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Question

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

Vedclass · Accepted Answer

$abcaa$

Vedclass · Answer

(B) To solve this,we count the total number of characters including blanks. The series is $a . . . b c c b . . . c a . . . c c a . . . b a a b . . . c$. Total characters = $24$.
We can divide the series into groups of $4$: $(a . . .) (b c c b) (. . . c) (a . . .) (c c a .) (. . . b) (a a b .) (. . . c)$.
Alternatively,looking at the pattern $a b c a / b c c b / c a b c / a c c a / b a a b / c a b c$,we can observe a repeating cycle.
By placing $abcaa$ in the blanks:
$a(b)(c)(a)bccb(a)(a)(c)ca(b)(c)cca(a)(b)(a)baab(c)(a)(b)c$.
Testing the sequence $a b c a b c c b c a b c a c c a b a a b c a b c$,the pattern follows a repeating structure of $abc$ and $bcca$.
Thus,the correct sequence is $abcaa$.

Choose the correct alternative to fill in the blanks in the series: $a....bccb....ca...cca....baab....c$

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