Indicate a general rule for the n^(th) term of this sequence. -6a, -3a, 0a, 3a, 6a. . . A.] a_(n)=3an+9a B.] a_(n)=-3an-9a C.] a_(n)=-3an+9a D.] a_(n)=3an-9a

Question

Indicate a general rule for the
`n^(th)` term of this sequence.

-6a, -3a, 0a, 3a, 6a. . .
A.]
`a_(n)=3an+9a`
B.]
`a_(n)=-3an-9a`
C.]
`a_(n)=-3an+9a`
D.]
`a_(n)=3an-9a`

Answer 1

Answer: option D) An = 3an - 9a

Justification:

1) That is an arithmetic sequence which you prove by determining the difference between two consecutive terms (which shall be constant if it indeed is an arithmetic sequence):

2)

Second term - first term = -3a - (-6a) = -3a + 6a = 3a
Third term - second term = 0a - (-3a) = 3a
Fourth term - third term = 3a - 0a = 3a
Fith term - fourth term = 6a - 3a = 3a

Therefore, it is an arithmetic sequence with difference d = 3a.

3) the general rule for the nth term of an arithmetic sequence is given by the formula:

An = Ao + d (n - 1)

Where Ao = -6a and d, as determined above, is 3a

=> An = -6a + 3a (n - 1)

Expand the parenthesis (distributive property) =>

An = -6a + 3an - 3a = -9a + 3an = 3an - 9a = option D. <------- answer.