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Question

(B) To solve the series $ab . . aa . . bbb . . . aaa . . bbaa$,we look for a repeating pattern.
Counting the total number of characters (including bl

Vedclass · Accepted Answer

$baab$

Vedclass · Answer

(B) To solve the series $ab . . aa . . bbb . . . aaa . . bbaa$,we look for a repeating pattern.
Counting the total number of characters (including blanks),we have $20$ positions.
Let's divide the series into groups of $4$: $(ab . .) (aa . .) (bbb .) (. . aa) (. . bbaa)$. This does not seem to fit.
Let's try a pattern based on increasing counts of $a$ and $b$: $ab, aab, aaabb, aaaabb, aaaaabb$.
Looking at the sequence: $ab$ | $ba$ | $aa$ | $bb$ | $b$ | $bb$ | $a$ | $aa$ | $a$ | $bb$ | $aa$.
Actually,the pattern is $ab, ba, aa, bb, bbb, aaa, bbaa$.
By filling the blanks with $baab$,the sequence becomes: $ab(ba)aa(ab)bbb(a)aaa(b)bbaa$.
Thus,the correct sequence is $baab$.

Choose the correct alternative to complete the series: $ab . . aa . . bbb . . . aaa . . bbaa$

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