Exercise 124 Page 484 - 2. Section 8.2 - Core Connections Integrated I, 2013

a The first term is already given, so we can use it to find the second term. Then we will use the second term

t (2)

to find the third one

t (3),

and so on.

t (1) = 12 t (2) = t (1) - 5 = 12 - 5 = 7 t (3) = t (2) - 5 = 7 - 5 = 2 t (4) = t (3) - 5 = 2 - 5 = - 3 t (5) = t (4) - 5 = - 3 - 5 = - 8

To find an explicit equation for the sequence notice that each term is obtained by subtracting

5

from the previous one, starting with

12 .

Therefore, our explicit equation becomes the following.

t (n) = 12 - 5 (n - 1)

b Just as we did in Part A, we will use the first given term

t (1)

to find the second term

t (2) .

Then, using

t (2)

we will find the third term

t (3),

and so on.

t (1) = 32 t (2) = \frac{1}{2} t (1) = \frac{1}{2} (32) = 16 t (3) = \frac{1}{2} t (2) = \frac{1}{2} (16) = 8 t (4) = \frac{1}{2} t (3) = \frac{1}{2} (8) = 4 t (5) = \frac{1}{2} t (4) = \frac{1}{2} (4) = 2

To find an explicit equation for the sequence, notice that each term is obtained by multiplying the previous one by

\frac{1}{2},

starting from

32 .

Therefore, our explicit equation becomes the following.

t (n) = 32 (\frac{1}{2})^{n - 1}