Key Concepts and Formulas

• The $n$-th term of an Arithmetic Progression (A.P.) is given by $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference.
• The difference between consecutive terms in an A.P. is constant: $a_{k+1} - a_k = d$.
• Rationalizing the denominator of a fraction of the form $\frac{1}{\sqrt{X}+\sqrt{Y}}$ is done by multiplying the numerator and denominator by the conjugate $\sqrt{X}-\sqrt{Y}$, leading to $\frac{\sqrt{X}-\sqrt{Y}}{X-Y}$.
• A telescoping sum is a series where most of the intermediate terms cancel out, simplifying the sum to a difference of the first and last remaining terms.

Step-by-Step Solution

Step 1: Express the given conditions using the A.P. formula. We are given that $a_1, a_2, \ldots, a_n$ are in A.P. We are provided with two conditions: $a_5 = 2a_7$ and $a_{11} = 18$. Using the formula $a_n = a_1 + (n-1)d$: $a_5 = a_1 + (5-1)d = a_1 + 4d$ $a_7 = a_1 + (7-1)d = a_1 + 6d$ $a_{11} = a_1 + (11-1)d = a_1 + 10d$

Step 2: Use the condition $a_5 = 2a_7$ to find a relationship between $a_1$ and $d$. Substitute the expressions from Step 1 into the given equation: $a_1 + 4d = 2(a_1 + 6d)$ $a_1 + 4d = 2a_1 + 12d$ Rearranging the terms to solve for $a_1$: $a_1 - 2a_1 = 12d - 4d$ $-a_1 = 8d$ $a_1 = -8d$

Step 3: Use the condition $a_{11} = 18$ to find the values of $a_1$ and $d$. Substitute the expression for $a_{11}$ from Step 1 and the relationship $a_1 = -8d$ from Step 2 into the equation $a_{11} = 18$: $a_1 + 10d = 18$ $(-8d) + 10d = 18$ $2d = 18$ $d = \frac{18}{2} = 9$ Now, substitute the value of $d$ back into the equation $a_1 = -8d$: $a_1 = -8(9) = -72$ So, the first term is $a_1 = -72$ and the common difference is $d = 9$.

Step 4: Simplify the general term of the summation. The summation involves terms of the form $\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}$. We will rationalize this expression. Multiply the numerator and denominator by the conjugate $\sqrt{a_k}-\sqrt{a_{k+1}}$: $\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} = \frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} \times \frac{\sqrt{a_k}-\sqrt{a_{k+1}}}{\sqrt{a_k}-\sqrt{a_{k+1}}}$ Using the difference of squares formula $(X+Y)(X-Y) = X^2-Y^2$ in the denominator: $= \frac{\sqrt{a_k}-\sqrt{a_{k+1}}}{(\sqrt{a_k})^2 - (\sqrt{a_{k+1}})^2} = \frac{\sqrt{a_k}-\sqrt{a_{k+1}}}{a_k - a_{k+1}}$ Since $a_k$ and $a_{k+1}$ are terms in an A.P., we know that $a_{k+1} - a_k = d$. Therefore, $a_k - a_{k+1} = -d$. Substituting this into the expression: $= \frac{\sqrt{a_k}-\sqrt{a_{k+1}}}{-d} = \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}$

Step 5: Evaluate the summation. The given summation is: $S = \frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots+\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}$ This can be written in summation notation as: $S = \sum_{k=10}^{17} \frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}$ Using the simplified general term from Step 4: $S = \sum_{k=10}^{17} \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d} = \frac{1}{d} \sum_{k=10}^{17} (\sqrt{a_{k+1}}-\sqrt{a_k})$ Now, let's expand the summation to observe the telescoping nature: $S = \frac{1}{d} \left[ (\sqrt{a_{11}}-\sqrt{a_{10}}) + (\sqrt{a_{12}}-\sqrt{a_{11}}) + (\sqrt{a_{13}}-\sqrt{a_{12}}) + \ldots + (\sqrt{a_{18}}-\sqrt{a_{17}}) \right]$ The intermediate terms cancel out: $S = \frac{1}{d} \left[ -\sqrt{a_{10}} + \sqrt{a_{18}} \right] = \frac{1}{d} (\sqrt{a_{18}} - \sqrt{a_{10}})$

Step 6: Calculate the values of $a_{10}$ and $a_{18}$. Using $a_1 = -72$ and $d = 9$: $a_{10} = a_1 + (10-1)d = a_1 + 9d = -72 + 9(9) = -72 + 81 = 9$ $a_{18} = a_1 + (18-1)d = a_1 + 17d = -72 + 17(9) = -72 + 153 = 81$

Step 7: Substitute the calculated values into the simplified sum and find the final answer. We have $d=9$, $a_{10}=9$, and $a_{18}=81$. $S = \frac{1}{9} (\sqrt{81} - \sqrt{9})$ $S = \frac{1}{9} (9 - 3)$ $S = \frac{1}{9} (6) = \frac{6}{9} = \frac{2}{3}$ The problem asks for the value of $12 \times S$: $12S = 12 \times \frac{2}{3}$ $12S = 4 \times 2 = 8$

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with algebraic manipulations, especially when solving for $a_1$ and $d$. A small mistake in this step will affect the entire solution.
• Sign Convention: Pay close attention to signs during rationalization. The denominator $a_k - a_{k+1}$ is equal to $-d$, not $d$.
• Telescoping Sum Indices: Ensure the correct terms remain after cancellation in the telescoping sum, based on the starting and ending indices of the summation.

Summary

This problem required us to first determine the first term ($a_1$) and common difference ($d$) of an Arithmetic Progression using the given conditions. We then simplified the general term of the series by rationalizing the denominator, which revealed a telescoping sum. By evaluating this telescoping sum and using the calculated values of $a_1$ and $d$, we found the value of the given expression.

The final answer is \boxed{8}.