Question

Suppose an=3an-1 -5an-2 -4an-3. and a4= -7, a5=-36, a6=-85 , find a1 , a2, a3​

Answer 1

`a_1=4\\\\a_2=0\\\\a_3=3`

Step-by-step explanation:

The equation is:

`a_n=3a_(n-1)-5a_(n-2)-4a_(n-3)`

and

`a_4=-7, a_5=-36, a_6=-85`

1. Subsittute n = 6 into the equation:

`a_6=3a_(5)-5a_(4)-4a_(3)`

Now subsititute the known values:

`-85=3(-36)-5(-7)-4a_(3)`

You can solve for
`a_3`

`-85=-108+35-4a_(3)\\\\4a_3=12\\\\a_3=12/4\\\\a_3=3`

2. Substitute n = 5 into the equation:

`a_5=3a_(4)-5a_(3)-4a_(2)`

Substitute the known values and solve for
`a_2`

`-36=3(-7)-5(3)-4a_(2)\\\\4a_2=-21-15+36=0\\\\a_2=0`

3. Substitute n = 4 into the equation, subsitute the known values and solve for
`a_1`

`a_4=3a_(3)-5a_(2)-4a_(1)`

`-7=3(3)-5(0)-4a_(1)\\\\4a_1=9+7=16\\\\a_1=4`