Latest TCS Aptitude Question SOLUTION: a bb ccc ddd eeeee 133th Term what is the 133th Term in the series ?

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(#M40164918) TCS Numerical Ability question a bb ccc ddd eeeee 133th Term Keep an EYE

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

Asked In TCS Sourab Roy (6 years ago)

Unsolved Read Solution (15)

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Read Solution (15) :

JAYASHREE  (6 years ago)

n(n+1)/2=133
N=15.8~16
16 th letter is p

Is this solution Helpfull? Yes (13) | No (3)  

venugopal  (6 years ago)

a-1,bb-2 like that 26-z
26*5=130
131-A
132-B
133-C

Is this solution Helpfull? Yes (7) | No (6)  

Raimo  (6 years ago)

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.

Is this solution Helpfull? Yes (5) | No (1)  

sourav1  (6 years ago)

question should be "a bb ccc dddd eeeee ..... "

Is this solution Helpfull? Yes (2) | No (2)  

abhinav jirapure  (6 years ago)

c 133 times

Is this solution Helpfull? Yes (1) | No (5)  

Abhirami  (6 years ago)

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

Is this solution Helpfull? Yes (1) | No   

Ranjan Debnath  (6 years ago)

o
n(n+1)/2 < 133 for n=15 and the 15th alphabet is o

Is this solution Helpfull? Yes | No (5)  

gowthami  (6 years ago)

n(n+1)/2

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YUVAN  (6 years ago)

133/26=130+3
3rd letter is "c"
ANS: C

Is this solution Helpfull? Yes | No (3)  

ashwini giri  (6 years ago)

Z=26, 133/26=130+3=c letter

Is this solution Helpfull? Yes | No (3)  

veeranjaneyulu  (6 years ago)

26*5=130(Z)
131=a
132=b
133=c

Is this solution Helpfull? Yes | No (1)  

keerthana  (5 years ago)

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

Is this solution Helpfull? Yes | No   

Sumit tripathi  (5 years ago)

Total no. Of alphabets is 26.
26*5=130(z)
A=131
B=132
C=133

Is this solution Helpfull? Yes | No   

Amartya  (4 years ago)

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

Is this solution Helpfull? Yes | No   

Sneha P  (4 years ago)

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

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