Question

what is the 14th term of the given sequence
64,-32,16,-8….

Answer 1

Answer:

$-\frac{1}{128}$

Step-by-step explanation:

1) Identify the type of sequence this is; whether it is geometric progression, quadratic sequence or linear sequence.

= Geometric progression

2) Find the $n^{th}$ term of this sequence. Every sequence of geometric progression is written in $u_{n} = u_{1} * r^{n-1}$ form, where $u_{1}$ is the first term of the sequence and $r$ is the common ratio. To find the $n^{th}$ term of this sequence, we need to find the common ratio ($r$) first.

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

$r = \frac{-32}{64}$

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

2.1) Write it in $u_{n} = u_{1} * r^{n-1}$ form.

$u_{n} = 64 * (-\frac{1}{2} )^{n-1}$

3) Find the $14^{th}$ term by substituting 14 into the $n$'s.

$u_{14} = 64 * (-\frac{1}{2} )^{14-1}$

$u_{14} = 64 * (-\frac{1}{2} )^{13}$

$u_{14} = 64 * -\frac{1}{8192}\\u_{14} = -\frac{1}{128}$