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Which recursive formula can be used to define this sequence for n>1n > 1?\newline8,4,16,28,40,52,-8, 4, 16, 28, 40, 52, \ldots\newlineChoices:\newline(A) an=an1+an1+12a_{n} = a_{n-1} + a_{n-1} + 12\newline(B) an=112an1a_{n} = \frac{1}{12}a_{n-1}\newline(C) an=an1+12a_{n} = a_{n-1} + 12\newline(D) an=7an1a_{n} = 7a_{n-1}

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Q. Which recursive formula can be used to define this sequence for n>1n > 1?\newline8,4,16,28,40,52,-8, 4, 16, 28, 40, 52, \ldots\newlineChoices:\newline(A) an=an1+an1+12a_{n} = a_{n-1} + a_{n-1} + 12\newline(B) an=112an1a_{n} = \frac{1}{12}a_{n-1}\newline(C) an=an1+12a_{n} = a_{n-1} + 12\newline(D) an=7an1a_{n} = 7a_{n-1}
  1. Analyze Sequence Type: Analyze the sequence to determine if it is arithmetic or geometric.\newlineThe sequence given is 8,4,16,28,40,52,-8, 4, 16, 28, 40, 52, \ldots\newlineTo determine if it is arithmetic, we check if the difference between consecutive terms is constant.\newline4(8)=124 - (-8) = 12, 164=1216 - 4 = 12, 2816=1228 - 16 = 12, 4028=1240 - 28 = 12, 5240=1252 - 40 = 12\newlineSince the difference is constant, the sequence is arithmetic with a common difference of 1212.
  2. Identify Recursive Formula: Identify the recursive formula for the arithmetic sequence.\newlineSince the common difference is 1212, the recursive formula will be of the form an=a(n1)+da_n = a_{(n-1)} + d, where dd is the common difference.\newlineSubstituting 1212 for dd, we get the recursive formula: an=a(n1)+12a_n = a_{(n-1)} + 12
  3. Match with Choices: Match the recursive formula with the given choices.\newlineThe correct recursive formula is an=an1+12a_n = a_{n-1} + 12.\newlineComparing this with the given choices, we find that choice (C) an=an1+12a_n = a_{n-1} + 12 matches our formula.

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