Question: "What is the eighth term given the sequence:..."

High School
Algebra
Pre-Algebra
Equations, expressions, and inequalitie

Lyric Morris

Answered question

2022-02-05

What is the eighth term given the sequence:

+ 896, - 448, + 224, - 112, \dots

Answer & Explanation

vhudzeniy2n

Beginner2022-02-06Added 16 answers

Step 1
Let any position count be i
Let any term be $a_{i}$
So
$a_{1} \to$ first term
$a_{2} \to$ second term
$a_{i} \to$ its term
Two things I notice:
Step 1
Alternates between positive and negative starting at positive.
So this can be achieved by
$a_{i} \to a_{1} \times {(- 1)}^{2}$
$a_{i} \to a_{2} \times {(- 1)}^{3}$
$a_{i} \to a_{3} \times {(- 1)}^{4}$
$\Rightarrow a_{i} \to a_{i} \times {(- 1)}^{i + 1}$
Step 2
Each term is half the previous
So we have
${(- 1)}^{2} \times a_{1}$
$a_{2} = {(- 1)}^{3} \times \frac{1}{2} \times a_{1}$
$a_{3} = {(- 1)}^{4} \times \frac{1}{2} \times \frac{1}{2} \times a_{1}$
This implies $a_{1} = {(- 1)}^{i + 1} \times {(\frac{1}{2})}^{i - 1} \times a_{1}$
Step 3
$a_{i} = 896 {(- 1)}^{i + 1} {(\frac{1}{2})}^{i - 1}$
Thus
$a_{8} = 896 {(- 1)}^{8 + 1} {(\frac{1}{2})}^{8 - 1} = - 7$

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