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Which recursive formula can be used to define this sequence for $n > 1$ ? $\newline$ $1, 12, 23, 34, 45, 56, \ldots$ $\newline$ Choices: $\newline$ $a_n = \frac{17}{6}a_{n - 1}$ $\newline$ $a_n = a_{n - 1} + a_{n - 2} + 11$ $\newline$ $a_n = a_{n - 1} - 11$ $\newline$ $a_n = a_{n - 1} + 11$

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Find a recursive formula

Full solution

Q. Which recursive formula can be used to define this sequence for

n > 1

?

\newline

1, 12, 23, 34, 45, 56, \ldots

\newline

Choices:

\newline

a_n = \frac{17}{6}a_{n - 1}

\newline

a_n = a_{n - 1} + a_{n - 2} + 11

\newline

a_n = a_{n - 1} - 11

\newline

a_n = a_{n - 1} + 11

Identify Pattern in Sequence: To find the recursive formula, we need to look at the pattern in the sequence. Let's examine the difference between consecutive terms. $\newline$ $12 - 1 = 11$ $\newline$ $23 - 12 = 11$ $\newline$ $34 - 23 = 11$ $\newline$ $45 - 34 = 11$ $\newline$ $56 - 45 = 11$ $\newline$ We can see that each term is $11$ more than the previous term.
Express $n$ th Term: Based on the pattern, we can express the $n$ th term ( $a_n$ ) as the sum of the ( $n-1$ )th term ( $a_{n-1}$ ) and $11$ . This gives us the recursive formula: $\newline$ $a_n = a_{n-1} + 11$
Check Given Choices: Now let's check the given choices to see which one matches our formula: $\newline$ Choices: $\newline$ [a] $a_n = \frac{17}{6}a_{n - 1}$ (This is not correct as it does not represent the pattern we found.) $\newline$ [b] $a_n = a_{n - 1} + a_{n - 2} + 11$ (This is not correct as it involves two previous terms and does not represent the simple addition of $11$ to the previous term.) $\newline$ [c] $a_n = a_{n - 1} - 11$ (This is not correct as it subtracts $11$ instead of adding it.) $\newline$ [d] $a_n = a_{n - 1} + 11$ (This is correct as it matches the pattern we found.)

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