# What is the seventh term of the geometric sequence where a1=-4,096 a4=64 ?

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Feb 9, 2018

The seventh term of the geometric sequence is
$a 7 = - 1$

#### Explanation:

Given:
$a 1 = - 4096$
$a 4 = 64$
To find:
a7=?

We know that,

In geometric progression, the successive terms are in constant ratio.

Let r be the constant ratio.

1st term $a 1 = - 4096$
2nd term $a 2 = - 4096 r$
3rd term $a 3 = - 4096 {r}^{2}$
4th term $a 4 = - 4096 {r}^{3}$

Equating $a 4 = 64$

Simplifying,
$- 4096 {r}^{3} = 64$
${r}^{3} = \frac{64}{-} 4096$
$64 = {4}^{3}$
$- 4096 = {\left(- 16\right)}^{3}$

Thus,
${r}^{3} = {4}^{3} / {\left(- 16\right)}^{3}$

By the law of indices
${4}^{3} / {\left(- 16\right)}^{3} = {\left(\frac{4}{- 16}\right)}^{3}$

Now,
${r}^{3} = {\left(\frac{4}{- 16}\right)}^{3}$

Simplifying
$r = \frac{4}{-} 16 = \frac{1}{-} 4 = - \frac{1}{4}$

The common ratio is thus, $- \frac{1}{4}$
5th term $a 5 = 64 \left(- \frac{1}{4}\right) = - 16$
6th term$a 6 = - 16 \left(- \frac{1}{4}\right) = 4$
7th term $a 7 = 4 \left(- \frac{1}{4}\right) = - 1$

Listing the series
$a 1 = - 4096$
$a 2 = - 4096 \left(- \frac{1}{4}\right) = 1024$
$a 3 = 1024 \left(- \frac{1}{4}\right) = - 256$
$a 4 = - 256 \left(- \frac{1}{4}\right) = 64$, checked
$a 5 = 64 \left(- \frac{1}{4}\right) = - 16$
$a 6 = - 16 \left(- \frac{1}{4}\right) = 4$
$a 7 = 4 \left(- \frac{1}{4}\right) = - 1$

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