The first three terms of a geometric sequence are 3,9,27. Find the first term in the sequence which exceeds 500

Question

Now do you have to use trial and error for this kind of problem?

Answer 1

The terms of the sequence are $3, 9, 27, 81, 243, \color{red}{729},..., $

but you could say they are $3^1, 3^2, 3^3, 3^4, ... $

and then the answer will be $3^{\lfloor\log 500/\log 3\rfloor+1}=3^6=729.$

Answer 2

You can write down the general term as $3^n$. Then you want to find the smallest integer $n$ such that $3^n>500$, i.e., $n >\frac{\ln 500}{\ln 3}$. If you have access to a calculator, then no trial/error is required.

Answer 3

$3^n\gt 500$ hence $ n \gt log_3{500} $ and $n \in N $ $$ 3^5 \lt 500\lt 3^6. $$ hence n=6 for 1st term greater than 500 is $3^6=729$