Latest Elitmus Aptitude Question SOLUTION: Given series- azzbbyyycccc.... find the 201st term????

if question is azzbbbyyyyccccc...............
and need to find 201st term then ans is as follow:
As we know first ten number add. is 55
and from 11th to 20th add. is 155.
55+155=210
but we want 201th term thus we need to exclude 20th term
then upto 10th term(1+2+3+....+10) add. is 55 and from 11th to 19th add. is 135 (55+135=190 and 190+20=210)
azbycxdwevfugthsirjqkplomn
then 201st term is q

i think the answer should be Q .

201th term is k
given series is -azzbbyyycccc.....
abcdefghijklmnopqrstuvwxyz
zyxwvut.....

here every even term takes 2 alphabets i.e 1st even place takes z and b and the 2nd even place takes c and x .....so j in the position173 and the next term R and k as it is even place each alphabet repeats 24 times 173+24 times of R=197 and 197+24 times of k=217
therefore the 201 position is k
therefore 173+24=197 and the next term is k as it is

ans is K
201 addition till 20th term ,and 20th term is k

the right question is azzbbbyyyyccccc

there is a mistake in the question...
series will be.

azzbbbyyyy......

If given series is azzbbyyycccc.....

Then,
The series can be separated into two parts:- Ascending alphabet series (abbcccc...) and Descending alphabet series (zzyyyxxxx....)
Here, we can clearly see that,
In ascending alphabet series,
count of 1st alphabet (a) = 1 , count of 2nd alphabet (b) = 2 , count of 3rd alphabet (c) = 4
So, the series of count of alphabets in ascending alphabet series becomes
1, 2, 2^2, 2^4, 2^5 and so on , which is a G.P. -------> 1

Now, similarly for descending alphabet series,
The count series becomes
2, 3, 4, 5, 6 and so on which is a A.P. -------> 2

Now , if we want to find 201st term, we need to do the sum of series that is obtained by grouping above 1 and 2 series.

Let n be the number of pair of the phrases in (series 1, series 2) when the combined series is formed by alternately taking each series phrases. By phrases, here it is meant the combination of similar alphabets occurring consecutively.

1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + ................... +2^(n-1) +
2 + 3 + 4 + 5 + 6 + 7 + .................. + (n+1)

=> [2^n - 1] + [n(n+3)/2]

Now, the value of n in this above formed summation should be such that it results into a number that is near to 201, so that we get a range of n in which 201st term comes.
So, on hit and trial, if n=7, the result is 155. Also, if n=8, the result is 291.

So, the 201th term will be in 8th pair in the pair of phrases.
In 8th pair, the first phrase (which is from ascending series) ends on 155 + 2^7 = 248th term.
So, 201th term comes in the first phrase of the 8th pair, which contains alphabet 'i' in it.

Therefore, 'i' is the answer.

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If given series is azzbbbyyyyccccc...............
and need to find 201st term then ans is as follow:
As we know first ten number add. is 55
and from 11th to 20th add. is 155.
55+155=210
but we want 201th term thus we need to exclude 20th term
then upto 10th term(1+2+3+....+10) add. is 55 and from 11th to 19th add. is 135 (55+135=190 and 190+20=210)
azbycxdwevfugthsirjqkplomn
then 201st term is q

if it is azzbbbyyyy.......
then r will be the answer

1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20=210 so it will be 20th element term and considering the series azzbbbcccc so first is 1 ,26,2,25,3,24,4,23,5,22,6,21,7,21,8,20,9,19,10,18,11,17,12,16,13,15,14,13.............it will be 20th element so ans is q

if series is: azzbbyyycccc... den 201th term will be K.
if series is: azzbbbyyyy... den 201th term will be Q

q is the right answer