Math Problem Statement

Create a sequence satisfying the following: Prove that it satisfies the given condition. 2. The first term is 12.

Solution

We need to create a sequence such that it satisfies a given condition (though the condition isn't explicitly stated), but the first term of the sequence is 12. I'll make an assumption that we are to construct an arithmetic or geometric sequence, and I will guide through both approaches.

Let's explore two common sequences:

1. Arithmetic Sequence

An arithmetic sequence is defined by each term increasing or decreasing by a constant value dd. The general formula for an arithmetic sequence is: an=a1+(n1)da_n = a_1 + (n-1)d where a1a_1 is the first term, dd is the common difference, and nn is the term number.

Since the first term is 12, we have: a1=12a_1 = 12 Now, suppose d=3d = 3, the common difference is 3. Then the sequence will be: a2=12+3=15,a3=12+2(3)=18,a4=12+3(3)=21,a_2 = 12 + 3 = 15, \quad a_3 = 12 + 2(3) = 18, \quad a_4 = 12 + 3(3) = 21, \ldots The sequence will look like: 12,15,18,21,24,12, 15, 18, 21, 24, \dots.

Verifying the Arithmetic Property:

To check if it satisfies the arithmetic condition:

  • The difference between consecutive terms is constant: a2a1=1512=3,a3a2=1815=3,a4a3=2118=3,a_2 - a_1 = 15 - 12 = 3, \quad a_3 - a_2 = 18 - 15 = 3, \quad a_4 - a_3 = 21 - 18 = 3, \dots Thus, the sequence satisfies the condition for being arithmetic with common difference d=3d = 3.

2. Geometric Sequence

A geometric sequence is defined by each term being multiplied by a constant ratio rr. The general formula for a geometric sequence is: an=a1rn1a_n = a_1 \cdot r^{n-1} where a1a_1 is the first term and rr is the common ratio.

Since the first term is 12, we have: a1=12a_1 = 12 Let r=2r = 2, the common ratio is 2. Then the sequence will be: a2=122=24,a3=1222=48,a4=1223=96,a_2 = 12 \cdot 2 = 24, \quad a_3 = 12 \cdot 2^2 = 48, \quad a_4 = 12 \cdot 2^3 = 96, \ldots The sequence will look like: 12,24,48,96,12, 24, 48, 96, \dots.

Verifying the Geometric Property:

To check if it satisfies the geometric condition:

  • The ratio between consecutive terms is constant: a2a1=2412=2,a3a2=4824=2,a4a3=9648=2,\frac{a_2}{a_1} = \frac{24}{12} = 2, \quad \frac{a_3}{a_2} = \frac{48}{24} = 2, \quad \frac{a_4}{a_3} = \frac{96}{48} = 2, \dots Thus, the sequence satisfies the condition for being geometric with common ratio r=2r = 2.

Conclusion

Both sequences satisfy the conditions of being arithmetic and geometric, depending on which progression we choose. For the arithmetic sequence, we assumed d=3d = 3, and for the geometric sequence, we assumed r=2r = 2. In both cases, the first term is 12, and they meet the respective conditions.

Do you have a specific condition or would you like further details on any of these sequences?

Follow-up Questions

  1. What if the common difference dd or the common ratio rr is negative?
  2. How would the sequence change if the first term were a different number?
  3. Can we have a sequence where the difference or ratio is not constant?
  4. What is the general form of a recursive sequence with the same first term?
  5. How would you find the sum of the first nn terms of these sequences?

Tip

For arithmetic sequences, the sum of the first nn terms can be quickly calculated using the formula Sn=n2(2a1+(n1)d)S_n = \frac{n}{2} \cdot (2a_1 + (n-1)d).

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Math Problem Analysis

Mathematical Concepts

Arithmetic Sequences
Geometric Sequences

Formulas

Arithmetic sequence formula: a_n = a_1 + (n-1)d
Geometric sequence formula: a_n = a_1 * r^(n-1)

Theorems

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Suitable Grade Level

Grades 9-12