Question

In the problem, show that each sequence is arithmetic. Find the common difference and write out the first four terms.

$\left\{a_n\right\}=\left\{4-2n\right\}$

Short answer

The solution is $d=-2$; first four terms$=2,0,-2,-4$.

Step-by-step solution

Step 1. Given information 

$\left\{a_n\right\}=\left\{4-2n\right\}$

Step 2. Calculation 

The first term is$a_1=4-2·1=2$.

The $n$th term and the $\left(n-1\right)$st term of the sequence are-

$a_n=4-2n$ and $a_{n-1}=4-2\left(n-1\right)=4-2n+2=6-2n$

The common difference is -

$d=a_n-a_{n-1}=\left(4-2n\right)-\left(6-2n\right)=-2$

Since the difference of any two successive terms is the constant $-2$, the sequence $\left\{a_n\right\}$is arithmetic.

The first four terms are$2,0,-2,-4$.