find the first five terms in a1= 1/6 and d=1/2

To find the first five terms of an arithmetic sequence, we use the formula:

\[
a_n = a_1 + (n - 1) \cdot d
\]

where:
- \( a_1 \) is the first term,
- \( d \) is the common difference, and
- \( n \) is the term number.

Given:
- \( a_1 = \frac{1}{6} \)
- \( d = \frac{1}{2} \)

Let's calculate the first five terms:

1. **First term (\( a_1 \))**:
   \[
   a_1 = \frac{1}{6}
   \]

2. **Second term (\( a_2 \))**:
   \[
   a_2 = a_1 + d = \frac{1}{6} + \frac{1}{2} = \frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3}
   \]

3. **Third term (\( a_3 \))**:
   \[
   a_3 = a_1 + 2d = \frac{1}{6} + 2\left(\frac{1}{2}\right) = \frac{1}{6} + \frac{6}{6} = \frac{7}{6}
   \]

4. **Fourth term (\( a_4 \))**:
   \[
   a_4 = a_1 + 3d = \frac{1}{6} + 3\left(\frac{1}{2}\right) = \frac{1}{6} + \frac{9}{6} = \frac{10}{6} = \frac{5}{3}
   \]

5. **Fifth term (\( a_5 \))**:
   \[
   a_5 = a_1 + 4d = \frac{1}{6} + 4\left(\frac{1}{2}\right) = \frac{1}{6} + \frac{12}{6} = \frac{13}{6}
   \]

So, the first five terms of the sequence are:

\[
\frac{1}{6}, \frac{2}{3}, \frac{7}{6}, \frac{5}{3}, \frac{13}{6}
\]

Would you like further details or have any questions?

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Here are some related questions you might find useful:

1. How do you find the \( n \)-th term in an arithmetic sequence?
2. How do you determine the sum of the first \( n \) terms in an arithmetic sequence?
3. What is the difference between arithmetic and geometric sequences?
4. How do you find the common difference if only two terms of the sequence are given?
5. What is the formula for the sum of an infinite arithmetic series?

**Tip:** For arithmetic sequences, the difference between consecutive terms remains constant, making it easy to predict future terms once the first term and the common difference are known.

Learn how to find the first five terms of an arithmetic sequence given the first term a1 = 1/6 and the common difference d = 1/2. This problem covers arithmetic sequence formulas and calculations. Suitable for students in Grades 7-9.

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