Chapter 12.2, Problem 9AYU

In Problems 7-16, show that each sequence is arithmetic. Find the common difference and write out the first four terms.

$\left\{c_n\right\}=\left\{6-2n\right\}$

To determine

To find: Each sequence is arithmetic and the common difference. Write out the first four terms $\left\{c_n\right\}=\left\{6-2n\right\}$.

Answer to Problem 9AYU

$d=-2$, $c_1=4$, $c_2=2$, $c_3=0$, $c_4=-2$

Explanation of Solution

Given:

$\left\{c_n\right\}=\left\{6-2n\right\}$

Calculation:

The $n^{\text{th}}$ and ${\left(n-1\right)}^{\text{th}}$ term of the sequence $\left\{c_n\right\}$ are

$c_n=6-2n\text{ and }{c}_{n-1}=6-2n+2=8-2n$

$d=c_n-c_{n-1}=\text{ 6}-\text{ 2<mi>n</mi> }-\text{ 8}+\text{ 2<mi>n</mi> }=\text{ <mo>−</mo>2}$   (a constant)

Since the difference of any two successive term is the constant, the given sequence is arithmetic and the common difference is $-2$.

The first four terms are,

$c_1=6-\text{ 2 }=4$

$c_2=6-\text{ 4 }=2$

$c_3=6-\text{ 6 }=0$

$c_4=6-\text{ 8 }=-2$