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cinderofsoulsss

Answer:

4) a₁₅ = 128

5) B. [tex]a_n=a_{n-1}+6[/tex]

Step-by-step explanation:

4)

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

[tex]a_n=a_1+(n-1)d[/tex]

Here, the common difference is 9.

[tex]11-2=9\\20-11=9[/tex]

Use that to write the equation and solve for a₁₅:

[tex]a_n=2+(n-1)9\\\\a_{15}=2+(15-1)9\\a_{15}=2+(14)9\\a_{15}=2+126\\a_{15}=128[/tex]

5)

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

[tex]a_n=a_{n-1}+d[/tex]

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

Here, the common difference is 6:

[tex]2--4=6\\8-2=6\\14-8=6[/tex]

And the equation for this is B. [tex]a_n=a_{n-1}+6[/tex]

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