Key Concepts and Formulas

• Reduction Formulae for Definite Integrals: A formula that relates an integral of a certain form to an integral of a similar form with a lower exponent. This simplifies the evaluation of integrals involving powers of trigonometric functions.
• Trigonometric Identities: Specifically, the identity $\cot^2 x = \csc^2 x - 1$ is crucial for transforming the integrand.
• Definite Integral Substitution: When performing substitution in a definite integral, the limits of integration must be changed according to the substitution variable.
• Arithmetic Progression (A.P.): A sequence of numbers such that the difference between consecutive terms is constant. If $a, b, c$ are in A.P., then $2b = a+c$.

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Step-by-Step Solution

Step 1: Derive the Reduction Formula for $I_n$ We are given $I_n = \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^n}x\,dx}$. To find a reduction formula, we aim to relate $I_n$ to $I_{n-2}$. We can rewrite the integrand as: ${\cot }^n x = {\cot^{n-2}x} \cdot {\cot^2 x}$ Using the trigonometric identity ${\cot^2 x = \csc^2 x - 1}$: ${\cot }^n x = {\cot^{n-2}x} (\csc^2 x - 1)$ Now, substitute this back into the integral: $I_n = \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{\cot^{n-2}x (\csc^2 x - 1)\,dx}}$ Split the integral into two parts: $I_n = \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{\cot^{n-2}x \csc^2 x\,dx}} - \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{\cot^{n-2}x\,dx}}$ The second integral is clearly $I_{n-2}$. For the first integral, let $u = \cot x$. Then, $du = -\csc^2 x\,dx$, so $\csc^2 x\,dx = -du$. We must change the limits of integration: When $x = {\pi \over 4}$, $u = \cot({\pi \over 4}) = 1$. When $x = {\pi \over 2}$, $u = \cot({\pi \over 2}) = 0$. So, the first integral becomes: $\int\limits_{1}^{0} {u^{n-2}(-du)} = -\int\limits_{1}^{0} {u^{n-2}\,du} = \int\limits_{0}^{1} {u^{n-2}\,du}$ Evaluating this integral: $\left[ \frac{u^{n-1}}{n-1} \right]_{0}^{1} = \frac{1^{n-1}}{n-1} - \frac{0^{n-1}}{n-1} = \frac{1}{n-1}$ (This is valid for $n-1 > 0$, i.e., $n>1$. For this problem, $n \ge 2$, so this is valid.) Therefore, the reduction formula is: $I_n = \frac{1}{n-1} - I_{n-2}$ Rearranging this, we get: $I_n + I_{n-2} = \frac{1}{n-1}$

Step 2: Calculate the required sums of $I_n$ Using the reduction formula $I_n + I_{n-2} = \frac{1}{n-1}$, we can compute the sums required for the options:

• For $n=4$: $I_4 + I_2 = \frac{1}{4-1} = \frac{1}{3}$.
• For $n=5$: $I_5 + I_3 = \frac{1}{5-1} = \frac{1}{4}$.
• For $n=6$: $I_6 + I_4 = \frac{1}{6-1} = \frac{1}{5}$.

Step 3: Analyze the Options

We need to check which of the given sequences is in an Arithmetic Progression (A.P.). A sequence $a, b, c$ is in A.P. if $2b = a+c$.

Option (A): Consider the sequence $\frac{1}{I_2 + I_4}, \frac{1}{I_3 + I_5}, \frac{1}{I_4 + I_6}$. Substituting the values calculated in Step 2: The sequence is $\frac{1}{1/3}, \frac{1}{1/4}, \frac{1}{1/5}$, which simplifies to $3, 4, 5$. Let's check if this sequence is in A.P.: The difference between the second and first term is $4 - 3 = 1$. The difference between the third and second term is $5 - 4 = 1$. Since the differences are constant, the sequence $3, 4, 5$ is in A.P.

Option (B): Consider the sequence $I_2 + I_4, I_3 + I_5, I_4 + I_6$. Substituting the values calculated in Step 2: The sequence is $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}$. Let's check if this sequence is in A.P.: The difference between the second and first term is $\frac{1}{4} - \frac{1}{3} = \frac{3-4}{12} = -\frac{1}{12}$. The difference between the third and second term is $\frac{1}{5} - \frac{1}{4} = \frac{4-5}{20} = -\frac{1}{20}$. Since $-\frac{1}{12} \neq -\frac{1}{20}$, this sequence is not in A.P.

Option (C): Consider the sequence $\frac{1}{I_2 + I_4}, \frac{1}{I_3 + I_5}, \frac{1}{I_4 + I_6}$. This is the same sequence as in Option (A), which is $3, 4, 5$. Let's check if this sequence is in G.P. A sequence $a, b, c$ is in G.P. if $b^2 = ac$. Here, $b^2 = 4^2 = 16$. And $ac = 3 \times 5 = 15$. Since $16 \neq 15$, this sequence is not in G.P.

Option (D): Consider the sequence $I_2 + I_4, (I_3 + I_5)^2, I_4 + I_6$. Substituting the values calculated in Step 2: The sequence is $\frac{1}{3}, (\frac{1}{4})^2, \frac{1}{5}$. This is $\frac{1}{3}, \frac{1}{16}, \frac{1}{5}$. Let's check if this sequence is in G.P.: The ratio of the second term to the first term is $\frac{1/16}{1/3} = \frac{3}{16}$. The ratio of the third term to the second term is $\frac{1/5}{1/16} = \frac{16}{5}$. Since $\frac{3}{16} \neq \frac{16}{5}$, this sequence is not in G.P.

Step 4: Conclude the Correct Option Based on the analysis in Step 3, only Option (A) presents a sequence that is in an Arithmetic Progression.

Common Mistakes & Tips

• Incorrectly Changing Limits: When using substitution in definite integrals, failing to change the limits of integration is a common error. Always update the limits based on the substitution variable.
• Algebraic Errors in Reduction Formula: Ensure careful algebraic manipulation when deriving the reduction formula. A small mistake here can lead to incorrect sums.
• Distinguishing A.P. and G.P. Conditions: Remember the conditions for A.P. ($2b = a+c$) and G.P. ($b^2 = ac$). Applying the wrong condition will lead to an incorrect answer.

Summary

The problem requires deriving a reduction formula for the given integral $I_n$. The reduction formula $I_n + I_{n-2} = \frac{1}{n-1}$ was successfully derived by using the identity $\cot^2 x = \csc^2 x - 1$ and substitution. This formula allowed us to calculate the values of $I_n + I_{n-2}$ for specific values of $n$. By substituting these values into the sequences given in the options, we found that the sequence $\frac{1}{I_2 + I_4}, \frac{1}{I_3 + I_5}, \frac{1}{I_4 + I_6}$ results in $3, 4, 5$, which is an arithmetic progression.

The final answer is \boxed{A}.