If `x_1,x_2 …,x_(20)` are in H.P and `x_1,2,x_(20)` are in G.P then `Sigma_(r=1)^(19)x_rr_(x+1)`

Options

A. 76
B. 80
C. 84
D. none of these

Explanation

To solve the problem step by step, we will analyze the given conditions and use properties of harmonic progression (H.P.) and geometric progression (G.P.).

**Step 1: Understanding the Progressions**
- We know that \( x_1, x_2, \ldots, x_{20} \) are in H.P. This means that their reciprocals \( \frac{1}{x_1}, \frac{1}{x_2}, \ldots, \frac{1}{x_{20}} \) are in A.P. (Arithmetic Progression).
- We also know that \( x_1, x_2, \ldots, x_{20} \) are in G.P., which means there is a common ratio \( r \) such that \( x_{n} = x_1 \cdot r^{(n-1)} \).

**Step 2: Finding the Common Difference in A.P.**
- Let the common difference of the A.P. formed by the reciprocals be \( d \).
- Thus, we can express the terms as:
\[ \frac{1}{x_n} = \frac{1}{x_1} + (n-1)d \]
for \( n = 1, 2, \ldots, 20 \).

**Step 3: Expressing the Terms in Terms of \( x_1 \) and \( d \)**
- From the above equation, we can express \( x_n \) as:
\[ x_n = \frac{1}{\frac{1}{x_1} + (n-1)d} \]

**Step 4: Finding the Relationship Between \( x_1 \) and \( x_{20} \)**
- Since \( x_1, x_2, \ldots, x_{20} \) are in G.P., we can write:
\[ x_{20} = x_1 r^{19} \]
- We also know that the product of the first and last terms in a G.P. is equal to the square of the geometric mean:
\[ x_1 x_{20} = x_1^2 r^{19} \]

**Step 5: Setting Up the Summation**
- We need to evaluate:
\[ \sum_{r=1}^{19} x_r x_{r+1} \]
- Using the relationship from the G.P., we can express \( x_r \) and \( x_{r+1} \):
\[ x_r = x_1 r^{(r-1)} \quad \text{and} \quad x_{r+1} = x_1 r^r \]
- Therefore,
\[ x_r x_{r+1} = x_1^2 r^{(r-1) + r} = x_1^2 r^{(2r-1)} \]

**Step 6: Evaluating the Summation**
- The summation becomes:
\[ \sum_{r=1}^{19} x_r x_{r+1} = x_1^2 \sum_{r=1}^{19} r^{(2r-1)} \]

**Step 7: Using the Given Conditions**
- Since \( x_1, x_{20} \) are in G.P., we have:
\[ x_1 x_{20} = 4 \]
- Therefore, substituting \( x_1 \) and \( x_{20} \) into our earlier relationship gives us:
\[ 19 x_1 x_{20} = 19 \cdot 4 = 76 \]

**Final Answer:**
Thus, the value of \( \sum_{r=1}^{19} x_r x_{r+1} \) is \( 76 \).