Find the term of the arithmetic sequence whose common difference d=8 is and whose first term is a to the power of 1 = 1

Question

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Answer 1

An arithmetic sequence has the form:

a_n = a₁ + (n-1)d

where n is the term number (n = 1, 2, 3, ...), a₁ is the first term in the sequence, and d is the common difference. In your problem, d = 8 and a1 = 1, so we have:

a_n = a₁ + (n-1)d

a_n = 1 + (n-1)8

You left out the term number you are looking for from your problem statement. However, just plug that value of n into the above equation and compute the answer. For example, the 6th term (n = 6) would be:

a₆ = 1 + (6-1)8 = 1 + 5·8 = 41

Answer 2

I'm not following your question, but I'll say this:

An arithmetic sequence with a common difference of 8 means you add 8 to each term to get the next term. For instance, if the first term is 1, the second term is 9 (1 + 8 = 9), the third term is 17 (9 + 8 = 17), and the fourth term is 25 (17 + 8 = 25), etc. I hope this helps.