Question

Indicate a general rule for the nth term of this sequence. 6a, 3a, 0, -3a, -6a, . . .

Answer 1

6a-(n-1)(3a) is the nth term
Expanding we get 6a-3an+3a=9a-3an=3a(3-n)
SOLUTION: nth term is 3a(3-n)

Answer 2

Answer:

The answer is $a_n=6a-3a(n-1)$

Step-by-step explanation:

In order to determine the general rule, we have to identify the difference between each term. We have to subtract every two continuous terms:

$3a-6a=-3a\\0-3a=-3a\\-3a-0=-3a\\-6a-(-3a)=-6a+3a=-3a$

Then, we know that we have to subtract -3a to every next term. The first term of the sequence is 6a and after that we have to subtract -3a since the second to the nth term.

Thus the nth term of the sequence is:

$a_n=6a-3a(n-1)$

To check the expression:

$n=1\\a_1=6a-3a(1-1)=6a-3a(0)=6a\\\\n=3\\a_3=6a-3a(3-1)=6a-3a(2)=6a-6a=0$

So, the expression is right.