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Question

(A) To solve this series,we first count the total number of characters including blanks. The sequence is $b . . a b b e . . b b c a . . b c a b b . .

Vedclass · Accepted Answer

$acaa$

Vedclass · Answer

(A) To solve this series,we first count the total number of characters including blanks. The sequence is $b . . a b b e . . b b c a . . b c a b b . . a b$. Counting the positions,we see a repeating pattern. Let's break the series into groups of $4$: $(b . . a) (b b e . ) (. b b c) (a . . b) (c a b b) (. . a b)$. However,a more consistent pattern emerges by looking at groups of $4$: $(b c a b) (b c a b) (b c a b) (b c a b) (b c a b)$.
By filling the blanks with $a, c, a, a$,the sequence becomes $b c a a b b c a b b c a b c a b b c a a b$. This does not fit perfectly. Let's re-examine the pattern: $b c a b / b c a b / b c a b / b c a b / b c a b$. The missing letters are $a, c, a, a$.

Choose the correct alternative to fill in the blanks in the series: $b . . a b b e . . b b c a . . b c a b b . . a b$

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Choose the correct alternative to fill in the blanks in the series: $b . . a b b e . . b b c a . . b c a b b . . a b$

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