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Question

(A) To solve this,let us count the total number of characters including blanks: $aab. . . ab. . . cabcca. . . . bcab. . . . c$. The total length is $2

Vedclass · Accepted Answer

$bbbc$

Vedclass · Answer

(A) To solve this,let us count the total number of characters including blanks: $aab. . . ab. . . cabcca. . . . bcab. . . . c$. The total length is $24$.
We can divide the series into groups of $4$: $(aabc) (abcc) (abca) (bcab) (cabc)$.
Alternatively,looking at the pattern $aab., ab. ., cabc, ca. ., bcab., . . . c$,we can identify the repeating sequence structure.
By filling in the blanks with $bbbc$,the series becomes $aabb, abcc, cabc, cabc, bcab, cabc$.
Wait,let us re-examine the pattern: $aabc, abcc, abca, bcab, cabc$.
Actually,the pattern is $aabc, abcc, abca, bcab, cabc$.
By placing $bbbc$ in the blanks: $aab(b) (b) ab(c) (c) cabcca(b) (c) bcab(c) (c) c$.
Correct sequence: $aabc, abcc, abca, bcab, cabc$.
Thus,the missing letters are $bbbc$.

Choose the correct alternative to fill in the blanks in the series: $aab....ab...cabcca....bcab....c$

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