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