Exercise 1 Page 439

Explicit equations for arithmetic sequences follow a specific format. a_n= a_1+( n-1) d In this form, a_1 is the first term in a given sequence, d is the common difference from one term to the next, and a_n is the nth term in the sequence. For this exercise, the first term is a_1= 3. Let's observe the other terms to determine the common difference d. 3+6 →9+6 →15+6 →21... By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence.

a_n=a_1+(n-1)d

SubstituteII

d= 6, a_1= 3

a_n= 3+(n-1)( 6)

Distr

Distribute 6

a_n=3+6n-6

SubTerm

Subtract term

a_n=6n-3

This equation can be used to find any term in the given sequence. To find a_(50), the 50th term in the sequence, we substitute 50 for n.

a_n=6n-3

Substitute

n= 50

a_(50)=6( 50)-3

Multiply

Multiply

a_(50)=300-3

SubTerm

Subtract term

a_(50)=297

The 50th term in the sequence is 297.