MATHCOUNTS State - Geometric Sequence

Let the first term and the common ratio of the geometric sequence be $a$ and $r$ respectively. Then, we have:

$\begin{cases} ar^{4014-1} = 135 \\ ar^{14-1} = 375 \end{cases} \implies \dfrac {ar^{4013}}{ar^{13}} = r^{4000} = \dfrac {135}{375} = \dfrac 9{25}$

Then, the 2014th term:

$\begin{aligned} ar^{2013} & = {\color{#3D99F6}ar^{13}} \times r^{2000} & \small \color{#3D99F6} \text{The 14th term} \\ & = {\color{#3D99F6}375} \times \sqrt{\color{#D61F06}r^{4000}} & \small \color{#D61F06} \text{Note that } r^{4000} = \frac 9{25} \\ & = 375 \times \sqrt{\color{#D61F06}\frac 9{25}} \\ & = 375 \times \frac 35 \\ & = \boxed{225} \end{aligned}$

Let ${a}$ equal the 1st term and $b$ equal the common ratio. Then:

$ab^{4013} = 135$

$ab^{13} = 375$

$ab^{2013} = x$

We notice $\frac{9}{25} \dot\ {375} = {135}$ . Replacing, we get:

$\frac{9}{25} \dot\ ab^{13} = ab^{4013}$ .

$\frac{9}{25} = b^{4000}$

$\frac{3}{5} = b^{2000}$

Looking at our original equations, we see:

$\frac{ab^{2013}}{ab^{13}} = \frac{x}{375}$

$b^{2000} = \frac{x}{375}$

$\frac{3}{5} = \frac{x}{375}$

$x =$ 225