Question

A geometric sequence has first term 1/9 and common ratio 3. Which is the first term of the sequence which exceeds 1000?

Answer 1

$a_{n}= \frac{1}{9} (3)^{n-1}$
(We know this from a=1/9 and r=3)
Simplifying this, we get:
$\frac{1}{9} (3)^{-1} (3)^n$

Since we're finding the first term that exceeds 1000, let's set it equal to 1000.

$\frac{1}{27}(3)^n=1000$
Multiplying both sides by 27
$3^n=27000$

$log_{3}27000=n$

n≈9.2

We have to round n up, since if n=9, the value would be <1000.
Therefore n=10. Substituting n=10,
$\frac{1}{27}3^{10}$
=2187

Therefore the first term that exceeds 1000 is 2187, and it is the 10th term