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3, 9, 27... In the sequence above, each term after the first is found [#permalink]
Shouldn't the answer be 6?
bcz sum would be 1092 when n=6 (i.e 3+9+27+81+243+729)
or in other words 1092 = 0.5*(2187-3) --> 2187 is the (n+1)th term --> n+1 = 7 --> n = 6
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3, 9, 27... In the sequence above, each term after the first is found [#permalink]
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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.

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Re: 3, 9, 27... In the sequence above, each term after the first is found [#permalink]
answer is 6
2187 is n+1 th term...

1.3
2.9
3.27
4.81
5.243
6.729
7.2187

so n=6
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