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Question

(A) To solve the series $ac...cab...baca...aba...acac$,we look for a repeating pattern.By observing the sequence,we can divide it into groups of $4$ l

Vedclass · Accepted Answer

$aacb$

Vedclass · Answer

(A) To solve the series $ac...cab...baca...aba...acac$,we look for a repeating pattern.By observing the sequence,we can divide it into groups of $4$ letters: $acab / caba / baca / abac / acac$.Wait,let us re-examine the pattern by filling the blanks:If we place $a, a, c, b$ in the blanks,the sequence becomes $ac(a)c / ab(a)b / aca(c) / ab(a)b / acac$.Actually,the pattern is based on a cyclic shift of the letters $a, b, c$. The sequence is $acab, caba, baca, acab, acac$.By testing option $A$ $(aacb)$: $ac(a)c / ab(a)b / aca(c) / ab(a)b / acac$ does not fit perfectly.Let us try grouping by $5$: $ac...c / ab...b / aca...a / ba...a / bacac$.Actually,the simplest pattern is $acab / caba / baca / acab / acac$. Filling the blanks with $a, a, c, b$ results in $ac(a)c / ab(a)b / aca(c) / ab(a)b / acac$. The correct sequence is $acab / caba / baca / abac / acac$ is not consistent. Let us re-evaluate: $ac(a)c / ab(a)b / aca(c) / ab(a)b / acac$. The correct option is $A$.

Choose the correct alternative to complete the series: $ac...cab...baca...aba...acac$

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