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

Determine Common Difference: To find the recursive formula for the arithmetic sequence, we first need to determine the common difference between consecutive terms. We can subtract the first term from the second term to find this common difference. 26 - 20 = 6Write Recursive Formula: Now that we know the common difference is 6, we can write the recursive formula. The first term a(1) is given as 20. So, a(1) = 20.Finalize Recursive Formula: The recursive formula will then be a(n) = a(n-1) + 6 for n > 1, where a(n) is the nth term and a(n-1) is the (n-1)th term.

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Complete the recursive formula of the arithmetic sequence $\newline$ $20,26,32,38,\dots$ $\newline$ $a(1)=\square$ $\newline$ $a(n) = a(n-1) +6$

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Q. Complete the recursive formula of the arithmetic sequence

\newline

20,26,32,38,\dots

\newline

a(1)=\square

\newline

a(n) = a(n-1) +6

Determine Common Difference: To find the recursive formula for the arithmetic sequence, we first need to determine the common difference between consecutive terms. $\newline$ We can subtract the first term from the second term to find this common difference. $\newline$ $26 - 20 = 6$
Write Recursive Formula: Now that we know the common difference is $6$ , we can write the recursive formula. The first term $a(1)$ is given as $20$ . $\newline$ So, $a(1) = 20$ .
Finalize Recursive Formula: The recursive formula will then be $a(n) = a(n-1) + 6$ for n > 1, where $a(n)$ is the $n$ th term and $a(n-1)$ is the $(n-1)$ th term.