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Question

(C) To solve this series,let us count the total number of characters including the blanks.
The series is: $ba . . . . b . . . aabb . . . . a . . . .

Vedclass · Accepted Answer

$ababab$

Vedclass · Answer

(C) To solve this series,let us count the total number of characters including the blanks.
The series is: $ba . . . . b . . . aabb . . . . a . . . . . . a . . . bb$.
Counting the total positions,we find there are $24$ characters.
We can divide the series into groups of $4$ or $6$.
Let us try dividing into groups of $4$: $(ba . . ) (b . . .) (aabb) (. . . .) (a . . .) (. . bb)$.
This does not seem to follow a clear pattern.
Let us try groups of $6$: $(ba . . . b) (. . aabb) (. . . . a .) (. . . a . bb)$.
Alternatively,observe the repeating pattern $baab$. If we fill the blanks to form $baab$ repeatedly:
$ba(ab)b(a) (aabb) (a) (ba) (ab) (a) (ba) (ab) b$.
By testing option $C$ $(ababab)$:
$ba(a)b(b)b(a)a(b)aabb(a)a(b)a(a)b(b)a(a)bb$.
Actually,the pattern is $baab$ repeated: $baab, baab, baab, baab, baab, baab$.
Filling the blanks with $ababab$ results in $ba(a)b(b)b(a)a(b)aabb(a)a(b)a(a)b(b)a(a)bb$.
Correct sequence is $ababab$.

Choose the correct alternative to fill in the blanks in the series: $ba....b...aabb....a........a...bb$

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