To find the 5 terms of the arithmetic series where the 3rd term is 18 and the 10th term is 24, we'll follow these steps:

Step 1: Define the formula for the nth term of an arithmetic sequence.

The nth term $a_n$ of an arithmetic sequence can be expressed as: $a_n = a_1 + (n-1) \cdot d$ where:

• $a_1$ is the first term,
• $d$ is the common difference,
• $n$ is the term number.

Step 2: Set up equations based on the given terms.

We know:

• The 3rd term $a_3 = 18$,
• The 10th term $a_{10} = 24$.

Using the formula: $a_3 = a_1 + 2d = 18 \quad \text{(Equation 1)}$ $a_{10} = a_1 + 9d = 24 \quad \text{(Equation 2)}$

Step 3: Solve the system of equations.

Subtract Equation 1 from Equation 2 to eliminate $a_1$ and solve for $d$: $(a_1 + 9d) - (a_1 + 2d) = 24 - 18$ $7d = 6$ $d = \frac{6}{7}$

Now, substitute $d = \frac{6}{7}$ into Equation 1 to find $a_1$: $a_1 + 2 \left(\frac{6}{7}\right) = 18$ $a_1 + \frac{12}{7} = 18$ $a_1 = 18 - \frac{12}{7} = \frac{126 - 12}{7} = \frac{114}{7} \approx 16.29$

Step 4: Calculate the 5 terms of the series.

Using $a_1 = \frac{114}{7}$ and $d = \frac{6}{7}$:

• $a_1 = \frac{114}{7}$
• $a_2 = \frac{114}{7} + \frac{6}{7} = \frac{120}{7}$
• $a_3 = \frac{114}{7} + 2 \cdot \frac{6}{7} = \frac{126}{7} = 18$
• $a_4 = \frac{114}{7} + 3 \cdot \frac{6}{7} = \frac{132}{7}$
• $a_5 = \frac{114}{7} + 4 \cdot \frac{6}{7} = \frac{138}{7}$

So, the 5 terms of the arithmetic series are approximately: $\frac{114}{7}, \frac{120}{7}, 18, \frac{132}{7}, \frac{138}{7}$

Would you like more details or have any other questions?

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Here are five related questions:

1. How do you find the sum of the first n terms of an arithmetic series?
2. Can you derive the nth term formula from the sum formula of an arithmetic series?
3. How does the common difference affect the growth of the sequence?
4. What happens if the common difference $d$ is zero?
5. How do arithmetic sequences compare to geometric sequences?

Tip: Always double-check your algebraic manipulation when solving systems of equations to avoid errors in finding terms or the common difference.