Problem 78 - Beginning and Intermediate Algebra

Question

Find the sum of the terms of the infinite geometric sequence, if possible. $$-12,8,-\frac{16}{3}, \frac{32}{9}, \dots$$

Step-by-Step Solution

Answer

The sum of the terms of the given infinite geometric sequence is \(S_\infty = -\frac{36}{5}\).

Step 1: Find the common ratio

Divide any term by its predecessor to find the common ratio: $$ r = \frac{a_2}{a_1} = \frac{8}{-12} = -\frac{2}{3} $$ Since \(|-\frac{2}{3}| < 1\), we can continue to find the sum of the infinite geometric sequence.

Step 2: Find the sum of the infinite sequence using the formula

Now we plug the values of \(a_1\) and \(r\) into the formula \(S_\infty = \frac{a_1}{1-r}\): $$ S_\infty = \frac{-12}{1 - (-\frac{2}{3})} $$

Step 3: Simplify the expression

Next, simplify the expression in the denominator and numerator: $$ S_\infty = \frac{-12}{1 + \frac{2}{3}} = \frac{-12}{\frac{5}{3}} $$ Now, multiply both the numerator and denominator by 3 to remove the fraction: $$ S_\infty = \frac{-12 \times 3}{5} = \frac{-36}{5} $$ So, the sum of the terms of the given infinite geometric sequence is: $$ S_\infty = -\frac{36}{5} $$

Key Concepts

Common Ratio

The common ratio is a crucial component when dealing with geometric sequences and series. It is the factor by which we multiply each term to get the next term in the sequence.

For the geometric sequence

• \(-12, 8, -\frac{16}{3}, \frac{32}{9}, \dots\),
• the common ratio \(r\) is found by dividing any term in the sequence by the term before it:
• For instance, dividing the second term \(8\) by the first term \(-12\), giving the common ratio \(r = \frac{8}{-12} = -\frac{2}{3}\).

The common ratio for this sequence is \(-\frac{2}{3}\). If the absolute value of this ratio, \(|r|\), is less than 1, the sequence forms an infinite geometric series that converges. If it is greater than 1, the series would diverge, and the sum could not be found.

Geometric Sequence

A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio.

To better understand:

• Suppose the first term of the sequence is \(a_1\), denoted as \(a_1\),
• and the common ratio is \(r\), then the second term \(a_2\) is \(a_1 \times r\),
• the third term \(a_3\) is \(a_1 \times r^2\), and so forth.

In the sequence we discuss:

• \(-12, 8, -\frac{16}{3}, \frac{32}{9}, \dots\), we can see this ratio pattern continues throughout the sequence.
• Understanding this structure helps in quickly identifying the pattern and calculating further terms or sums of the sequence.

Series Sum Formula

When dealing with infinite geometric sequences where the common ratio's absolute value is less than one, we can calculate the series' sum using the series sum formula for infinite geometric series.

This formula is given by:

• \[ S_\infty = \frac{a_1}{1-r} \]
• where \(S_\infty\) represents the sum of all terms in the infinite sequence,
• \(a_1\) is the first term, and
• \(r\) is the common ratio.

Applying this formula to our sequence:

• The first term \(\, a_1\) is \(-12\)
• and the common ratio \( r \) is \(-\frac{2}{3}\).

Plug these into the formula:

• \[ S_\infty = \frac{-12}{1 - (-\frac{2}{3})} = \frac{-12}{1+\frac{2}{3}} = \frac{-12}{\frac{5}{3}} \]

After simplifying the expression,

• the calculation results in a series sum of \(-\frac{36}{5}\).

This showcases how we can effectively determine the sum of an infinite geometric series, provided the series converges.