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Complete the recursive formula of the arithmetic sequence 20,26,32,38,dots. {:[a(1)=◻],[a(n)=a(n-1)+]:}

Complete the recursive formula of the arithmetic sequence 20,26,32,38,… 20,26,32,38, \ldots .\newlinea(1)=□a(n)=a(n−1)+ \begin{array}{l} a(1)=\square \\ a(n)=a(n-1)+ \end{array}

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Q. Complete the recursive formula of the arithmetic sequence 20,26,32,38,… 20,26,32,38, \ldots .\newlinea(1)=□a(n)=a(n−1)+ \begin{array}{l} a(1)=\square \\ a(n)=a(n-1)+ \end{array}
  1. Identify first term and common difference: Identify the first term of the sequence and the common difference between consecutive terms. The first term is 2020, and by subtracting the first term from the second term, we can find the common difference: 26−20=626 - 20 = 6. The common difference is 66.
  2. Write recursive formula: Write the recursive formula for the arithmetic sequence using the first term a1 a_1 and the common difference d d . The recursive formula has two parts: the first part states the first term, and the second part describes how to find any term from the previous term using the common difference an=an−1+d a_n = a_{n-1} + d .
  3. First term: a(1)=20a(1) = 20: The first term is given as a(1)=20a(1) = 20. This is the starting point of the sequence.
  4. Second part of recursive formula: a(n)=a(n−1)+6a(n) = a(n-1) + 6: The second part of the recursive formula will be a(n)=a(n−1)+6a(n) = a(n-1) + 6, which means that any term in the sequence is found by adding 66 to the previous term.

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