Question

Consider the following series : A B C D .... X Y Z | Y X .... B A | B C D .... Y Z. | Y X .... C B A | B C .... YZ .... Which letter occupies the 1000th position in theabove series ? ​

Answer 1

Answer:

1000th position will be occupied by B

Step-by-step explanation:

Consider the following series : A B C D .... X Y Z | Y X .... B A | B C D .... Y Z. | Y X .... C B A | B C .... YZ .... Which letter occupies the 1000th position in the above series ? ​

A B C D .... X Y Z  - 26 Letters

Y X .... B   -  24 Letters

A B C D .... X Y Z  | Y X .... B    (26 + 24 = 50 Letters)

here B is 50th Letter

Then This sequence keep repeating

50 * 20 = 1000

=> 1000th position will be occupied by B

Answer 2

Answer:

See the first series contains all 26 letters so I subtracted 26 from 1000.

Next all the series contain 25 letters so the closest multiple of 25 to 974 was 38 I.e 950

Now there remains 24 letters. The last series is bound to be reverse so 24th letter will be B

Step-by-step explanation:

We have 3 patterns:

I: A B C D ... X Y Z, which occurs only once.

Y X ... B A, which repeats alternately.

B C ... Y Z, which repeats alternately.

Now, I: has 26 terms.

So, number of terms before the desired term = (999 - 26) = 973.

Each of the patterns which occurs after I: has 25 letters.

Now, 973 ÷ 25 gives quotient = 38 and remainder = 23.

Thus, the 1000th trem of the given series is the 24th term of the 39th pattern after I:.

Clearly, the 39th pattern is II: and its 24th term is B.

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