MATH SOLVE

4 months ago

Q:
# What are the explicit equation and domain for an arithmetic sequence with a first term of 6 and a second term of 2? an = 6 − 2(n − 1); all integers where n ≥ 1 an = 6 − 2(n − 1); all integers where n ≥ 0 an = 6 − 4(n − 1); all integers where n ≥ 0 an = 6 − 4(n − 1); all integers where n ≥ 1

Accepted Solution

A:

First term of 6→when n=1→an=a1=6

In the expressions we have a term that contains n-1, then n ≥ 1, then it can be options 1 and 4.

A second term of 2→when n=2→an=a2=2

With the first option when n=2 we get:

a2=6-2(2-1)=6-2(1)=6-2→a2=4 different of 2. This equation doesn't work.

With the fourth option when n=2 we get:

a2=6-4(2-1)=6-4(1)=6-4→a2=2. This equation works.

Answer: Fourth option an = 6 − 4(n − 1); all integers where n ≥ 1

In the expressions we have a term that contains n-1, then n ≥ 1, then it can be options 1 and 4.

A second term of 2→when n=2→an=a2=2

With the first option when n=2 we get:

a2=6-2(2-1)=6-2(1)=6-2→a2=4 different of 2. This equation doesn't work.

With the fourth option when n=2 we get:

a2=6-4(2-1)=6-4(1)=6-4→a2=2. This equation works.

Answer: Fourth option an = 6 − 4(n − 1); all integers where n ≥ 1