1,2,1,2,2,1,2,2,2,1,2,2,2,2,1.........................
Find the Sum of 1230 terms of the series..?

• series =1,2,1,2,2,1,2,2,2
  first term 1
  second term 2,1
  third term 2,2,1
  ...so on
  no of terms is 1+2+3+4+.....
  after checking the sum almost equal to 1230 is 49 terms (you can check 49*(49+1)/2=1225)
  and from given series 1+(2+1)+(2+2+1)+(2+2+2+1)+.....(1+2*49)
  sum is 1+3+5+7.......97 it means sum of 97 odd nos.
  =49*49=2401
  last 5 terms (2+2+2+2+2) part of 50th terms 2,2,2,2,2...........1
  ans is 2401+10=2411
  
• 9 years agoHelpfull: Yes(23) No(0)
• ans>2411
  
  each series has one more term then previous
  i.e
  1,2 ..2term
  1,2,2...3 term
  so for 1230 term there will be 48 complete series
  and last six term will be 1,2,2,2,2,2,
  
  now each of the series addition is 2 more then previous series
  i.e
  1+2=3
  1+2+2=5
  so on
  so its form an A.P
  so sum of 48 term will be
  S=n/2(2*a+(n-1)d)
  48/2(2*3+47*2)
  =2400
  now
  sum of remaining last eight term
  1,2,2,2,2,2=11
  hence answer is 2400+15=2411
  Read more at http://www.m4maths.com/placement-puzzles.php#qloSY1L2GlStYzTo.99
• 9 years agoHelpfull: Yes(5) No(0)
• ans=2421
  1,2,1,2,2,1,2,2,2,1,2,2,2,2,...................
  we are going to pair the number firstly as
  first term =1
  second term =2,1
  third term =2,2,1
  fourth term=2,2,2,1
  in first term no of term =1
  in second term no of term =2.............
  so
  1+2+3+4.............
  we have to check whether this sum is approximately equal to 1230
  so by taking 49 terms we are getting sum as 1225
  in all the terms we have a unity.
  and from second terms we have (n-1) numbers of .
  so sum=49+(2+4+6+8+.............98)=2411
  but we tis is the sum of 1225 terms only
  next five terms =2,2,2,2,2
  sum of these terms=10
  total sum=2421
  
• 9 years agoHelpfull: Yes(4) No(4)
• series is 1,2,1,2,2,1,2,2,2
  first term 1
  second term 2,1
  third term 2,2,1
  fourth term 2,2,2,1
  and so on......
  thus no of terms= 1+2+3+4+........
  we have to calculate the sum of 1230 terms.
  after hit nd trial we will find sum of 49 terms is 1225. after that there will be 2,2,2,2,2.(as we require 1230 terms hence (1230-1225)=5 term of next sequence.)
  now,
  sum of upto 1st term is 1 (i.e- 1)
  sum of upto 2nd term is 4 (i.e- 1+(2,1))
  sum of upto 3rd term is 9 (i.e- 1+(2,1)+(2+2+1))
  sum of upto 4th term is 16 (i.e- 1+(2,1)+(2+2+1)+(2+2+2+1))
  sum of upto 5th term is 25 (i.e- 1+(2,1)+(2+2+1)+(2+2+2+1)+(2+2+2+2+1)) and so on....
  hence sum of upto 49 terms will be 49*49 =2401
  so required sum =2401*(2+2+2+2+2)=2411
• 9 years agoHelpfull: Yes(4) No(0)
• there would be in total 48 complete series and 6 extra terms which include 1,2,2,2,2,2
  
  so suum of 48 tems is equal to 2400
  and plus 6 terms that is 11
  so ans is 2411
• 9 years agoHelpfull: Yes(2) No(0)
• ans>2415
  
  each series has one more term then previous
  i.e
  1,2 ..2term
  1,2,2...3 term
  so for 1230 term there will be 48 complete series
  and last eight term will be 1,2,2,2,2,2,2,2
  
  now each of the series addition is 2 more then previous series
  i.e
  1+2=3
  1+2+2=5
  so on
  so its form an A.P
  so sum of 48 term will be
  S=n/2(2*a+(n-1)d)
  48/2(2*3+47*2)
  =2400
  now
  sum of remaining last eight term
  1,2,2,2,2,2,2,2=15
  hence answer is 2400+15=2415
  
• 9 years agoHelpfull: Yes(0) No(4)