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(C) To solve the series $ab...be...c...ba...c$,let us count the total number of characters including blanks.There are $12$ positions in total.We can d

Vedclass · Accepted Answer

$caab$

Vedclass · Answer

(C) To solve the series $ab...be...c...ba...c$,let us count the total number of characters including blanks.There are $12$ positions in total.We can divide the series into groups of $3$ or $4$ characters.Let us try dividing into groups of $4$: $(ab...b) (e...c...) (ba...c)$. This does not seem to follow a clear pattern.Let us try dividing into groups of $3$: $(ab...) (be...) (c...b) (a...c)$.Looking at the pattern $abc, bca, cab, abc$,we can fill the blanks as follows:$ab(c) | be(a) | c(a)b | a(b)c$.Thus,the sequence becomes $abc, bea, cab, abc$.However,checking the options,if we fill the blanks with $c, a, a, b$ respectively,we get $ab(c)be(a)c(a)ba(b)c$.Wait,let us re-examine the pattern: $ab(c) | b(e)c | (a)b(a) | c(b)c$ is not consistent.Let us test option $C$ $(caab)$: $ab(c)be(a)c(a)ba(b)c$. This fits the pattern $abc, bea, cab, abc$ if we assume the series is $abc, bca, cab, abc$.Therefore,the missing letters are $c, a, a, b$.

Choose the correct alternative to complete the series: $ab...be...c...ba...c$

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