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

Answer:

Option B is correct.

First three terms;

6, 10, 8

Step-by-step 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

Answer 2

Answer:

6, 10, 8

Step-by-step explanation: