To solve the first question, we are tasked with finding the 6th term of the series:

$\sum_{n=1}^{\infty} 15 \left(\frac{1}{2}\right)^n$

Solution

1. General Term of the Series: The general term for the series is: $a_n = 15 \left(\frac{1}{2}\right)^n$

2. Find the 6th Term ($n = 6$): Substitute $n = 6$ into the formula: $a_6 = 15 \left(\frac{1}{2}\right)^6$

3. Simplify: $\left(\frac{1}{2}\right)^6 = \frac{1}{64}$ So: $a_6 = 15 \cdot \frac{1}{64} = \frac{15}{64}$

Answer:

The 6th term of the series is:

$\boxed{\frac{15}{64}}$

Would you like assistance with the second question or any further clarification?

Related Questions:

1. How do you determine the convergence of geometric series?
2. Can you derive a general formula for the sum of a finite geometric series?
3. What happens to the series if the ratio is greater than 1?
4. How does the convergence test work for infinite series like $\frac{1}{n^p}$?
5. Could you compute other terms in the series using the same approach?

Tip: In geometric series, always verify if the ratio lies between $-1$ and $1$ to determine convergence.