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In a sequence of numbers,
a_(2)=16,a_(3)=22,a_(4)=28,a_(5)=34. Which equation can be used to find the
n^("th ") term in the sequence,
a_(n) ?

a_(n)=6n+10quada_(n)=-6n+22
A.
B.

a_(n)=6n+4quada_(n)=6n+16
c.
D.

$1$ . In a sequence of numbers, $a_{2}=16, a_{3}=22, a_{4}=28, a_{5}=34$ . Which equation can be used to find the $n^{\text {th }}$ term in the sequence, $a_{n}$ ? $\newline$ $a_{n}=6 n+10 \quad a_{n}=-6 n+22$ $\newline$ A. $\newline$ B. $\newline$ $a_{n}=6 n+4 \quad a_{n}=6 n+16$ $\newline$ c. $\newline$ D.

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Convert a recursive formula to an explicit formula

Full solution

Q.

1

. In a sequence of numbers,

a_{2}=16, a_{3}=22, a_{4}=28, a_{5}=34

. Which equation can be used to find the

n^{\text {th }}

term in the sequence,

a_{n}

?

\newline

a_{n}=6 n+10 \quad a_{n}=-6 n+22

\newline

A.

\newline

B.

\newline

a_{n}=6 n+4 \quad a_{n}=6 n+16

\newline

c.

\newline

D.

Find Common Difference: To find the pattern, let's subtract each term from the next one to see the common difference. $22 - 16 = 6$ , $28 - 22 = 6$ , $34 - 28 = 6$ .
Use Arithmetic Sequence Formula: The common difference is $6$ , so the sequence is arithmetic and the nth term can be found using the formula $a_n = a_1 + (n - 1)d$ , where $d$ is the common difference.
Calculate $a_1$ : We need to find $a_1$ . Since $a_2 = 16$ and $d = 6$ , we can find $a_1$ by subtracting the common difference from $a_2$ : $a_1 = 16 - 6 = 10$ .
Write Formula for nth Term: Now we can write the formula for the nth term: $a_n = 10 + (n - 1) \times 6$ .
Simplify Formula: Simplify the formula: $a_n = 10 + 6n - 6$ .
Combine Like Terms: Combine like terms: $a_n = 6n + 4$ .

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