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Question

(A) To solve the series $a . c a . b e . b c c . . . b c a$,we look for a repeating pattern.
By observing the sequence,we can divide it into groups o

Vedclass · Accepted Answer

$bbaa$

Vedclass · Answer

(A) To solve the series $a . c a . b e . b c c . . . b c a$,we look for a repeating pattern.
By observing the sequence,we can divide it into groups of $4$ letters: $(a b c a) (b c b e) (b c c a) (b c a)$.
However,a more consistent pattern emerges by filling the blanks: $a(b)c a(b)b e(c)b c c(a)(b)(a)b c a$.
Let's test the sequence: $a b c a, b b e b, c b c c, a b a b, c a$.
Actually,the pattern is based on repeating groups: $a b c a, b b c a, b b c a, b b c a$.
Filling the blanks with $b, b, a, a, b$ completes the sequence as $a b c a, b b c a, b b c a, b b c a$.
Thus,the missing letters are $b, b, a, a, b$.

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

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