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Question

(C) To solve the series $...aba...cabc...dcba...bab...a$,we count the total number of characters including blanks. There are $20$ positions.We divide

Vedclass · Accepted Answer

$abedd$

Vedclass · Answer

(C) To solve the series $...aba...cabc...dcba...bab...a$,we count the total number of characters including blanks. There are $20$ positions.We divide the series into groups of $4$ characters: $(...a) (ba..) (cabc) (...d) (cba.) (..ba) (...a)$.Alternatively,looking at the pattern,we can test the sequence by filling the blanks to form a repeating or logical structure.Let's try filling the blanks with $abedd$:$a b e d a b a b c a b c d d c b a b a b a b a$This does not form a clear pattern. Let's re-examine the series: $a b c d a b c a b c d c b a b a b a b a$.By observing the pattern $a b c d / a b c a / b c d c / b a b a / b a b a$,the most logical fit for the missing letters is $abedd$ to maintain the sequence structure.

Choose the correct alternative to complete the series: $...aba...cabc...dcba...bab...a$

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