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(D) To solve this series,we count the total number of letters and blanks: $18$ positions.
We can divide the series into groups of $3$ or $6$.
Let

Vedclass · Accepted Answer

$bcabc$

Vedclass · Answer

(D) To solve this series,we count the total number of letters and blanks: $18$ positions.
We can divide the series into groups of $3$ or $6$.
Let's try groups of $3$: $(cab), (abc), (abc), (abc), (abc), (abc)$.
Looking at the pattern: $c a b, a b c, a b c, b e c, b e c, b a b$.
This does not seem consistent. Let's try groups of $6$: $(cababc), (abcabc), (abcabc)$.
Wait,let's re-examine the sequence: $c a b . a . c . b e . b e . b . a b$.
Actually,the pattern is $c a b, a b c, a b c, b e c, b e c, b a b$ is incorrect.
Let's test option $B$: $a c b c b$.
Substituting: $c a b (a) a (c) c (b) b e (c) b e (b) b (c) a b$.
This is also not forming a clear pattern.
Let's re-evaluate the sequence: $c a b / a b c / a b c / b c a / b c a / b c a$.
If we fill the blanks with $b, c, a, b, c$,we get $c a b, b a b, c a c, b e b, b e b, b c a b$.
Correct pattern is $c a b, a b c, a b c, b c a, b c a, b c a$.
Thus,the missing letters are $a, b, c, b, c$ which corresponds to option $D$.

Choose the correct alternative to fill in the blanks for the series: $cab....a....c....be....be....b....ab$

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