Math Problem Statement

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

Solution

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

an=a1+(n1)da_n = a_1 + (n - 1) \cdot d

where:

  • a1a_1 is the first term,
  • dd is the common difference, and
  • nn is the term number.

Given:

  • a1=16a_1 = \frac{1}{6}
  • d=12d = \frac{1}{2}

Let's calculate the first five terms:

  1. First term (a1a_1): a1=16a_1 = \frac{1}{6}

  2. Second term (a2a_2): a2=a1+d=16+12=16+36=46=23a_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 (a3a_3): a3=a1+2d=16+2(12)=16+66=76a_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 (a4a_4): a4=a1+3d=16+3(12)=16+96=106=53a_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 (a5a_5): a5=a1+4d=16+4(12)=16+126=136a_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:

16,23,76,53,136\frac{1}{6}, \frac{2}{3}, \frac{7}{6}, \frac{5}{3}, \frac{13}{6}

Would you like further details or have any questions?


Here are some related questions you might find useful:

  1. How do you find the nn-th term in an arithmetic sequence?
  2. How do you determine the sum of the first nn 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.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Sequences

Formulas

Arithmetic sequence formula

Theorems

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Suitable Grade Level

Grades 7-9