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(A) To solve this,we count the total number of blanks and letters in the series: $....abb....bb....a....bbab....ba$. There are $20$ positions in total

Vedclass · Accepted Answer

$bababa$

Vedclass · Answer

(A) To solve this,we count the total number of blanks and letters in the series: $....abb....bb....a....bbab....ba$. There are $20$ positions in total. We can divide the series into groups of $4$ letters each: $(....) (abb.) (..bb) (..a.) (bbab) (....ba)$.Alternatively,looking at the pattern $a b b a$,we can try to fill the blanks to form a repeating sequence. If we test the pattern $abba$,the series becomes:$a b b a | a b b a | a b b a | a b b a | a b b a$.Comparing this with the original series $....abb....bb....a....bbab....ba$,we fill the blanks as follows:$a b b a | a b b a | a b b a | a b b a | a b b a$.The letters filled in the blanks are $a, b, a, b, a, b$. However,checking the options,we look for a sequence that fits the structure. By testing option $B$ $(bbabbb)$,it does not fit. Let's re-examine the pattern: $a b b a | a b b a | a b b a | a b b a | a b b a$. The missing letters are $a, b, a, b, a, b$. Since this is not an option,let's re-evaluate the grouping. If the pattern is $abab$,the series becomes $a b a b | a b a b | a b a b | a b a b | a b a b$. The blanks filled would be $a, b, a, b, a, b$. Given the options,the most logical sequence that completes the pattern is $bababa$.

Choose the correct alternative to complete the series: $....abb....bb....a....bbab....ba$

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