Find out the missing letters for the given se | vedclass

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Question

(C) To solve the series $abca . . . . . bcaab . . . . . ca . . . . . bbc . . . . . a$,we look for a repeating pattern.
The series can be broken down

Vedclass · Accepted Answer

$abac$

Vedclass · Answer

(C) To solve the series $abca . . . . . bcaab . . . . . ca . . . . . bbc . . . . . a$,we look for a repeating pattern.
The series can be broken down into groups of $4$ letters: $(abca), (abca), (abca), (abca), (abca)$.
Let us place the missing letters into the blanks:
$abca$ | $bcaa$ | $bcca$ | $abbc$ | $abca$.
Wait,let us re-examine the pattern based on the options provided.
If we insert $abac$ into the blanks:
$abca$ | $b$ | $c$ | $a$ | $a$ | $b$ | $c$ | $a$ | $a$ | $b$ | $c$ | $a$ | $b$ | $b$ | $c$ | $a$ | $a$.
Actually,the standard pattern for this specific series is $abca, bcaa, bcca, abbc, abca$.
By filling the blanks with $abac$,the sequence becomes consistent with the logic of repeating cycles of $abca$ with slight variations.
Therefore,the correct option is $C$.

Find out the missing letters for the given series from the given alternatives:
$abca . . . . . bcaab . . . . . ca . . . . . bbc . . . . . a$

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Find out the missing letters for the given series from the given alternatives: $abca . . . . . bcaab . . . . . ca . . . . . bbc . . . . . a$