Question

what is a formula for the nth term of the given sequence? 3, 8, 13... answer attempt 1 out of 2 \\( a_n = 8 + 5(n - 1) \\) \\( a_n = 3 + 5n \\) \\( a_n = 8 - 5n \\) \\( a_n = -7 + 5(n + 1) \\) submit answer

Explanation:

Step1: Identify Sequence Type

The sequence \(3, 8, 13, \dots\) is arithmetic. Common difference \(d = 8 - 3 = 5\).

Step2: Arithmetic Sequence Formula

The formula for the \(n\)-th term of an arithmetic sequence is \(a_n = a_1 + (n - 1)d\), where \(a_1 = 3\), \(d = 5\). So \(a_n = 3 + 5(n - 1)\). Let's simplify other options:
- Option 1: \(a_n = 8 + 5(n - 1)\) (uses \(a_2 = 8\) as first term, valid for arithmetic sequence with \(a_2\) as start, simplifies to \(3 + 5n\)? Wait, no: \(8 + 5(n - 1)=8 + 5n - 5 = 3 + 5n\)? Wait, no, \(8 + 5(n - 1)=5n + 3\)? Wait, let's test \(n = 1\): \(8 + 5(0)=8\), but first term is 3. Wait, no, maybe I made a mistake. Wait, the first term \(a_1 = 3\), so let's check each option for \(n = 1\), \(n = 2\), \(n = 3\).
- Option 1: \(n = 1\): \(8 + 5(0)=8\) (wrong, should be 3). Wait, no, maybe the option is written with \(a_2\) as the first term? Wait, no, the sequence starts at 3. Wait, let's re - evaluate.
Wait, the arithmetic sequence formula can also be written as \(a_n=a_k+(n - k)d\). For \(k = 2\), \(a_2 = 8\), \(d = 5\), so \(a_n=8 + 5(n - 2)\)? No, the given option 1 is \(a_n = 8+5(n - 1)\). Wait, maybe I miscalculated. Let's check option 4: \(a_n=-7 + 5(n + 1)\). For \(n = 1\): \(-7+5(2)=-7 + 10 = 3\) (correct). For \(n = 2\): \(-7+5(3)=-7 + 15 = 8\) (correct). For \(n = 3\): \(-7+5(4)=-7 + 20 = 13\) (correct). Wait, and option 1: \(a_n = 8+5(n - 1)\). For \(n = 1\): \(8+5(0)=8\) (wrong). Option 2: \(a_n = 3+5n\). For \(n = 1\): \(3 + 5(1)=8\) (wrong). Option 3: \(a_n=8 - 5n\). For \(n = 1\): \(8 - 5 = 3\) (correct for \(n = 1\)), \(n = 2\): \(8-10=-2\) (wrong). Option 4: \(a_n=-7 + 5(n + 1)\). For \(n = 1\): \(-7+10 = 3\), \(n = 2\): \(-7 + 15 = 8\), \(n = 3\): \(-7+20 = 13\) (correct). Also, let's go back to the standard formula. The standard formula \(a_n=a_1+(n - 1)d\) with \(a_1 = 3\), \(d = 5\) gives \(a_n=3 + 5(n - 1)=5n-2\). Wait, \(3+5(n - 1)=5n - 2\). Now let's simplify option 4: \(-7 + 5(n + 1)=-7+5n + 5=5n - 2\), which is the same as \(3 + 5(n - 1)\). And option 1: \(8 + 5(n - 1)=5n+3\) (for \(n = 1\), 8; \(n = 2\), 13; but first term is wrong). Wait, maybe the question has a typo in the option, but when we check the values:
- For option 1: \(a_n = 8+5(n - 1)\), when \(n = 1\), \(a_1 = 8\) (incorrect, should be 3). But when \(n = 2\), \(a_2=8 + 5(1)=13\) (correct), \(n = 3\), \(a_3=8 + 5(2)=18\) (but our sequence's third term should be 13? No, wait our sequence is \(3,8,13\), third term is 13. Wait, \(n = 3\) for option 1: \(8+5(2)=18\) (wrong). Wait, I am confused. Wait, let's recalculate the common difference. \(8 - 3 = 5\), \(13 - 8 = 5\), so common difference \(d = 5\), first term \(a_1 = 3\). So the formula is \(a_n=3 + 5(n - 1)=5n - 2\). Now let's simplify each option:
- Option 1: \(a_n=8 + 5(n - 1)=5n+3\) (for \(n = 1\), 8; \(n = 2\), 13; \(n = 3\), 18)
- Option 2: \(a_n=3 + 5n\) (for \(n = 1\), 8; \(n = 2\), 13; \(n = 3\), 18)
- Option 3: \(a_n=8 - 5n\) (for \(n = 1\), 3; \(n = 2\), \(-2\); \(n = 3\), \(-7\))
- Option 4: \(a_n=-7 + 5(n + 1)=5n - 2\) (for \(n = 1\), \(5(1)-2 = 3\); \(n = 2\), \(5(2)-2 = 8\); \(n = 3\), \(5(3)-2 = 13\))

Wait, I see my mistake earlier. Option 4 simplifies to \(a_n=5n - 2\), which matches the formula \(a_n=3 + 5(n - 1)=5n - 2\). And option 1: \(a_n=8 + 5(n - 1)=5n + 3\), which is different. Wait, but when \(n = 1\), option 4 gives \(a_1=-7 + 5(2)=3\) (correct), \(n = 2\): \(-7+5(3)=8\) (correct), \(n = 3\): \(-7+5(4)=13\) (correct). Option 1: when \(n = 1\), \(a_1 = 8\) (incorrect), \(n = 2\), \(a_2=13\) (correct), \(n = 3\), \(a_3=18\) (incorrect). So the correct formula should satisfy \(n = 1\) gives 3, \(n = 2\) gives 8, \(n = 3\) gives 13. Let's check option 4:
- \(n = 1\): \(-7+5(1 + 1)=-7 + 10 = 3\)
- \(n = 2\): \(-7+5(2 + 1)=-7 + 15 = 8\)
- \(n = 3\): \(-7+5(3 + 1)=-7 + 20 = 13\)

It works. Also, let's check the formula \(a_n=a_1+(n - 1)d\) with \(a_1 = 3\), \(d = 5\), we get \(a_n=3+5(n - 1)=5n - 2\). Now, let's rewrite option 4: \(-7 + 5(n + 1)=5n+5 - 7=5n - 2\), which is the same as \(3 + 5(n - 1)\).

Wait, but there is also option 1: \(a_n=8 + 5(n - 1)\). Let's expand it: \(8+5n - 5=5n + 3\). For \(n = 1\), it gives 8, which is the second term. So if we consider the sequence starting from the second term, but the sequence given starts from 3 (first term). So the correct formula should have \(a_1 = 3\). So option 4 is correct. Also, let's check option 1 again. If we use \(n = 1\) in option 1, we get 8, which is the second term. So if we adjust the formula to start at \(n = 1\) with the second term as the "first" term in the formula, but the problem says "the \(n\)-th term of the given sequence" where the sequence starts at 3 ( \(n = 1\) is 3). So option 4 is correct. Also, let's check the formula \(a_n=-7 + 5(n + 1)\):

\(a_n=-7+5n + 5=5n - 2\), and \(3+5(n - 1)=3 + 5n-5=5n - 2\), so they are equivalent.

Answer:

\(a_n = -7 + 5(n + 1)\) (the fourth option)