In an arithmetic sequence, a = 8 and d = 3. The first term in this sequence to exceed 30 is the:

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BlackTom AI · Accepted Answer

To tackle this, I’ll start by identifying the general formula for the nth term in an arithmetic sequence. With a1 = 8 and common difference d = 3, the nth term is a_n = a1 + (n-1)d = 8 + (n-1)·3 = 3n + 5.

Option a: 6th term. If n = 6, a_6 = 3(6) + 5 = 18 + 5 = 23, which is not greater than 30. Therefore, this term does not exceed 30.

Option b: 7th term. If n = 7, a_7 = 3(7) + 5 = 21 + 5 = 26, still below 30, so this is not the first term over 30.

Option c: 8th term. If n = 8, a_8 = 3(8) + 5 = 24 + 5 = 29, which remains under 30, meaning this isn’t the first exceeding term.

Option d: 9th term. If n = 9, a_9 = 3(9) + 5 = 27 + 5 = 32, which is indeed the first term to surpass 30.

Putting it together, the inequality 3n + 5 > 30 occurs when n > 25/3, so the smallest integer n satisfying this is n = 9. Therefore, the first term to exceed 30 is the 9th term.

MUF0141 Fund. Mathematics Unit 1 - Semester 2, 2025 5.6 Arithmetic Sequences Quiz (Formative)

In an arithmetic sequence, a = 8 and d = 3. The first term in this sequence to exceed 30 is the:

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