Question

NO LINKS!!!

a. Identify each sequence as arithmetic, geometric, or neither.
b. If it is arithmetic or geometric, describe the sequence generator

n t(n)
0 16
1 8
2 4
3 2
4 1
5 1/2
6 1/4

Answer 1

a) Geometric sequence

`\textsf{b)} \quad t(n)=8 \left((1)/(2)\right)^(n-1)`

Explanation:

An Arithmetic Sequence has a constant difference between each consecutive term.

A Geometric Sequence has a constant ratio (multiplier) between each consecutive term.

Part (a)

From inspection of the given table, t(n) halves each time n increases by 1.

Therefore, the sequence is geometric with a common ratio of 1/2.

Part (b)

`\boxed{\begin{minipage}{5.5 cm}\underline{Geometric sequence}\\\\$a_n=ar^(n-1)$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the first term. \\\phantom{ww}$\bullet$ $r$ is the common ratio.\\\phantom{ww}$\bullet$ $a_n$ is the $n$th term.\\\phantom{ww}$\bullet$ $n$ is the position of the term.\\\end{minipage}}`

Given:

• a = 8
• r = 1/2
• aₙ = t(n)

Substitute the initial value (when n = 1) and the common ratio into the formula to create an equation for the nth term:

`\implies t(n)=8 \left((1)/(2)\right)^(n-1)`

Answer 2

• a) The sequence is geometric,
• b)
  `t(n) = 2^(4-n)`

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Given sequence from zeros to 6th terms:

• 16, 8, 4, 2, 1, 1/2, 1/4

It is a GP with the common ratio of 1/2 as each term is half the previous one.

Use the nth term equation, considering the first term is t(1) = 8, and common ratio r = 1/2:

• 
  `t(n) = t(1)*r^(n-1)`
• 
  `t(n) = 8*(1/2)^(n-1)=8*2^(1-n)=2^3*2^(1-n)=2^(4-n)`