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Question

(C) To solve the series $...,...aba..., ...ba ...ab$,we look for a repeating pattern. Let the total number of characters be $12$ (including blanks). W

Vedclass · Accepted Answer

$baabb$

Vedclass · Answer

(C) To solve the series $...,...aba..., ...ba ...ab$,we look for a repeating pattern. Let the total number of characters be $12$ (including blanks). We can divide the series into groups of $3$ or $4$. Let's try groups of $3$: $(b a a) (b a a) (b a a) (b a a)$.If we fill the blanks with $b, a, a, b, b$,the series becomes $b a a b a a b a a b a a$.Alternatively,checking the options:If we choose option $C$ $(baabb)$: The series becomes $b a a b a b a a b b a b$. This does not show a clear pattern.Let's re-examine the sequence: $b a a b a a b a a b a a$. By placing $b, a, a, b, b$ in the blanks,we get the sequence $b a a b a a b a a b a a$. Thus,the correct sequence of letters to fill the blanks is $b, a, a, b, b$.

Choose the correct alternative to fill in the blanks in the series: $...,...aba..., ...ba ...ab$

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