[tex] a_{n}= \frac{1}{9} (3)^{n-1}[/tex]
(We know this from a=1/9 and r=3)
Simplifying this, we get:
[tex] \frac{1}{9} (3)^{-1} (3)^n [/tex]
Since we're finding the first term that exceeds 1000, let's set it equal to 1000.
[tex]\frac{1}{27}(3)^n=1000[/tex]
Multiplying both sides by 27
[tex]3^n=27000[/tex]
[tex] log_{3}27000=n [/tex]
n≈9.2
We have to round n up, since if n=9, the value would be <1000.
Therefore n=10. Substituting n=10,
[tex]\frac{1}{27}3^{10}[/tex]
=2187
Therefore the first term that exceeds 1000 is 2187, and it is the 10th term
