Exercise 12 Page 325

We are given the first two terms of the sequence: a_1=3 and a_2=- 12. We know that a_3 is the third term when the sequence is arithmetic, and b_3 is the third term when the sequence is geometric. Let's recall the basic facts about these two types of sequences.

• A sequence is arithmetic if the difference between consecutive terms is constant. This difference is usually called the common difference, and is denoted by d.
• A sequence is geometric if the ratio between consecutive terms is constant. This ratio is usually called the common ratio, and is denoted by r.

Finding a_3

Since a_1=3, a_2=- 12, and a_3 form an arithmetic sequence, let's find its common difference d by finding the difference between the first and the second term.

d=a_2-a_1

SubstituteII

a_2= - 12, a_1= 3

d= - 12- 3

SubTerms

Subtract terms

d=- 15

Each subsequent term in the arithmetic sequence can be obtained by adding d to the previous one. Knowing this, we can find a_3.

a_3=a_2+d

SubstituteII

a_2= - 12, d= - 15

a_3= - 12+( - 15)

AddTerms

Add terms

a_3=- 27

Finding b_3

Since a_1=3, a_2=- 12, and b_3 form a geometric sequence, let's find its common ratio d by finding the ratio between the first and the second term.

r=a_2/a_1

SubstituteII

a_2= - 12, a_1= 3

r=- 12/3

CalcQuot

Calculate quotient

r=- 4

Each subsequent term in the geometric sequence can be obtained by multiplying the previous one by r. Knowing this, we can find b_3.

b_3=a_2* r

SubstituteII

a_2= - 12, r= - 4

b_3= - 12* ( - 4)

Multiply

Multiply

b_3=48

Finding a_3-b_3

Finally, we only have to substitute the appropriate values and calculate a_3-b_3.

a_3-b_3

SubstituteII

a_3= - 27, b_3= 48

- 27-48

SubTerms

Subtract terms

- 75