Question

What are the first three terms of the sequence defined by the recursive function

an=an-1-(an-2-4)
a5=-2
a6=0

A)-14,14,-4
B)6,10,8
C)2,0,0
D)2,8,10

Answer 1

6, 10, 8

Explanation:

Answer 2

Option B is correct.

First three terms;

6, 10, 8

Explanation:

Given the recursive function:

`a_n = a_(n-1)-(a_(n-2)-4)` .....[1]

`a_5 = -2`

`a_6=0`

Put n = 6 in [1] we have;

`a_6= a_(6-1)-(a_(6-2)-4)`

Simplify:

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

Substitute the given values;

`0 = -2-(a_4-4)`

Add 2 to both sides we have;

`2 =-(a_4-4)`

or

`2 =-a_4+4`

Subtract 4 from both sides we have;

`-2=-a_4`

or

`a_4 = 2`

Put n = 5 in [1] we get;

`a_5= a_(5-1)-(a_(5-2)-4)`

Simplify:

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

Substitute the given values;

`-2= 2-(a_3-4)`

Subtract 2 from both sides we have;

`-4 =-(a_3-4)`

or

`4 =a_3 - 4`

Add 4 to both sides we have;

`8=-a_3`

or

`a_3 = 8`

Put n = 4 in [1] we get;

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

Simplify:

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

Substitute the given values;

`2= 8-(a_2-4)`

Subtract 8 from both sides we have;

`-6 =-(a_2-4)`

or

`6 =a_2 - 4`

Add 4 to both sides we have;

`10=a_2`

or

`a_2 = 10`

Put n = 3 in [1] we get;

`a_3= a_(3-1)-(a_(3-2)-4)`

Simplify:

`a_3= a_(2)-(a_(1)-4)`

Substitute the given values;

`8= 10-(a_1-4)`

Subtract 10 from both sides we have;

`-2 =-(a_1-4)`

or

`2 =a_1 - 4`

Add 4 to both sides we have;

`6=a_1`

or

`a_1 = 6`

therefore, the first three terms of the given sequence are; 6, 10, 8