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

Look at the following set of numbers and determine if there is any property, if so, what is it?

\( 94,96,98,100,102,104 \)

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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