Find the Variable Expression n for Sequence 12, 18, 24, 30, 36

Arithmetic Sequences with Variable Expressions

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

12,18,24,30,36 12,18,24,30,36

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
00:06 This is the first term according to the given data
00:15 Let's observe the change between terms (D) according to the given data
00:27 This is the constant difference in the sequence (D)
00:30 Let's use the formula to describe the sequence
00:39 Let's substitute appropriate values and solve to find the sequence formula
00:52 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:

12,18,24,30,36 12,18,24,30,36

Describe the property using the variable n n

2

Step-by-step solution

To solve this problem, we first recognize that the sequence 12,18,24,30,36 12, 18, 24, 30, 36 is an arithmetic sequence.

  • Step 1: Identify the first term a1 a_1 .

    The first term a1 a_1 of the sequence is 12 12 .

  • Step 2: Determine the common difference d d .

    The difference between consecutive terms is consistent: 1812=6 18 - 12 = 6 . Hence, the common difference d=6 d = 6 .

  • Step 3: Formulate the expression for the general term.

    The general term of an arithmetic sequence is given by a(n)=a1+(n1)×d a(n) = a_1 + (n-1) \times d .

  • Step 4: Substitute the identified values into the formula.

    Substituting the known values, we get a(n)=12+(n1)×6 a(n) = 12 + (n-1) \times 6 .

Thus, the expression for the general term of the given sequence is a(n)=12+(n1)×6 a(n) = 12 + (n-1) \times 6 , which corresponds to choice 3.

3

Final Answer

a(n)=12+(n1)×6 a(n)=12+(n-1)\times6

Key Points to Remember

Essential concepts to master this topic
  • Formula: General term uses a(n)=a1+(n1)×d a(n) = a_1 + (n-1) \times d
  • Technique: First term is 12, common difference is 6
  • Check: Substitute values: a(3)=12+(31)×6=24 a(3) = 12 + (3-1) \times 6 = 24

Common Mistakes

Avoid these frequent errors
  • Using n instead of (n-1) in the formula
    Don't write a(n)=12+n×6 a(n) = 12 + n \times 6 = gives wrong values! This shifts every term by one position. Always use a(n)=12+(n1)×6 a(n) = 12 + (n-1) \times 6 to match the position correctly.

Practice Quiz

Test your knowledge with interactive questions

12 ☐ 10 ☐ 8 7 6 5 4 3 2 1

Which numbers are missing from the sequence so that the sequence has a term-to-term rule?

FAQ

Everything you need to know about this question

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

+

Because when n = 1, we want the first term (12), not 12 + 6 = 18. Using (n1) (n-1) makes the formula work: a(1)=12+(11)×6=12 a(1) = 12 + (1-1) \times 6 = 12 .

How do I find the common difference?

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Subtract any term from the next term: 1812=6 18 - 12 = 6 , 2418=6 24 - 18 = 6 . The difference should be the same between all consecutive pairs.

What if I want to find the 10th term?

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Use the formula with n = 10: a(10)=12+(101)×6=12+54=66 a(10) = 12 + (10-1) \times 6 = 12 + 54 = 66 . The pattern continues beyond the given terms!

Can I check my formula with the given numbers?

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Yes! Test each position: a(1)=12 a(1) = 12 , a(2)=18 a(2) = 18 , a(3)=24 a(3) = 24 . If your formula gives these values, it's correct.

What makes this an arithmetic sequence?

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The constant difference of 6 between consecutive terms. This equal spacing is what defines an arithmetic sequence, making the formula predictable.

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