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Question

(A) To solve the series $aa . . ab . . . . . . aaa . . . a$,we look for a repeating pattern.
By observing the sequence,we can divide it into groups o

Vedclass · Accepted Answer

$aaab$

Vedclass · Answer

(A) To solve the series $aa . . ab . . . . . . aaa . . . a$,we look for a repeating pattern.
By observing the sequence,we can divide it into groups of $4$ letters: $(aaab), (aaab), (aaab)$.
Let's fill in the blanks:
$aa(a)(b)ab(a)(b)(a)(b)aaa(b)(a)(b)a$.
Wait,let's re-examine the pattern: $aaab, aaab, aaab$.
The given series is $aa . . ab . . . . . . aaa . . . a$.
If we place $a, b, a, b, a, b, a, b$ in the blanks,we get $aa(a)(b)ab(a)(b)(a)(b)aaa(b)(a)(b)a$.
Actually,the simplest pattern is $aaab$ repeated. The missing letters are $a, b, a, b, a, b, a, b$.
However,looking at the options provided,the most logical fit for the first blank is $aaab$.

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

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