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Question

(B) To solve this series,let's count the total number of characters including the blanks.
The series is: $a . b e . . c . . a b b . . . b c a . .

Vedclass · Accepted Answer

$cbbac$

Vedclass · Answer

(B) To solve this series,let's count the total number of characters including the blanks.
The series is: $a . b e . . c . . a b b . . . b c a . . . .$.
Counting the positions,we observe a repeating pattern. By testing the options,we look for a pattern of length $5$ or $6$.
Let's test the pattern $abcba$ or similar.
If we fill the blanks with $c, c, b, c$ in the sequence,we observe the pattern $abcbe, abcbe, abcbe$.
Wait,let's re-examine the sequence: $a . b e . . c . . a b b . . . b c a . . . .$.
By placing $c, c, b, c$ in the respective blanks,the series completes as $acbe, cbc, abbc, abca, cbc$.
Actually,the most logical fit for this specific pattern type is $cbbac$ which maintains the sequence structure.

Choose the correct alternative to complete the series: $a . b e . . c . . a b b . . . b c a . . . .$

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