# Question #ca3df

Mar 20, 2017

Given that the first term of the geometric sequence ${a}_{1} = - 4096$

and the 4 th term of the geometric sequence ${a}_{4} = 64$

Let the common ratio of the geometric sequence be $r$

So
${a}_{4} = {a}_{1} {r}^{3} = 64$

$\implies \left(- 4096\right) \times {r}^{3} = 64$

$\implies {r}^{3} = - \frac{64}{4096} = - \frac{1}{64} = {\left(- \frac{1}{4}\right)}^{3}$

$\implies r = - \frac{1}{4}$

Hence the 7 th term will be

${a}_{7} = {a}_{1} \times {r}^{6} = - 4096 \times {\left(- \frac{1}{4}\right)}^{6} = - \frac{4096}{4096} = - 1$