# How do you find a_7 for the geometric sequence 1/32, 1/16, 1/8,...?

Dec 7, 2016

$\frac{3}{8}$

#### Explanation:

Let the ith term be ${a}_{i}$

Note that:$\text{ } \frac{1}{32} < \frac{1}{16} < \frac{1}{8}$

$\textcolor{b l u e}{\text{First thoughts}}$

So the first term is: $\text{ } {a}_{i} \to {a}_{1} = \frac{1}{32}$

Notice that $16 = \frac{32}{2} \text{ and that } 8 = \frac{16}{2}$

To change $\frac{1}{32}$ into $\frac{1}{16}$ we multiply by 2 in that:

$\frac{1}{32} \times 2 = \frac{2}{32} = \frac{2 \div 2}{32 \div 2} = \frac{1}{16}$

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$\textcolor{b l u e}{\text{Determine the pattern}}$

${a}_{1} \text{ "->" } \frac{1}{32}$

${a}_{2} \text{ "->" } \frac{1}{32} \times 2 = \frac{1}{16}$

${a}_{3} \text{ "->" } \frac{1}{32} \times 2 \times 2 = \frac{1}{8}$

By observation we spot that for any $i$ we have:

${a}_{i} = \frac{1}{32} \times 2 \left(i - 1\right)$

Thus $\text{ "a_i->a_7=1/32xx2(7-1)" "=" "1/32xx12" "=" } \frac{12}{32}$

$\frac{12}{32} \text{ "-=" "(12-:4)/(32-:4) " "=" } \frac{3}{8}$