HSF.BF.A.2 Math Practice Questions — Sports Academy Games

Practice playeasy

The first five terms of an arithmetic sequence are $12, 18, 24, 30, 36$ . What is the $11$ th term?

The common difference is $d = 6$ . With $a_1 = 12$ , the $11$ th term is $a_{11} = 12 + 10(6) = 72$ .

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In an arithmetic sequence, the fifth term is $20$ and the sixth term is $14\textsf{.}$ What is the ninth term?

The common difference is $14 - 20 = -6\textsf{.}$ From the sixth term to the ninth term there are $3$ steps, so the ninth term is $14 + (-6) \times 3 = -4\textsf{.}$

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The first term of a sequence is $2$ . Each term after the first is $4$ times the preceding term. If $a_n$ represents the $n$ th term, which equation gives $a_n$ in terms of $n$ ?

Following the pattern:
1st term, or $n = 1\textsf{:}$ $2 \cdot 4^{0}$
2nd term, or $n = 2\textsf{:}$ $2 \cdot 4^{1}$
3rd term, or $n = 3\textsf{:}$ $2 \cdot 4^{2}$
4th term, or $n = 4\textsf{:}$ $2 \cdot 4^{3}$
Since each term has an exponent that is $1$ less than the term number, the $n$ th term is $a_n = 2 \cdot 4^{n-1}$ .

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The first five terms of an arithmetic sequence are $1, 4, 7, 10, 13$ . What is the $16$ th term?

The common difference is $d = 3$ . With $a_1 = 1$ , the $16$ th term is $a_{16} = 1 + 15(3) = 46$ .

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Find the $5$ th term: $a_1 = 3$ , $d = 4$ .

$a_5 = 3 + 4(4) = 3 + 16 = 19$ .

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The first term of a sequence is $8$ . Each term after the first is $2$ times the preceding term. If $f(n)$ represents the $n$ th term, which equation gives $f(n)$ in terms of $n$ ?

Following the pattern:
1st term, or $n = 1\textsf{:}$ $8 \cdot 2^{0}$
2nd term, or $n = 2\textsf{:}$ $8 \cdot 2^{1}$
3rd term, or $n = 3\textsf{:}$ $8 \cdot 2^{2}$
4th term, or $n = 4\textsf{:}$ $8 \cdot 2^{3}$
Since each term has an exponent that is $1$ less than the term number, the $n$ th term is $f(n) = 8 \cdot 2^{n-1}$ .