- Define the sums of the geometric series using \(S_{\infty} = \frac{a}{1-r}\) and \(S_6 = \frac{a(1-r^6)}{1-r}\).
- Set up the equation \(\frac{a}{1-r} = \frac{8}{7} \times \frac{a(1-r^6)}{1-r}\) and simplify to \(r^6 = \frac{1}{8}\).
- Relate \(r^6 = \frac{1}{8}\) to the form \(r = \pm \frac{1}{\sqrt{k}}\), which implies \(r^6 = \frac{1}{k^3}\).
- Identify that \(k^3 = 8\), so \(k = 2\), resulting in \(\boxed{k=2}\).
Explanation
- Analysis of the Geometric Series
We are given a geometric series with first term \(a\) and common ratio \(r\) (where \(|r| < 1\)). We know the sum to infinity \(S_{\infty}\) and the sum of the first \(n\) terms \(S_n\) are given by:
\[S_{\infty} = \frac{a}{1 - r}, \quad S_n = \frac{a(1 - r^n)}{1 - r}.\]
The problem provides the condition \(S_{\infty} = \frac{8}{7} \times S_6\). We aim to find the integer \(k\) such that \(r = \pm \frac{1}{\sqrt{k}}\).
- Setting Up and Simplifying the Equation
Substitute the formulas for \(S_{\infty}\) and \(S_6\) into the given equation:
\[\frac{a}{1 - r} = \frac{8}{7} \times \frac{a(1 - r^6)}{1 - r}.\]
Assuming \(a
eq 0\) and \(|r| < 1\), we can divide both sides by \(\frac{a}{1-r}\):
\[1 = \frac{8}{7} \times (1 - r^6).\]
- Solving for the Constant k
To isolate \(r\), we solve the simplified equation for \(r^6\):
\[1 = \frac{8}{7} - \frac{8}{7}r^6\]
\[\frac{8}{7}r^6 = \frac{8}{7} - 1 = \frac{1}{7}.\]
Wait, let's recheck the algebra.
\[1 = \frac{8}{7}(1 - r^6) \implies \frac{7}{8} = 1 - r^6 \implies r^6 = 1 - \frac{7}{8} = \frac{1}{8}.\]
This matches the calculation. Now we need to express \(r^6 = \frac{1}{8}\) in the form \(r = \pm \frac{1}{\sqrt{k}}\).
Squaring both sides of the target expression \(r = \pm \frac{1}{\sqrt{k}}\) gives \(r^2 = \frac{1}{k}\).
Then \(r^6 = (r^2)^3 = (\frac{1}{k})^3 = \frac{1}{k^3}\).
Comparing \(\frac{1}{k^3} = \frac{1}{8}\), we find \(k^3 = 8\), which means \(k = 2\).
- Conclusion
Having found \(k = 2\), we can write the common ratio as \(r^6 = (\frac{1}{\sqrt{2}})^6 = (\frac{1}{2})^3 = \frac{1}{8}\). Thus, \(r = \pm \frac{1}{\sqrt{2}}\) is the correct form, with \(k=2\). The integer \(k\) is \(\boxed{2}\).
Examples
This problem relates to the concept of geometric decay, often found in fields like finance and physics. For example, if a radioactive substance loses a fixed proportion of its mass each day, the remaining amount follows a geometric series. Knowing the sum to infinity allows scientists to predict the total accumulation of waste over time or, in finance, to determine the total present value of a series of future cash flows. Understanding these series helps model growth and decline accurately.