Find the general term for these sequence.

Question

12,−6,3,−32,34,−38,316,...

5,−4,3,−2,1,0,−1,2,−3,...

Answer 1

5,-4,3,-2,1,0,-1,2,-3, ...

a1 = 5

a2 = 1-a1= (n-1)-a1

a3 = a1 -2 =a1-(n-1)

a4 = 3-a1= (n-1)-a1

an = a1-n+1 when n is odd

an = 6-n when n is odd

an = n-1-a1 when n is even

an= n-6 when n is even

n-6 =-(6-n)

use (-1)^n to alternate the signs

an = (-1)^n][n-6] for all nth terms

a9 = -(9-6) = -3

a10 = (10-6) = 4

for the other problem

12, -6, 3, -32, 34, -38, 316

you could also use a similar coefficient (-1)^(n+1) or (-1)^(n-1) to alternate the signs of the sequence.

12, -6, 3 is a geometric sequence with common ratio r = -1/2 which replaces the need for (-1)^(n+1), at least for the first three terms.

infinite possibilities for the other terms, none obvious