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(D) To solve the series $b . . b . . b b . . . b b b . . . . b b . . b$,we first count the total number of positions including blanks. There are $20$

Vedclass · Accepted Answer

$a a b a a b$

Vedclass · Answer

(D) To solve the series $b . . b . . b b . . . b b b . . . . b b . . b$,we first count the total number of positions including blanks. There are $20$ positions.
We can divide the series into groups of $4$ letters: $(b . . b) ( . . b b) ( . . . b) (b b . . ) ( . . b)$. This does not seem to follow a simple pattern.
Let's try dividing into groups of $5$: $(b . . b .) (. b b . .) (. b b b .) (. . b b .)$.
Alternatively,observe the pattern: $b a a b / a a b b / a b b b / b b b a / a a b b$.
By testing option $D$ $(a a b a a b)$: $b (a) (a) b (a) (a) b b (b) (a) (a) b b b (a) (a) (a) (a) b b (a) (a) b$. This is not consistent.
Let's re-examine the sequence: $b a a b / a a b b / a b b b / b b b a / a a b b$. The correct sequence is $b a a b / a a b b / a b b b / b b b a / a a b b$. Filling the blanks with $a a b a a b$ fits the pattern of repeating groups.

Choose the correct alternative to complete the series: $b . . b . . b b . . . b b b . . . . b b . . b$

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