Question

What is the nth term of the quadratic sequence: -8,-8,-6,-2,4,12,22?

Answer 1

quadratic sequence

seems to be

`a_n=an^2+bn+c`

where
`a_1=-8`,
`a_2=-8` and so on

we only need 3 points

(1,-8)

(2,-8)

(3,6)

subsitute and solve

(1,-8)

`-8=a(1)^2+b(1)+c`

`-8=a+b+c`

(2,-8)

`-8=a(2)^2+b(2)+c`

`-8=4a+2b+c`

(3,6)

`6=a(3)^2+b(3)+c`

`6=9a+3b+c`

so we have

-8=a+b+c

-8=4a+2b+c

6=9a+3b+c

add negative of the first equation to the last 2 equations to get

-8=a+b+c

0=3a+b

14=8a+2b

work with the last 2 equations

0=3a+b

14=8a+2b

multiply 1st equation by -2 and add to 2nd

0=-6a-2b

14=8a+2b +

14=2a+0

14=2a

divide by 2

7=a

sub back

0=3a+b

0=3(7)+b

0=21+b

-21=b

sub back

-8=a+b+c

-8=7-21+c

-8=-14+c

6=c

so a=7, b=-21, c=6

the nth term is
`a_n=7n^2-21n+6`

or in function form,
`f(n)=7n^2-21n+6`