Write a recursive rule for the sequence. Then find {eq}a_7 {/eq}.

{eq}a_1 = 16; r = \frac{1}{4} {/eq}

Write a recursive rule for the sequence. Then find {eq}a_7 {/eq}.

{eq}a_1 = 16; r = \frac{1}{4} {/eq}

For the case of a geometric series, we have the recurrence relation:

{eq}\displaystyle {a_{n + 1}} = {a_n} \cdot r \Rightarrow {a_{n + 1}} = {a_n} \cdot \frac{1}{4} = \frac{1}{4}{a_n}. {/eq}

Obtaining the seventh term, it results:

{eq}\displaystyle {a_1} = 16\\ \displaystyle {a_2} = \frac{1}{4} \cdot 16 = 4\\ \displaystyle {a_3} = \frac{1}{4} \cdot 4 = 1\\ \displaystyle {a_4} = \frac{1}{4} \cdot 1 = \frac{1}{4}\\ \displaystyle {a_5} = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{{16}}\\ \displaystyle {a_6} = \frac{1}{4} \cdot \frac{1}{{16}} = \frac{1}{{64}}\\ \displaystyle {a_7} = \frac{1}{4} \cdot \frac{1}{{64}} = \frac{1}{{256}} {/eq}

Alternatively, the recursive relation implies

{eq}\displaystyle a_{n + 1} = {a_n} \cdot r \Rightarrow a_{n} = {a_1} \cdot r^{n-1} \\ \displaystyle a_n = 16 \frac{1}{4^{n-1}} \\ \displaystyle a_7 = 16 \frac{1}{4^{6}} = \frac{1}{4^{4}} = \frac{1}{256} {/eq}