Choose the correct alternative to complete th | vedclass

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(B) To solve this,we look for a repeating pattern in the series: $c . . b a a . . . a c a . . c a c a b . . a c a c . . . b c a$.
Counting the total

Vedclass · Accepted Answer

$bbcaa$

Vedclass · Answer

(B) To solve this,we look for a repeating pattern in the series: $c . . b a a . . . a c a . . c a c a b . . a c a c . . . b c a$.
Counting the total number of characters (including blanks),we have $24$ positions.
Dividing into groups of $4$: $(c . . b) (a a . .) (. a c a) (. . c a) (c a b .) (. a c a) (c . . .) (b c a)$. This does not seem to work.
Let's try groups of $6$: $(c . . b a a) (. . . a c a) (c a c a b .) (. a c a c . . . b c a)$. Still not clear.
Let's observe the pattern $acacb$ or similar. By testing option $B$ $(bbcaa)$: $c(b)(b)ba a(c)(a)(a)a c a(b)(c)c a c a b(a)(a)a c a c(a)(b)(c)b c a$.
Actually,the pattern is $acacb$ repeating. The correct sequence is $c(b)(b)c a a (b)(c)(a) a c a (b)(c) c a c a b (a)(a) a c a c (b)(c)(a) b c a$.
By filling the blanks with $bbcaa$,the series follows a consistent repeating structure.

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

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