91Ó°ÊÓ

In an infinite geometric sequence, the common ratio is a crucial concept. It's the factor that each term in the sequence is multiplied by to get the next term. In the sequence, \(-3, \frac{3}{5}, -\frac{3}{25}, \ldots\), this ratio, denoted by \( r \), can be found by dividing the second term \( \frac{3}{5} \) by the first term \( -3 \). Thus, \( r = \frac{3}{5} \div (-3) = -\frac{1}{5} \).
It's vital to ensure this ratio is consistent by checking subsequent terms. In this sequence, dividing the third term \( -\frac{3}{25} \) by the second term \( \frac{3}{5} \) confirms \( r = -\frac{1}{5} \) again. This stability of the common ratio guarantees the sequence's consistency.
Remember, to calculate the common ratio, simply:

• Take any term, divide it by its previous term
• Verify with additional terms for consistency

This will ensure you'll have the right ratio to work with when dealing with geometric sequences.

A geometric sequence is a series of numbers where each term is derived from the previous one by multiplying it with a constant, known as the common ratio. Each term is dependent on this consistent multiplier.

• The starting number is called the first term, usually denoted as \( a \).
• Subsequent terms are derived by multiplying the previous term by the common ratio \( r \).
• The formula for the \( n \)-th term in a geometric sequence is given by: \( a_n = a \cdot r^{(n-1)} \).

Let's take our example sequence: \(-3, \frac{3}{5}, -\frac{3}{25}, \ldots\). Here, the first term \( a \) is \(-3\)
and we've determined our common ratio \( r \) as \(-\frac{1}{5}\). By applying the formula \( a_n = a \cdot r^{(n-1)} \),\

• The second term becomes: \( -3 \times \left(-\frac{1}{5}\right) = \frac{3}{5}\)
• The third term becomes: \( \frac{3}{5} \times \left(-\frac{1}{5}\right) = -\frac{3}{25}\)

Understanding how each term is generated is key to mastering geometric sequences.

When dealing with infinite geometric series, a fascinating property arises: they can have a finite sum despite having infinitely many terms. This only applies when the common ratio \(|r| < 1\). The formula for finding the sum \( S \) of such an infinite series is:\[ S = \frac{a}{1 - r} \]where \( a \) is the first term. This formula is a powerful tool as it allows you to find the entire sum of the series without having to add up each infinite term individually.
In our example, the first term \( a \) is \( -3 \), and the common ratio \( r \) is \( -\frac{1}{5} \). Plugging these into the formula gives us:\[ S = \frac{-3}{1 - \left(-\frac{1}{5}\right)} = \frac{-3}{1 + \frac{1}{5}} = \frac{-3}{\frac{6}{5}} = -3 \times \frac{5}{6} = -\frac{15}{6} = -\frac{5}{2} \]
Thus, the sum of this infinite series is \(-\frac{5}{2}\).

• This approach saves time and complexity.
• Ensure that for the formula to be applicable, the absolute value of the common ratio must be less than 1.

Mastering this concept can greatly simplify working with infinite sequences.