# What is the eighth term given the sequence: +896, -448, +224,-112,....... ?

Sep 12, 2016

the eighth term is $\text{ } - 7$

#### Explanation:

Let any position count be $i$
Let any term be ${a}_{i}$

So
${a}_{1} \to \text{ first term}$
${a}_{2} \to \text{ second term}$
${a}_{i} \to \text{ "ith" term}$

Two things I notice:
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Point 1:}} \to$Alternates between positive and negative starting at positive.

So this can be achieved by

${a}_{i} \to {a}_{1} \times {\left(- 1\right)}^{2}$

${a}_{i} \to {a}_{2} \times {\left(- 1\right)}^{3}$

${a}_{i} \to {a}_{3} \times {\left(- 1\right)}^{4}$

$\implies {a}_{i} \to {a}_{i} \times {\left(- 1\right)}^{i + 1}$

,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Point 2:}} \to$Each term is half the previous

So we have

${\left(- 1\right)}^{2} \times {a}_{1}$
${a}_{2} = {\left(- 1\right)}^{3} \times \frac{1}{2} \times {a}_{1}$
${a}_{3} = {\left(- 1\right)}^{4} \times \frac{1}{2} \times \frac{1}{2} \times {a}_{1}$

This implies ${a}_{i} = {\left(- 1\right)}^{i + 1} \times {\left(\frac{1}{2}\right)}^{i - 1} \times {a}_{1}$

Where ${a}_{1} = + 896$
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

color(blue)(a_i=896(-1)^(i+1)(1/2)^(i-1)

Thus

$\text{ } \textcolor{p u r p \le}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} {a}_{8} = 896 {\left(- 1\right)}^{8 + 1} {\left(\frac{1}{2}\right)}^{8 - 1} = - 7 \textcolor{w h i t e}{\frac{2}{2}} |}}}$