Exercise 5 Page 326

We are asked to write a recursive rule and an explicit rule for the given sequence. We know that the second term of the given sequence is 7. Furthermore, each term of the sequence is 10 more than the preceding term. Therefore the sequence is an arithmetic sequence, and its common difference is equal to d=10. This means that we can fill in one blank already. Let's do it! a_1= , a_n=a_(n-1)+10 [0.3em] a_n= n- Since the second term is 7 and it is 10 more than the first term, we get that the first term is 10 less than 7. This is 7-10=- 3. Therefore, we can fill in the first blank. a_1=- 3, a_n=a_(n-1)+10 [0.3em] a_n= n- Next, we will find the explicit rule of the arithmetic sequence. a_n=a_1+(n-1)d In our case, a_1= - 3 and d= 10. Let's substitute these values and find the rule!

a_n= a_1+(n-1) d

SubstituteII

a_1= - 3, d= 10

a_n= - 3 +(n-1)( 10)

▼

Simplify right-hand side

Distr

Distribute 10

a_n=- 3+n(10)-1(10)

Multiply

Multiply

a_n=- 3+10n-10

SubTerms

Subtract terms

a_n=10n-13

Finally, we can fill in all the blanks in the exercise! a_1=- 3, a_n=a_(n-1)+10 [0.3em] a_n=10 n-13