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Question

(B) To solve the series $c . . b b a . . . . c a b . . . . a c . . . . a b . . . . a c$,we first count the total number of characters including blanks

Vedclass · Accepted Answer

$a c b c b$

Vedclass · Answer

(B) To solve the series $c . . b b a . . . . c a b . . . . a c . . . . a b . . . . a c$,we first count the total number of characters including blanks.
Total characters = $20$.
We can divide the series into groups of $4$ characters each: $(c . . b) (b a . .) (. c a b) (. . a c) (. . a b) (. . a c)$.
Alternatively,looking at the pattern $c a b c b a c b a c a b c b a c a b a c$,the repeating sequence is $c a b c b a$.
By filling the blanks: $c (a) (b) b / a (c) (b) (c) / b (a) (c) (b) / a (c) (b) (c) / a (b) (c) (b) / a (c)$.
Checking the sequence $c a b b a c b c a b c b a c b c a b a c$,the pattern follows $a c b c b$ to complete the sequence logically.

Choose the correct alternative to fill in the blanks in the series: $c . . b b a . . . . c a b . . . . a c . . . . a b . . . . a c$

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