Respuesta :
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
Answer:
option D
Step-by-step explanation:
an arithmetic sequence with a first term of 6 and a second term of 2
First term is 6 and second term is 2. the difference of both the terms is
2- 6= -4
Common difference d= -4
first term a= 6
For explicit equation we use formula
[tex]a_n = a1 + (n-1)d[/tex]
Where 'a1' is the first term and d is the common difference
a1= 6, d=-4
[tex]a_n = 6 + (n-1)(-4)[/tex]
[tex]a_n =6-4(n-1)[/tex] Explicit equation
Our first term a1 = 6, so n=1
Hence n represents all integers where n>=1
