To solve the problem using John's method, we follow these steps:

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Step 1: General formula for the sum of a geometric series

The sum of the first $n$ terms of a geometric series is given by: $S_n = a \frac{r^n - 1}{r - 1}, \quad \text{if } r \neq 1$ where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.

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Step 2: Identify the series parameters

• $n = 10$ (the number of terms),
• The 10th term $t_{10} = 5859375$,
• Common ratio $r = \frac{5}{3}$.

From the formula for the nth term of a geometric series: $t_n = a \cdot r^{n-1}$ we substitute $t_{10} = 5859375$, $r = \frac{5}{3}$, and $n = 10$: $5859375 = a \cdot \left(\frac{5}{3}\right)^9.$

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Step 3: Solve for $a$ (the first term)

Rewriting the equation: $a = \frac{5859375}{\left(\frac{5}{3}\right)^9}.$

Calculate $\left(\frac{5}{3}\right)^9$: $\left(\frac{5}{3}\right)^9 = \frac{5^9}{3^9}.$ Thus: $a = \frac{5859375}{\frac{5^9}{3^9}} = 5859375 \cdot \frac{3^9}{5^9}.$ Simplify: $a = 5859375 \cdot \left(\frac{3}{5}\right)^9.$

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Step 4: Construct a new series with $r = \frac{3}{5}$

John’s method suggests considering a series with $r' = \frac{3}{5}$, which is the reciprocal of $r = \frac{5}{3}$. The sum of this new series can be calculated similarly. Notice that: $S_{10}' = a' \cdot \frac{\left(\frac{3}{5}\right)^{10} - 1}{\frac{3}{5} - 1},$ where $a'$ is related to the original $a$.

To compute efficiently, let's simplify using the symmetry of the series under $r$ and $r'$.

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Let me proceed with calculations to determine the exact sum. sympy import symbols, Rational, simplify

Define variables

r = Rational(5, 3) # Original common ratio r_reciprocal = Rational(3, 5) # Reciprocal common ratio t10 = 5859375 # 10th term of the original series n = 10 # Number of terms

Compute a (first term of the original series)

a = t10 / (r**(n-1))

Sum of the original series

S_original = a * (r**n - 1) / (r - 1)

New first term for the reciprocal series

a_reciprocal = a * (r_reciprocal**(n-1))

Sum of the reciprocal series

S_reciprocal = a_reciprocal * (r_reciprocal**n - 1) / (r_reciprocal - 1)

Simplify results

a_val = simplify(a) S_original_val = simplify(S_original) S_reciprocal_val = simplify(S_reciprocal)

a_val, S_original_val, S_reciprocal_valThe calculations yield the following:

1. First term of the original series ($a$): $a = 59049$

2. Sum of the original series ($S_{10}$): $S_{10} = 14559864$

3. Sum of the reciprocal series ($S_{10}'$): $S_{10}' = \frac{5640789630653496}{3814697265625} \approx 1478.963$

John’s method cleverly leverages the reciprocal relationship between the original and transformed series to verify computations or simplify certain parts of the process.

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Would you like a detailed breakdown of these calculations or additional clarifications? Here are related questions:

1. How does the reciprocal series relate to the original series' sum?
2. Why is the common ratio's reciprocal significant in John's method?
3. What are the implications of geometric series symmetry for sums?
4. Could we generalize this approach to other series transformations?
5. How does the reciprocal relationship simplify computation in practice?

Tip: Understanding how transformations like reciprocals affect series sums can simplify complex problems, especially in symmetry-driven scenarios!