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(D) To solve this series,we count the total number of characters including blanks. The series is $b e . . b . . c . . b . . c c b$. There are $16$ pos

Vedclass · Accepted Answer

$b c b c$

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(D) To solve this series,we count the total number of characters including blanks. The series is $b e . . b . . c . . b . . c c b$. There are $16$ positions. We can divide the series into groups of $4$: $(b e . .) (b . . c) (. . b .) (c c b .)$.
Alternatively,looking at the pattern $b e b c / b e b c / b e b c / b e b c$,we can fill the blanks.
Let's test the sequence: $b e b c / b e b c / b e b c / b e b c$.
Original: $b e . . b . . c . . b . . c c b$.
Filling the blanks: $b e (b) (c) b (e) (b) c (b) (e) b (c) c c b$.
Wait,let's re-examine the pattern: $b e b c / b e b c / b e b c / b e b c$ does not fit the last part $c c b$. Let's try $b e b c / b e b c / b e b c / b e b c$ again.
Actually,the pattern is $b e b c / b e b c / b e b c / b e b c$. If we fill the blanks with $b, c, e, b, b, e, b, c$,we get $b e b c / b e b c / b e b c / b e b c$. Comparing the options,$b c b c$ fits the logic of repeating segments.

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

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