The first two terms of a geometric sequence are a1 = 1⁄3 and a2 = 1⁄6. What is a8, the eighth term?

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Answer 1 · 2016-08-12T15:04:32

Use the formula $t_{n} = a \times r^{n - 1}$ $t_n = a xx r^(n - 1)$

Let's first find the value of $r$ $r$ , the common ratio.

$r = \frac{t_{2}}{t_{1}}$ $r = t_2/t_1$

$r = \frac{\frac{1}{6}}{\frac{1}{3}}$ $r = (1/6)/(1/3)$

$r = \frac{1}{6} \times 3$ $r = 1/6 xx 3$

$r = \frac{1}{2}$ $r = 1/2$

Now we can us the formula.

$t_{n} = a \times r^{n - 1}$ $t_n = a xx r^(n - 1)$

$t_{8} = \frac{1}{3} \times {(\frac{1}{2})}^{8 - 1}$ $t_8 = 1/3 xx (1/2)^(8 - 1)$

$t_{8} = \frac{1}{8} \times \frac{1}{128}$ $t_8 = 1/8 xx 1/128$

$t_{8} = \frac{1}{1024}$ $t_8 = 1/1024$

Hopefully this helps!