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Complete the recursive formula of the arithmetic sequence

4,22,40,58,dots".. "

{:[b(1)=],[b(n)=b(n-1)+]:}

Complete the recursive formula of the arithmetic sequence\newline4,22,40,58,..  4,22,40,58, \ldots \text {.. } \newlineb(1)=b(n)=b(n1)+ \begin{array}{l} b(1) = \square \\ b(n) = b(n-1)+\square\\ \end{array}

Full solution

Q. Complete the recursive formula of the arithmetic sequence\newline4,22,40,58,..  4,22,40,58, \ldots \text {.. } \newlineb(1)=b(n)=b(n1)+ \begin{array}{l} b(1) = \square \\ b(n) = b(n-1)+\square\\ \end{array}
  1. Identify first term: Identify the first term of the sequence. The first term, denoted as b(1)b(1), is the starting point of the sequence. In this case, the first term is 44.
  2. Determine common difference: Determine the common difference between consecutive terms. To find the common difference, subtract the first term from the second term: 224=1822 - 4 = 18. This common difference will be used to find each subsequent term from the previous term.
  3. Write recursive formula: Write the recursive formula using the first term and the common difference. The recursive formula for an arithmetic sequence is given by:\newlineb(1)=b(1) = first term,\newlineb(n)=b(n1)+db(n) = b(n-1) + d, where dd is the common difference and n > 1.\newlineSubstitute the known values into the formula:\newlineb(1)=4b(1) = 4,\newlineb(n)=b(n1)+18b(n) = b(n-1) + 18, for n > 1.

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