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

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Answer:

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

Explanation:

The equation is:

$a_n=3a_{n-1}-5a_{n-2}-4a_{n-3}$

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

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 $a_1$

$a_4=3a_{3}-5a_{2}-4a_{1}$

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

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