Key Concepts and Formulas

• Arithmetic Progression (A.P.): A sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference ($d$).
• $n$-th Term of an A.P.: If the first term is $a$ and the common difference is $d$, the $n$-th term is given by $a_n = a + (n-1)d$.

Step-by-Step Solution

Part 1: Analyzing the first A.P. ($x_n$)

• Step 1: Define and Set Up Equations for the First A.P. Let the first term of the A.P. $x_1, x_2, \dots, x_n$ be $a_x$ and its common difference be $d_x$. We are given $x_3 = 8$ and $x_8 = 20$. Using the formula $a_n = a + (n-1)d$: For $x_3 = 8$: $a_x + (3-1)d_x = 8 \implies a_x + 2d_x = 8$ (Equation 1) For $x_8 = 20$: $a_x + (8-1)d_x = 20 \implies a_x + 7d_x = 20$ (Equation 2) Explanation: We translate the given information into a system of linear equations involving the first term and common difference of the first A.P.

• Step 2: Solve for $a_x$ and $d_x$. Subtract Equation 1 from Equation 2 to eliminate $a_x$: $(a_x + 7d_x) - (a_x + 2d_x) = 20 - 8$ $5d_x = 12 \implies d_x = \frac{12}{5}$ Substitute $d_x$ back into Equation 1: $a_x + 2\left(\frac{12}{5}\right) = 8$ $a_x + \frac{24}{5} = 8 \implies a_x = 8 - \frac{24}{5} = \frac{40-24}{5} = \frac{16}{5}$ Explanation: We solve the system of equations to find the specific values of the first term and common difference for the first A.P.

• Step 3: Calculate $x_5$. Using the formula $a_n = a + (n-1)d$ for $n=5$: $x_5 = a_x + (5-1)d_x = a_x + 4d_x$ $x_5 = \frac{16}{5} + 4\left(\frac{12}{5}\right) = \frac{16}{5} + \frac{48}{5} = \frac{64}{5}$ Explanation: We use the determined values of $a_x$ and $d_x$ to find the 5th term of the first A.P.

Part 2: Analyzing the second A.P. ($1/h_n$)

• Step 4: Define and Set Up Equations for the Second A.P. We are told that ${1 \over {{h_1}}}$, ${1 \over {{h_2}}}$ , . . . , ${1 \over {{h_n}}}$ form an A.P. Let this sequence be $y_n = \frac{1}{h_n}$. Let its first term be $a_y$ and common difference be $d_y$. We are given $h_2 = 8$ and $h_7 = 20$. This means: $y_2 = \frac{1}{h_2} = \frac{1}{8}$ $y_7 = \frac{1}{h_7} = \frac{1}{20}$ Using the formula $a_n = a + (n-1)d$ for $y_n$: For $y_2 = \frac{1}{8}$: $a_y + (2-1)d_y = \frac{1}{8} \implies a_y + d_y = \frac{1}{8}$ (Equation 3) For $y_7 = \frac{1}{20}$: $a_y + (7-1)d_y = \frac{1}{20} \implies a_y + 6d_y = \frac{1}{20}$ (Equation 4) Explanation: It is crucial to recognize that the sequence of reciprocals, $1/h_n$, forms the A.P. We convert the given $h_n$ values into the corresponding terms of this A.P. and set up a system of equations.

• Step 5: Solve for $a_y$ and $d_y$. Subtract Equation 3 from Equation 4: $(a_y + 6d_y) - (a_y + d_y) = \frac{1}{20} - \frac{1}{8}$ $5d_y = \frac{2}{40} - \frac{5}{40} = -\frac{3}{40} \implies d_y = -\frac{3}{200}$ Substitute $d_y$ back into Equation 3: $a_y + \left(-\frac{3}{200}\right) = \frac{1}{8}$ $a_y = \frac{1}{8} + \frac{3}{200} = \frac{25}{200} + \frac{3}{200} = \frac{28}{200} = \frac{7}{50}$ Explanation: We solve the system of equations to find the first term and common difference of the A.P. formed by the reciprocals of $h_n$.

• Step 6: Calculate $h_{10}$. First, find the 10th term of the A.P. $y_n$: $y_{10} = a_y + (10-1)d_y = a_y + 9d_y$ $y_{10} = \frac{7}{50} + 9\left(-\frac{3}{200}\right) = \frac{7}{50} - \frac{27}{200}$ To subtract, find a common denominator (200): $y_{10} = \frac{28}{200} - \frac{27}{200} = \frac{1}{200}$ Since $y_{10} = \frac{1}{h_{10}}$, we have $\frac{1}{h_{10}} = \frac{1}{200}$, which means $h_{10} = 200$. Explanation: We calculate the 10th term of the $y_n$ sequence. Since $y_{10}$ is the reciprocal of $h_{10}$, we take the reciprocal of $y_{10}$ to find $h_{10}$.

Part 3: Final Calculation

• Step 7: Compute the Product $x_5 \cdot h_{10}$. We have $x_5 = \frac{64}{5}$ and $h_{10} = 200$. $x_5 \cdot h_{10} = \left(\frac{64}{5}\right) \cdot (200)$ $x_5 \cdot h_{10} = 64 \cdot \left(\frac{200}{5}\right) = 64 \cdot 40 = 2560$. Explanation: We multiply the previously calculated values of $x_5$ and $h_{10}$ to obtain the final answer.

Common Mistakes & Tips

• Reciprocal Confusion: The most common error is assuming $h_n$ itself forms an A.P. instead of $1/h_n$. Always verify which sequence is stated to be an A.P.
• Algebraic Accuracy: Pay close attention to calculations involving fractions and signs, as small errors can propagate and lead to an incorrect final answer.
• Systematic Approach: Break the problem into distinct parts for each A.P. This makes it easier to manage the variables and calculations.

Summary

The problem involved two sequences, $x_n$ and $h_n$. We identified that $x_n$ is an A.P. and used the given terms $x_3=8$ and $x_8=20$ to find its first term ($a_x = 16/5$) and common difference ($d_x = 12/5$). This allowed us to calculate $x_5 = 64/5$. We then recognized that the sequence $1/h_n$ forms an A.P. Using the given values $h_2=8$ and $h_7=20$, we found the corresponding terms of the reciprocal A.P. ($y_2 = 1/8$, $y_7 = 1/20$) and determined its first term ($a_y = 7/50$) and common difference ($d_y = -3/200$). From this, we calculated $y_{10} = 1/200$, which implies $h_{10} = 200$. Finally, the product $x_5 \cdot h_{10}$ was calculated as $(64/5) \cdot 200 = 2560$.

The final answer is \boxed{2560}, which corresponds to option (A).