The question has a a bit convoluted wording but it is a good question.
Your term is \(1092=\frac{1}{2}(a_{n+1} -3)\)
\(2184=(a_{n+1} -3)\)
\(2187=(a_{n+1})\)
Sit is the 7th term.
You can also use the cyclicity of number three.
All multiple of three has a cycle of 4 and start over again
3/9/7/1.....3/9/7/1
Of the answer choices A,B, and C are impossible
Between D and E: we do know that the number cannot end in 9 because the last but one number is 729 and the next one will end with 7. according to the cycle.
Considering we kick off with 1092 the next number in the cycle MUST end with 7
D is the answer
This approach is even faster. There are multiple ways to reach your answer or goal.
Regards
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