Question

HELP!!!!!!!!!!!!!!!!!!!

Answer 1

Answer:

4) a₁₅ = 128

5) B. $a_n=a_{n-1}+6$

Step-by-step explanation:

4)

This is an arithmetic sequence, where all terms have a common difference:

$a_n=a_1+(n-1)d$

Here, the common difference is 9.

$11-2=9\\20-11=9$

Use that to write the equation and solve for a₁₅:

$a_n=2+(n-1)9\\\\a_{15}=2+(15-1)9\\a_{15}=2+(14)9\\a_{15}=2+126\\a_{15}=128$

5)

This is also an arithmetic sequence, but it is a recursive arithmetic sequence in the form of:

$a_n=a_{n-1}+d$

This simply states that each term is d more than the term before it.

Here, the common difference is 6:

$2--4=6\\8-2=6\\14-8=6$

And the equation for this is B. $a_n=a_{n-1}+6$