Geometric Series

Hi there,

so I have attached a picture to a question that I am NOT stuck on, but I think I may have got the answer by chance?

So what I figured that the denominator is doubled each time, and as the nth term of the sequence is

(3/2) ^ (n-1)

So (3/2) ^ (n-1) = x/512

As the denominator is doubled each time, the denominator of the nth term affects this so I did

2^(n-1) = 512

(n-1)log(2) = log(512)

so: n-1 = log(512)/log(2)

n -1 = 9
n = 10

And then I did (3/2)^10-1 = 19683/512

I feel that the denominator pattern was just pure luck - Is there a proper method to solve this type of question? I don't think I will be this lucky in the real exam!
Thanks!

Sure, you're given two terms of a geometric series. The second term divided by the first term will be the common ratio. As will be the third term divided by the second term.

Here, the common ratio is $\frac{3}{2}$.

The first term is 1.

Your formula booklet then tells you the sequence is $u_n = 1 \left(\frac{3}{2}\right)^{n-1} = \left(\frac{3}{2}\right)^{n-1} = \frac{3^{n-1}}{2^{n-1}}$.

Now: $512 = 2^9$ so $n-1 = 9 \Rightarrow n=10$. Which gives $\left(\frac{3}{2}\right)^9 = \frac{19683}{512}$ as required.

Which, reading your post again, is more or less what you've done. It's all very formulaic though, there isn't any patter recognition, per se. Just common ratios, nth terms and equating the denominator to 512. The denominator comes straight from the common ratio.

$u_n = \frac{3^{n-1}}{2^{n-1}}$, i.e: the numerator is a power of 3 and the denominator is a power of 2.

Thanks a lot! Glad to see what I did was somewhat right. Guess I just overthink the question sometimes :P