Find the nth Term Formula for Sequence: 3,6,9,12,15

Arithmetic Sequences with Linear Term Formulas

Given the series whose difference between two jumped numbers is constant:

3,6,9,12,15 3,6,9,12,15

Describe the property using the variable n n

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Find the sequence formula
00:04 This is the first term according to the given data
00:13 Let's observe the difference between terms (D) according to the given data
00:22 We'll use the formula to describe the sequence
00:28 We'll substitute appropriate values and solve to find the sequence formula
00:42 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Given the series whose difference between two jumped numbers is constant:

3,6,9,12,15 3,6,9,12,15

Describe the property using the variable n n

2

Step-by-step solution

We'll solve the problem by following these steps:

  • Identify the first term (a1 a_1 ) and the common difference (d d ).
  • Use the arithmetic sequence formula a(n)=a1+(n1)×d a(n) = a_1 + (n - 1) \times d .
  • Substitute the known values into the formula.

Now, let's go through each step:

Step 1: Identify the given information:
The first term of the sequence (a1 a_1 ) is 3, and the common difference (d d ) is 3, as the difference between any two consecutive terms is constant and equal to 3.

Step 2: Use the formula for the n n -th term of an arithmetic sequence:
a(n)=a1+(n1)×d a(n) = a_1 + (n - 1) \times d .

Step 3: Plug in the known values:
- First term a1=3 a_1 = 3 .
- Common difference d=3 d = 3 .
Therefore, a(n)=3+(n1)×3 a(n) = 3 + (n - 1) \times 3 .

Check the formula by substituting values:
- For n=1 n = 1 : a(1)=3+(11)×3=3 a(1) = 3 + (1-1) \times 3 = 3
- For n=2 n = 2 : a(2)=3+(21)×3=6 a(2) = 3 + (2-1) \times 3 = 6
- Continue checking for other values.

Since the formula correctly generates the sequence values, the description of the series is a(n)=3+(n1)×3 a(n) = 3 + (n - 1) \times 3 .

Therefore, the correct answer is choice 3.

3

Final Answer

a(n)=3+(n1)×3 a(n)=3+(n-1)\times3

Key Points to Remember

Essential concepts to master this topic
  • Pattern: Arithmetic sequences have constant differences between consecutive terms
  • Formula: Use a(n)=a1+(n1)×d a(n) = a_1 + (n-1) \times d where first term is 3, difference is 3
  • Verify: Check formula works: a(2)=3+(21)×3=6 a(2) = 3 + (2-1) \times 3 = 6

Common Mistakes

Avoid these frequent errors
  • Using n instead of (n-1) in the formula
    Don't write a(n) = 3 + n × 3 = wrong values like a(1) = 6! This shifts every term up by one position. Always use (n-1) because the first term has zero added differences.

Practice Quiz

Test your knowledge with interactive questions

Is there a term-to-term rule for the sequence below?

18 , 22 , 26 , 30

FAQ

Everything you need to know about this question

Why do we use (n-1) instead of just n in the formula?

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Think about it: the first term (n=1) should have zero common differences added to it. Using (n-1) gives us (1-1) = 0, so we just get the first term by itself!

How do I find the common difference quickly?

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Just subtract any term from the next one: 6 - 3 = 3, or 12 - 9 = 3. In arithmetic sequences, this difference is always the same!

What if I want to find the 100th term?

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Use the same formula! a(100)=3+(1001)×3=3+297=300 a(100) = 3 + (100-1) \times 3 = 3 + 297 = 300 . The formula works for any position number.

Can I simplify the formula further?

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Yes! a(n)=3+(n1)×3=3+3n3=3n a(n) = 3 + (n-1) \times 3 = 3 + 3n - 3 = 3n . But keep the standard form during problem solving to avoid errors.

How do I check if my formula is correct?

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Test it with the given sequence! Plug in n=1, n=2, n=3 and see if you get 3, 6, 9. If all match, your formula is correct!

What makes this different from other sequence types?

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Arithmetic sequences add the same amount each time. Geometric sequences multiply by the same amount, and other sequences might have more complex patterns.

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