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(B) To solve the series $...aaba...bba...bba....abaa....b$,we count the total number of blanks and letters. There are $20$ positions in total. We can

Vedclass · Accepted Answer

$ababa$

Vedclass · Answer

(B) To solve the series $...aaba...bba...bba....abaa....b$,we count the total number of blanks and letters. There are $20$ positions in total. We can divide the series into groups of $4$ letters each: $(....) (....) (....) (....) (....)$.By observing the pattern,we can group them as: $(baab) (aaba) (bbaa) (abba) (baab)$.Alternatively,looking at the sequence $aaba, abba, abba, abaa, ab$,we can identify the repeating pattern by filling the blanks: $b, a, a, b, a, b$.Filling the blanks in the sequence: $b a a b a / a a b a / b b a a / a b b a / a b a a / b$.By testing option $A$ $(aabab)$: $a a b a b a a b a b b a a b a b a a b a b$. This does not form a consistent pattern.By testing option $B$ $(ababa)$: $a b a a b a a b b a a b b a a b a b a a b$. This forms the repeating sequence $aaba, abba, abba, abaa, ab$.The correct sequence is formed by filling the blanks with $ababa$.

Choose the correct alternative to complete the series: $...aaba...bba...bba....abaa....b$

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Choose the correct alternative to complete the series: $........babbba...a........$

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