a bb ccc ddd eeeee 133th Term what is the 133th Term in the series ?

• n(n+1)/2=133
  N=15.8~16
  16 th letter is p
• 7 years agoHelpfull: Yes(13) No(3)
• a-1,bb-2 like that 26-z
  26*5=130
  131-A
  132-B
  133-C
• 7 years agoHelpfull: Yes(7) No(6)
• A to O , sum is 120(1+2+3+4........+15) and the next letter is P, which on 16th so, 120+16=136(which comprise 133 too, so ans is P.
• 7 years agoHelpfull: Yes(5) No(1)
• question should be "a bb ccc dddd eeeee ..... "
• 7 years agoHelpfull: Yes(2) No(2)
• c 133 times
• 7 years agoHelpfull: Yes(1) No(5)
• after 130th term the cycle repeats as 130 is perfectly divisible by 26, then 131 term will be a again repeating for 131 times and b will be 132th term repeating for 132 times and hence 133rd term will be c repeating for 133 times
  an=a+(n-1)d
  a=1 and n =133 and d =1
• 7 years agoHelpfull: Yes(1) No(0)
• o
  n(n+1)/2 < 133 for n=15 and the 15th alphabet is o
• 7 years agoHelpfull: Yes(0) No(5)
• n(n+1)/2
• 7 years agoHelpfull: Yes(0) No(0)
• 133/26=130+3
  3rd letter is "c"
  ANS: C
• 7 years agoHelpfull: Yes(0) No(3)
• Z=26, 133/26=130+3=c letter
• 7 years agoHelpfull: Yes(0) No(3)
• 26*5=130(Z)
  131=a
  132=b
  133=c
• 6 years agoHelpfull: Yes(0) No(1)
• as in the ratio a bb ccc
  26*5=130
  then 130=z 131=a 132=b 133=b 134=c
  so the answer is b
• 6 years agoHelpfull: Yes(0) No(0)
• Total no. Of alphabets is 26.
  26*5=130(z)
  A=131
  B=132
  C=133
• 6 years agoHelpfull: Yes(0) No(0)
• N(N+1)/2=133
  N^2+N=266
  N^2+N-266=0
  N= [-1+(1-(4*1*266)^1/2]/2
  N=15.815
  N=16
  OR 16TH LETTER IS P
• 5 years agoHelpfull: Yes(0) No(0)
• 1+2+3+..... n = 133
  n = ?
  Sum of first n natural numbers = n(n+1)/2 = 133
  = (n^2+n) = 266
  = n2 + n -266 = 0 --> This cannot be factorized but the nth alphabet comes under the range of 133
  example
  a -1
  b b -2 (2+1=3)
  c c c -3 (1+2+3=6)
  ...................
  130 131 132 133 ...................156
  As the exact sum of nth term is not given, Apply Trial & Error method in all the given options
  a) O - 15th alphabet - 152+15-266 = -26 (it comes before the required alphabet)
  b) P - 16th alphabet - 162+16-266 = 6 (this is the immediate succeeding ie., the required series)
  c) Q - 17th alphabet - 172+17-266 = 40 (not the required series)
  d) N - 14th (14
• 5 years agoHelpfull: Yes(0) No(0)