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Chapter 5: Problem 65

Write the first six terms of the geometric sequence with the first term, \(a_{1}\), and common ratio, \(r\). \(a_{1}=\frac{1}{4}, r=\frac{1}{2}\)

Short Answer

Expert verified
The first six terms of the geometric series are: \(\frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, \frac{1}{64}, \frac{1}{128}\).

Step by step solution

01

Finding the First Term

The first term of the geometric sequence, \(a_{1}\), is given straightway as \(\frac{1}{4}\). So, the first term is \(\frac{1}{4}\).
02

Applying the Formula for the 2nd term

To find the second term of the sequence, apply the formula for the nth term of a geometric sequence \(a_{n} = a_{1} * r^{(n-1)}\). Substituting \(a_{1} = \frac{1}{4}\), \(r = \frac{1}{2}\), \(n = 2\) in the formula, we get: \(a_{2} = \frac{1}{4} * (\frac{1}{2})^{2-1} = \frac{1}{4} * \frac{1}{2} = \frac{1}{8}\).
03

Applying the Formula for the 3rd term

Likewise, for the third term, substitute \(a_{1} = \frac{1}{4}\), \(r = \frac{1}{2}\), \(n = 3\) in the formula, we get: \(a_{3} = \frac{1}{4} * (\frac{1}{2})^{3-1} = \frac{1}{4} * \frac{1}{4} = \frac{1}{16}\).
04

Applying the Formula for the 4th term

Continuing in this fashion, for the fourth term, substitute \(a_{1} = \frac{1}{4}\), \(r = \frac{1}{2}\), \(n = 4\) in the formula, we get: \(a_{4} = \frac{1}{4} * (\frac{1}{2})^{4-1} = \frac{1}{4} * \frac{1}{8} = \frac{1}{32}\).
05

Applying the Formula for the 5th term

For the fifth term, substitute \(a_{1} = \frac{1}{4}\), \(r = \frac{1}{2}\), \(n = 5\) in the formula, we get: \(a_{5} = \frac{1}{4} * (\frac{1}{2})^{5-1} = \frac{1}{4} * \frac{1}{16} = \frac{1}{64}\).
06

Applying the Formula for the 6th term

Finally, for the sixth term, substitute \(a_{1} = \frac{1}{4}\), \(r = \frac{1}{2}\), \(n = 6\) in the formula, we get: \(a_{6} = \frac{1}{4} * (\frac{1}{2})^{6-1} = \frac{1}{4} * \frac{1}{32} = \frac{1}{128}\).

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Most popular questions from this chapter

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