223;
  more 7 term are required (ie 1,2,2,2,2,2,2(sum=13))
  
  now therefore final result will be sum of the 2nd series(3,5,7,9...... upto 48th term)+13
  =2400+13
  =2413 ans
• 9 years agoHelpfull: Yes(1) No(0)
• Divide the series in two parts.
  1,1,1,1,1,.......
  2 , 2 2 , 2 2 2, 2 2 2 2, ...................
  
  i.e, sum is like
  1 2 3 4 5 6...... of 1st series
  2 4 6 8 10 .......... of 2nd series
  
  so in general it is like
  n
  n(n+1)
  i.e,
  n+n(n+1)
  =n²+2n
  
  We have find sum of 1230 term
  Just put n=1230
  sum= 1515360
  
• 9 years agoHelpfull: Yes(1) No(0)
• Solution:
  
  Group the no. As follows:
  
  X= {1, 2+1, 2+2+1, 2+2+2+1, 2+2+2+2+1,…}
  
  Now it gives you A.P. as follows:
  
  Y= {1, 3, 5, 7, 9,…}
  
  Number of terms contained in each Y is:
  
  Z= {1, 2, 3, 4, 5…}
  
  Given the sum to be found ulti 1230 terms but 1230 doesn't give you the exact number if you solve it by sum of n natural no.s formulae. So break 1230 into 1225+5.
  
  Now find the value of n given the sum is 1225.
  
  n(n+1)/2 = 1225
  
  This gives n=49
  
  Now use the sum of A.P. formulae in Y:
  
  Sum= n[2a+(n-1)d]/2
  
  Where n=49, a=1, d=2 we get:
  
  Sum = 2401
  
  Now this is the result for 1225 no.s
  
  To get the result of 1230 no.s we need to add the next 5 numbers in the sequence (which starts with 2). So the next 5 terms are 2,2,2,2,2.
  
  Therefore =2*5 = 10
  
  Final result is 2401+10 = 2411
• 7 years agoHelpfull: Yes(1) No(0)
• ans: 2353
  
  12
  122
  1222
  12222
  ......
  consider a square(with 2460 elements) sliced half diagonally. Sqrt(2460)= 49
  ==>49X49(outer square with 1)and 48X48(inner square with 2)
  ==>48^2=2304 elements, half diagonal ==> 2304/2= 1152 with each value of 2==> 1152*2=2304
  1152(2's)+49(1's(only one column))
  ==>2353
• 10 years agoHelpfull: Yes(0) No(3)
• In starting 10 terms 1212212221,
  the number of 1s is 4
  and the number of 2s is 6
  so 1s are 40% and 2s are 60% in the series.............
  so in 1230 there are 492 1s will be there
  and 738 2s....................
  so 2*732 + 492 = 1968..............
• 10 years agoHelpfull: Yes(0) No(1)