# How do you write a rule for the nth term of the geometric sequence, then find a6 given 2, 14, 98, 686,...?

May 4, 2016

33614

#### Explanation:

Geometric sequences are defined by the simple formula:
$f \left(n\right) = f \left(n - 1\right) \cdot x$
or some variation of it. Our task, most of the time is to find $x$ and then figure out how to write the formula so that we don't have to calculate each term in order to find the next one.

In this case, if you start dividing each of the known members by the previous one (eg. $\frac{686}{98}$) we get 7 as a result, so we know that:
$f \left(n\right) = f \left(n - 1\right) \cdot 7$

Looking from the begining (let's start at n=2) we have that:
$f \left(2\right) = f \left(1\right) \cdot 7 = 2 \cdot 7 = 14$
$f \left(3\right) = f \left(2\right) \cdot 7 = \left(f \left(1\right) \cdot 7\right) \cdot 7 = f \left(1\right) \cdot {7}^{2} = 2 \cdot {7}^{2} = 98$
$f \left(4\right) = f \left(3\right) \cdot 7 = \left[f \left(2\right) \cdot 7\right] \cdot 7 = \left[\left(f \left(1\right) \cdot 7\right) \cdot 7\right] \cdot 7 = f \left(1\right) \cdot {7}^{3} = 2 \cdot {7}^{3} = 686$

Notice that the last part of each line is $f \left(1\right) \cdot {7}^{y}$, where, conveniently, $y = n - 1$. Also remember that any number raised to 0 is equal to 1 and that $f \left(1\right) = 2$

This allows us to write:
$f \left(n\right) = 2 \cdot {7}^{n - 1}$
as the final formula for any number in the sequence.

Now I can open my calculator and find that $f \left(10\right) = 2 \cdot {7}^{9} = 80.707 .214$, or, as asked, $f \left(6\right) = 2 \cdot {7}^{5} = 33614$