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(A) To solve this, count the total number of characters including blanks. There are $15$ characters in total. We can divide the series into groups of

Vedclass · Accepted Answer

$cbaba$

Vedclass · Answer

(A) To solve this, count the total number of characters including blanks. There are $15$ characters in total. We can divide the series into groups of $3$ or $5$. Let's try groups of $3$: $(b _ a) (c _ c) (c _ c) (b _ a) (b _ a) (c)$. This does not seem to form a clear pattern. Let's try groups of $5$: $(b _ a c _) (c c _ c b) (_ a b _ a c)$. Looking at the pattern, if we fill the blanks with $b, b, a, a, c$, we get: $(b \mathbf{b} a c \mathbf{b}) (c c \mathbf{a} c b) (\mathbf{a} a b \mathbf{b} a c)$. This is not consistent. Let's re-examine the sequence: $b _ a c _ c c _ c b _ a b _ a c$. If we look at the pattern $b c a c b / c c a c b / b c a b a c$, it is also not clear. Let's try filling option $B$ $(bbaac)$: $b \mathbf{b} a c \mathbf{b} c c \mathbf{a} c b \mathbf{a} a b \mathbf{c} a c$. The pattern is $bcacb, ccacb, abac$. Actually, the correct sequence is $b \mathbf{c} a / c \mathbf{b} c / c \mathbf{a} c / b \mathbf{b} a / b \mathbf{a} c$. Wait, let's test option $A$ $(cbaba)$: $b \mathbf{c} a / c \mathbf{b} c / c \mathbf{a} c / b \mathbf{b} a / b \mathbf{a} c$. This forms a repeating pattern of $bca, cbc, cac, bba, bac$. The most logical fit for this specific series is $cbaba$.

Choose the correct alternative to complete the series: $b _ a c _ c c _ c b _ a b _ a c$

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