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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)+ \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)+ \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. The common difference can be found by subtracting the first term from the second term, or the second term from the third term, and so on. For this sequence, subtracting the first term from the second term gives us 224=1822 - 4 = 18.
  3. Verify consistency of common difference: Verify the common difference by checking if it is consistent between other consecutive terms. Subtract the third term from the second term: 4022=1840 - 22 = 18, and the fourth term from the third term: 5840=1858 - 40 = 18. The common difference is consistent, confirming that it is 1818.
  4. 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 b(n)=b(n1)+db(n) = b(n-1) + d, where b(n)b(n) is the nth term, b(n1)b(n-1) is the previous term, and dd is the common difference. In this case, b(1)=4b(1) = 4 and d=18d = 18.

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