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(C) To solve this,we look for a repeating pattern in the series: $a . c . a b b . . a . b e . b e . . a b$.
By observing the sequence,we can divide i

Vedclass · Accepted Answer

$b c c a a c$

Vedclass · Answer

(C) To solve this,we look for a repeating pattern in the series: $a . c . a b b . . a . b e . b e . . a b$.
By observing the sequence,we can divide it into groups of $4$ or $5$ letters.
Let's test the pattern $a b c a / b b c a / b b e a / b b e a / b a b$.
Alternatively,looking at the structure $a b c a, b b c a, b b e a, b b e a, b a b$,it seems the pattern involves repeating segments.
By filling the blanks with $b, c, c, a, a, c$,the series becomes: $a b c a, b b c a, b b e a, b b e a, b a b$.
Thus,the correct sequence of letters to fill the blanks is $b, c, c, a, a, c$.

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

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