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The problem involves a geometric progression (GP) where the first term is \( a = 20 \), and the common ratio is \( r = 2 \). We are tasked with determining the sum of the terms from the 10th to the 15th.

### Key steps to solve:

1. **General term of a geometric progression:**
   The \(n\)-th term of a GP is given by:
   \[
   T_n = a \cdot r^{n-1}
   \]
   where:
   - \(a = 20\) is the first term,
   - \(r = 2\) is the common ratio, and
   - \(n\) is the term number.

2. **Sum of terms from \(T_{10}\) to \(T_{15}\):**
   The sum of terms from the \(k\)-th to \(n\)-th term in a GP is given by:
   \[
   S = a \cdot \frac{r^k (r^m - 1)}{r - 1}
   \]
   where:
   - \( k = 10 - 1 \) for the \(10\)-th term
   - \( r

A series follows a geometric progression such that the first term is 20 and the common ratio is 2, determine the sum of the tenth (10th) to the fifteenth (15th) term.

Solve a geometric progression problem where the first term is 20 and the common ratio is 2. Learn how to calculate the sum from the 10th to the 15th term using the sum of a GP formula, suitable for Grades 9-12.

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