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

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 \(d\). The general formula for an arithmetic sequence is:
\[
a_n = a_1 + (n-1)d
\]
where \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the term number.

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

#### Verifying the Arithmetic Property:
To check if it satisfies the arithmetic condition:
- The difference between consecutive terms is constant:
\[
a_2 - a_1 = 15 - 12 = 3, \quad a_3 - a_2 = 18 - 15 = 3, \quad a_4 - a_3 = 21 - 18 = 3, \dots
\]
Thus, the sequence satisfies the condition for being arithmetic with common difference \(d = 3\).

---

### 2. Geometric Sequence
A geometric sequence is defined by each term being multiplied by a constant ratio \(r\). The general formula for a geometric sequence is:
\[
a_n = a_1 \cdot r^{n-1}
\]
where \(a_1\) is the first term and \(r\) is the common ratio.

Since the first term is 12, we have:
\[
a_1 = 12
\]
Let \(r = 2\), the common ratio is 2. Then the sequence will be:
\[
a_2 = 12 \cdot 2 = 24, \quad a_3 = 12 \cdot 2^2 = 48, \quad a_4 = 12 \cdot 2^3 = 96, \ldots
\]
The sequence will look like: \(12, 24, 48, 96, \dots\).

#### Verifying the Geometric Property:
To check if it satisfies the geometric condition:
- The ratio between consecutive terms is constant:
\[
\frac{a_2}{a_1} = \frac{24}{12} = 2, \quad \frac{a_3}{a_2} = \frac{48}{24} = 2, \quad \frac{a_4}{a_3} = \frac{96}{48} = 2, \dots
\]
Thus, the sequence satisfies the condition for being geometric with common ratio \(r = 2\).

---

### Conclusion
Both sequences satisfy the conditions of being arithmetic and geometric, depending on which progression we choose. For the arithmetic sequence, we assumed \(d = 3\), and for the geometric sequence, we assumed \(r = 2\). 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 \(d\) or the common ratio \(r\) 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 \(n\) terms of these sequences?

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

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

Explore how to construct an arithmetic or geometric sequence starting with the first term of 12. Understand the formulas used, verify the conditions for both types of sequences, and learn how to prove their properties. Suitable for students in Grades 9-12.

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