Question

100 POINTS! PLEASE HELP ASAP!

Answer 1

Question #1:

Since the common ratio is -2 that means each term is going to be multiplied by -2 to get the next term.

First term: -8

Second term: -8 * -2 = 16

Third term: 16 * -2 = -32

Fourth term: -32 * -2 = 64

Therefore, the first 4 terms of the geometric sequence are -8, 16, -32, 64

Question #2:

It gives an equation to follow to solve for each term of the sequence. We can solve for each term till we get to the 4th term.

Second term:

a2 = 2(5) + 3

a2 = 10 + 3

a2 = 13

Third term:

a3 = 2(13) + 3

a3 = 26 + 3

a3 = 29

Fourth term:

a4 = 2(29) + 3

a4 = 58 + 3

a4 = 61

Therefore, the fourth term of the sequence is 61.

Question #3:

a5 just means that we are going to find the fifth term of the sequence.

Second term:

a2 = -3(2) + 2

a2 = -6 + 2

a2 = -4

Third term:

a3 = -3(-4) + 2

a3 = 12 + 2

a3 = 14

Fourth term:

a4 = -3(14) + 2

a4 = -42 + 2

a4 = -40

Fifth term:

a5 = -3(-40) + 2

a5 = 120 + 2

a5 = 122

Therefore, the fifth term of the sequence is 122.

Best of Luck!

Answer 2

Answer:

PICTURE #1

$a_1 = 5\\a_2 = 2a__(n-1)$ $+ 3$

Step 1:  Solve for $a_2$

a1 = 5

$a_2 = 2a___n-1$ $+3$

$a_2=2a__(2-1)$ + 3

$a_2 = 2a_1 + 3$

$a_2 = 2(5) + 3$

$a_2 = 10 + 3$

$a_2 = 13$

Step 2:  Solve for $a_3$

$a_2 = 13$

$a_3 = 2a__(3-1)$ + 3

$a_3 = 2a_2 + 3$

$a_3 = 2(13) + 3$

$a_3 = 26 + 3$

$a_3 = 29$

Step 3:  Solve for $a_4$

$a_3 = 29$

$a_4 = 2a__(4-1)$ + 3

$a_4 = 2a_3 + 3$

$a_4 = 2(29) + 3$

$a_4 = 58 + 3$

$a_4 = 61$

The fourth terms is:  $a_4 = 61$

PICTURE #2

$a_1 = 2\\a_2 = -3a__(n - 1)$ $+ 2$

Step 1:  Find $a_2$

$a_2 = -3a_1 + 2$

$a_2 = -3(2) + 2$

$a_2 = -6 + 2$

$a_2 = -4$

Step 2:  Find $a_3$

$a_3 = -3a_2 + 2$

$a_3 = -3(-4) + 2$

$a_3 = 12 + 2$

$a_3 = 14$

Step 3:  Find $a_4$

$a_4 = -3a_3 + 2$

$a_4 = -3(14) + 2$

$a_4 = -42 + 2$

$a_4 = -40$

Step 4:  Find $a_5$

$a_5 = -3a_4 + 2$

$a_5 = -3(-40) + 2$

$a_5 = 120 + 2$

$a_5 = 122$

Answer:  The fifth term for the second picture is -> $a_5 = 122$