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Which recursive formula can be used to define this sequence for n > 1?\newline1,12,23,34,45,56,1, 12, 23, 34, 45, 56, \ldots\newlineChoices:\newline(A)an=an1+11a_n = a_{n-1} + 11\newline(B)an=an111a_n = a_{n-1} - 11\newline(C)an=176an1a_n = \frac{17}{6}a_{n-1}\newline(D)an=an1+an1+11a_n = a_{n-1} + a_{n-1} + 11

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Q. Which recursive formula can be used to define this sequence for n>1n > 1?\newline1,12,23,34,45,56,1, 12, 23, 34, 45, 56, \ldots\newlineChoices:\newline(A)an=an1+11a_n = a_{n-1} + 11\newline(B)an=an111a_n = a_{n-1} - 11\newline(C)an=176an1a_n = \frac{17}{6}a_{n-1}\newline(D)an=an1+an1+11a_n = a_{n-1} + a_{n-1} + 11
  1. Determine Pattern: We need to determine the pattern of the sequence to find the recursive formula. Let's look at the sequence:\newline1,12,23,34,45,56,1, 12, 23, 34, 45, 56, \ldots\newlineWe can see that each term after the first increases by 1111 from the previous term. This suggests that the sequence is arithmetic with a common difference of 1111.
  2. Confirm Common Difference: To confirm the common difference, we calculate the difference between consecutive terms: \newline121=1112 - 1 = 11\newline2312=1123 - 12 = 11\newline3423=1134 - 23 = 11\newline4534=1145 - 34 = 11\newline5645=1156 - 45 = 11\newlineSince the difference is consistently 1111, we can confirm that the sequence is arithmetic with a common difference of 1111.
  3. Express Recursive Formula: Now, we need to express this pattern as a recursive formula. A recursive formula for an arithmetic sequence is generally of the form:\newlinean=an1+da_n = a_{n-1} + d\newlinewhere ana_n is the nnth term, an1a_{n-1} is the previous term, and dd is the common difference.
  4. Match with Choices: Substituting the common difference d=11d = 11 into the recursive formula, we get:\newlinean=a(n1)+11a_n = a_{(n-1)} + 11\newlineThis formula will give us the next term in the sequence by adding 1111 to the previous term.
  5. Match with Choices: Substituting the common difference d=11d = 11 into the recursive formula, we get:\newlinean=a(n1)+11a_n = a_{(n-1)} + 11\newlineThis formula will give us the next term in the sequence by adding 1111 to the previous term.Now, let's match our recursive formula with the given choices:\newline(A) a=a+11a = a + 11 (This is incorrect; it lacks the subscript nn to denote the nnth term.)\newline(B) a=a11a = a - 11 (This is incorrect; it suggests subtracting 1111, which is not the pattern we found.)\newline(C) a=176aa = \frac{17}{6}a (This is incorrect; it suggests a geometric sequence with a ratio of 176\frac{17}{6}, which is not the case.)\newline(D) an=a(n1)+11a_n = a_{(n-1)} + 1100 (This is incorrect; it incorrectly suggests adding the term to itself before adding 1111.)\newlineThe correct recursive formula is not explicitly listed in the choices, but it is closest to choice (A) if we correct it to include the subscript nn.

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