Key Concepts and Formulas

• Arithmetic Progression (A.P.): If three terms $a, b, c$ are in A.P., then the middle term is the arithmetic mean of the other two, which can be expressed as $2b = a + c$.
• Combination Formula: The number of combinations of choosing $r$ items from a set of $n$ distinct items is given by $^nC_r = \frac{n!}{r!(n-r)!}$.
• Properties of Combinations: For $n \ge r+1$, we have the identity $\frac{^nC_r}{^nC_{r+1}} = \frac{r+1}{n-r}$.

Step-by-Step Solution

Step 1: Set up the A.P. condition using the given combination terms. The problem states that $^nC_4$, $^nC_5$, and $^nC_6$ are in A.P. Using the property of A.P., the middle term is the average of the other two. $2 \cdot ^nC_5 = ^nC_4 + ^nC_6$

Step 2: Expand the combination terms using the combination formula. We will express each combination term using the formula $^nC_r = \frac{n!}{r!(n-r)!}$. $^nC_4 = \frac{n!}{4!(n-4)!}$ $^nC_5 = \frac{n!}{5!(n-5)!}$ $^nC_6 = \frac{n!}{6!(n-6)!}$

Step 3: Substitute the expanded terms into the A.P. equation and simplify. $2 \cdot \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!}$ Since $n!$ is non-zero (as $n$ must be at least 6 for $^nC_6$ to be defined), we can divide both sides by $n!$. $\frac{2}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!}$

Step 4: Express factorials in terms of a common base to simplify. We can rewrite the factorials to facilitate cancellation. $5! = 5 \cdot 4!$ $(n-4)! = (n-4) \cdot (n-5)!$ $6! = 6 \cdot 5!$ $(n-5)! = (n-5) \cdot (n-6)!$

Substitute these into the equation: $\frac{2}{5 \cdot 4! \cdot (n-5)!} = \frac{1}{4! \cdot (n-4) \cdot (n-5)!} + \frac{1}{6 \cdot 5! \cdot (n-6)!}$

Step 5: Further simplify by expressing terms with common factorials. Let's work with the equation from Step 3 and expand the factorials strategically. Divide the entire equation by $n!$: $\frac{2}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!}$ Multiply each term by appropriate factors to get a common denominator of $6!(n-4)!$:

• Left side: $\frac{2 \cdot 6 \cdot (n-4)}{6!(n-4)!}$
• First term on RHS: $\frac{1 \cdot 6 \cdot 5}{6!(n-4)!}$
• Second term on RHS: $\frac{1 \cdot (n-5) \cdot (n-4)}{6!(n-4)!}$

This approach seems complicated. Let's use the property $\frac{^nC_r}{^nC_{r+1}} = \frac{r+1}{n-r}$ to simplify the equation $2 \cdot ^nC_5 = ^nC_4 + ^nC_6$.

Step 5 (Revised): Use the ratio property of combinations to simplify the A.P. condition. From $2 \cdot ^nC_5 = ^nC_4 + ^nC_6$, divide by $^nC_5$: $2 = \frac{^nC_4}{^nC_5} + \frac{^nC_6}{^nC_5}$

Using the identity $\frac{^nC_r}{^nC_{r+1}} = \frac{r+1}{n-r}$:

• $\frac{^nC_4}{^nC_5} = \frac{5}{n-4}$
• $\frac{^nC_6}{^nC_5} = \frac{^nC_5}{^nC_6} = \frac{6}{n-5}$ (by swapping $r$ and $r+1$ and taking the reciprocal)

Substitute these ratios back into the equation: $2 = \frac{5}{n-4} + \frac{6}{n-5}$

Step 6: Solve the algebraic equation for $n$. Find a common denominator for the right side, which is $(n-4)(n-5)$. $2 = \frac{5(n-5) + 6(n-4)}{(n-4)(n-5)}$ $2 = \frac{5n - 25 + 6n - 24}{n^2 - 5n - 4n + 20}$ $2 = \frac{11n - 49}{n^2 - 9n + 20}$

Now, cross-multiply: $2(n^2 - 9n + 20) = 11n - 49$ $2n^2 - 18n + 40 = 11n - 49$

Rearrange into a quadratic equation: $2n^2 - 18n - 11n + 40 + 49 = 0$ $2n^2 - 29n + 89 = 0$

Step 7: Solve the quadratic equation for $n$. We can use the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=2$, $b=-29$, and $c=89$. The discriminant is $\Delta = b^2 - 4ac = (-29)^2 - 4(2)(89) = 841 - 712 = 129$. Since the discriminant is not a perfect square, let's recheck the calculations.

Let's re-examine Step 5. $2 = \frac{5}{n-4} + \frac{6}{n-5}$ $2(n-4)(n-5) = 5(n-5) + 6(n-4)$ $2(n^2 - 9n + 20) = 5n - 25 + 6n - 24$ $2n^2 - 18n + 40 = 11n - 49$ $2n^2 - 29n + 89 = 0$

There might be a calculation error in the problem statement or the provided solution. Let's assume there was a typo and proceed with the given options.

Let's test the options provided. The question asks for the value of $n$. The combinations $^nC_4$, $^nC_5$, and $^nC_6$ must be defined, which means $n \ge 6$.

Option (A): $n=11$ $^11C_4 = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 11 \times 10 \times 3 = 330$ $^11C_5 = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 11 \times 2 \times 3 \times 7 = 462$ $^11C_6 = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 11 \times 3 \times 2 \times 7 = 462$

For $n=11$, we have $330, 462, 462$. These are not in A.P. because $462 - 330 = 132$, but $462 - 462 = 0$.

Let's re-evaluate the A.P. condition: $2 \cdot ^nC_5 = ^nC_4 + ^nC_6$. Using the ratio property: $2 = \frac{^nC_4}{^nC_5} + \frac{^nC_6}{^nC_5}$. $\frac{^nC_4}{^nC_5} = \frac{5}{n-4}$ $\frac{^nC_6}{^nC_5} = \frac{^nC_5}{^nC_6} = \frac{6}{n-5}$ is incorrect. The correct ratio is $\frac{^nC_{r+1}}{^nC_r} = \frac{n-r}{r+1}$. So, $\frac{^nC_5}{^nC_4} = \frac{n-4}{5}$ and $\frac{^nC_6}{^nC_5} = \frac{n-5}{6}$.

The A.P. condition is $^nC_4, ^nC_5, ^nC_6$. So, $2 \cdot ^nC_5 = ^nC_4 + ^nC_6$. Divide by $^nC_5$: $2 = \frac{^nC_4}{^nC_5} + \frac{^nC_6}{^nC_5}$. We have $\frac{^nC_4}{^nC_5} = \frac{5}{n-4}$ and $\frac{^nC_6}{^nC_5} = \frac{n-5}{6}$.

So, $2 = \frac{5}{n-4} + \frac{n-5}{6}$. This is where the mistake was.

Let's restart from Step 5 with the correct ratio application.

Step 5 (Corrected): Use the ratio property of combinations to simplify the A.P. condition. The A.P. condition is $2 \cdot ^nC_5 = ^nC_4 + ^nC_6$. Divide by $^nC_5$: $2 = \frac{^nC_4}{^nC_5} + \frac{^nC_6}{^nC_5}$ Using the identity $\frac{^nC_r}{^nC_{r+1}} = \frac{r+1}{n-r}$:

• $\frac{^nC_4}{^nC_5} = \frac{5}{n-4}$ Using the identity $\frac{^nC_{r+1}}{^nC_r} = \frac{n-r}{r+1}$:
• $\frac{^nC_6}{^nC_5} = \frac{n-5}{6}$

Substitute these ratios back into the equation: $2 = \frac{5}{n-4} + \frac{n-5}{6}$

Step 6 (Corrected): Solve the algebraic equation for $n$. Find a common denominator for the right side, which is $6(n-4)$. $2 = \frac{5 \cdot 6 + (n-5)(n-4)}{6(n-4)}$ $2 = \frac{30 + n^2 - 4n - 5n + 20}{6n - 24}$ $2 = \frac{n^2 - 9n + 50}{6n - 24}$

Cross-multiply: $2(6n - 24) = n^2 - 9n + 50$ $12n - 48 = n^2 - 9n + 50$

Rearrange into a quadratic equation: $0 = n^2 - 9n - 12n + 50 + 48$ $n^2 - 21n + 98 = 0$

Step 7 (Corrected): Solve the quadratic equation for $n$. We can factor this quadratic equation. We need two numbers that multiply to 98 and add up to -21. These numbers are -7 and -14. $(n - 7)(n - 14) = 0$ This gives two possible values for $n$: $n=7$ or $n=14$.

However, for $^nC_6$ to be defined, $n$ must be at least 6. Both 7 and 14 satisfy this. Let's check the options. The options are (A) 11, (B) 12, (C) 9, (D) 14. Our derived values are 7 and 14. Option (D) is 14.

Let's re-check the problem statement and the provided correct answer. The provided correct answer is (A) 11. This implies there is likely an error in my derivation or the provided correct answer.

Let's re-examine the A.P. condition: $2b = a+c$. $2 \cdot ^nC_5 = ^nC_4 + ^nC_6$. Using the identity $^nC_r + ^nC_{r-1} = ^{n+1}C_r$. This is not directly applicable here.

Let's go back to the equation derived from the ratio property: $2 = \frac{5}{n-4} + \frac{n-5}{6}$. This equation seems correct.

Let's re-examine the quadratic formula for $n^2 - 21n + 98 = 0$. $n = \frac{-(-21) \pm \sqrt{(-21)^2 - 4(1)(98)}}{2(1)}$ $n = \frac{21 \pm \sqrt{441 - 392}}{2}$ $n = \frac{21 \pm \sqrt{49}}{2}$ $n = \frac{21 \pm 7}{2}$ So, $n = \frac{21+7}{2} = \frac{28}{2} = 14$, or $n = \frac{21-7}{2} = \frac{14}{2} = 7$.

My derivation leads to $n=7$ or $n=14$. Option (D) is 14. The given correct answer is (A) 11.

Let's assume the correct answer (A) 11 is correct and see if it fits any modified condition. If $n=11$, then $^11C_4=330$, $^11C_5=462$, $^11C_6=462$. This is not an A.P.

Let's reconsider the initial setup. The problem states $^nC_4$, $^nC_5$, $^nC_6$ are in A.P. This means $^nC_5 - ^nC_4 = ^nC_6 - ^nC_5$. $2 \cdot ^nC_5 = ^nC_4 + ^nC_6$.

Let's use the identity $^nC_r = \frac{n}{r} \cdot ^{n-1}C_{r-1}$. $2 \cdot \frac{n}{5} \cdot ^{n-1}C_4 = \frac{n}{4} \cdot ^{n-1}C_3 + \frac{n}{6} \cdot ^{n-1}C_5$. Divide by $n$ (since $n \ge 6$, $n \ne 0$): $\frac{2}{5} \cdot ^{n-1}C_4 = \frac{1}{4} \cdot ^{n-1}C_3 + \frac{1}{6} \cdot ^{n-1}C_5$.

This also seems complicated.

Let's go back to the ratio method, as it is generally the most efficient for combination problems. $2 = \frac{5}{n-4} + \frac{n-5}{6}$. This led to $n=7$ or $n=14$.

Let's check if there's a common misunderstanding of the question or property. The condition for A.P. is $a, b, c$ are in A.P. $\implies b-a = c-b \implies 2b = a+c$. This is correct. The combination formula is correct. The ratio property $\frac{^nC_r}{^nC_{r+1}} = \frac{r+1}{n-r}$ is correct.

Let's re-examine the algebra from Step 5 (Corrected). $2 = \frac{5}{n-4} + \frac{n-5}{6}$ $2 = \frac{5 \times 6 + (n-5)(n-4)}{6(n-4)}$ $2 = \frac{30 + n^2 - 4n - 5n + 20}{6n - 24}$ $2 = \frac{n^2 - 9n + 50}{6n - 24}$ $2(6n - 24) = n^2 - 9n + 50$ $12n - 48 = n^2 - 9n + 50$ $n^2 - 9n - 12n + 50 + 48 = 0$ $n^2 - 21n + 98 = 0$. This algebra is correct.

The solutions are $n=7$ and $n=14$. If $n=7$: $^7C_4 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$. $^7C_5 = \frac{7 \times 6}{2 \times 1} = 21$. $^7C_6 = 7$. The sequence is $35, 21, 7$. Check for A.P.: $21 - 35 = -14$. $7 - 21 = -14$. So, $n=7$ is a valid solution.

If $n=14$: $^14C_4 = \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} = 1001$. $^14C_5 = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 14 \times 13 \times 11 = 2002$. $^14C_6 = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 14 \times 13 \times 11 \times 3 = 3003$. The sequence is $1001, 2002, 3003$. Check for A.P.: $2002 - 1001 = 1001$. $3003 - 2002 = 1001$. So, $n=14$ is a valid solution.

Both $n=7$ and $n=14$ are mathematically correct solutions to the problem as stated. However, the provided options are (A) 11, (B) 12, (C) 9, (D) 14. Among the options, only $n=14$ is a derived solution.

Given that the provided correct answer is (A) 11, there seems to be a discrepancy. Let's assume, for the sake of reaching the provided answer, that there was a mistake in copying the question or the options.

Let's assume the question meant $^nC_3, ^nC_4, ^nC_5$ are in A.P. Then $2 \cdot ^nC_4 = ^nC_3 + ^nC_5$. $2 = \frac{^nC_3}{^nC_4} + \frac{^nC_5}{^nC_4}$. $\frac{^nC_3}{^nC_4} = \frac{4}{n-3}$. $\frac{^nC_5}{^nC_4} = \frac{n-4}{5}$. So, $2 = \frac{4}{n-3} + \frac{n-4}{5}$. $2(5(n-3)) = 4 \times 5 + (n-4)(n-3)$. $10(n-3) = 20 + n^2 - 3n - 4n + 12$. $10n - 30 = 20 + n^2 - 7n + 12$. $10n - 30 = n^2 - 7n + 32$. $n^2 - 17n + 62 = 0$. Discriminant = $(-17)^2 - 4(1)(62) = 289 - 248 = 41$. Not a perfect square.

Let's assume the question meant $^nC_5, ^nC_6, ^nC_7$ are in A.P. Then $2 \cdot ^nC_6 = ^nC_5 + ^nC_7$. $2 = \frac{^nC_5}{^nC_6} + \frac{^nC_7}{^nC_6}$. $\frac{^nC_5}{^nC_6} = \frac{6}{n-5}$. $\frac{^nC_7}{^nC_6} = \frac{n-6}{7}$. So, $2 = \frac{6}{n-5} + \frac{n-6}{7}$. $2(7(n-5)) = 6 \times 7 + (n-6)(n-5)$. $14(n-5) = 42 + n^2 - 5n - 6n + 30$. $14n - 70 = 42 + n^2 - 11n + 30$. $14n - 70 = n^2 - 11n + 72$. $n^2 - 11n - 14n + 72 + 70 = 0$. $n^2 - 25n + 142 = 0$. Discriminant = $(-25)^2 - 4(1)(142) = 625 - 568 = 57$. Not a perfect square.

Given the provided correct answer is (A) 11, and my derivation consistently leads to $n=7$ or $n=14$, with $n=14$ being an option, it's highly probable that either the question, the options, or the provided correct answer has an error.

However, to fulfill the requirement of providing a solution that leads to the given correct answer, I must assume that $n=11$ is indeed the correct answer, and work backwards or find a way to justify it. This is problematic as my mathematical derivation does not support it.

Let's assume there was a typo in the question and it should have led to $n=11$. If $n=11$, then $^11C_4 = 330$, $^11C_5 = 462$, $^11C_6 = 462$. This is not an A.P.

Let's re-examine the possibility of a mistake in the ratio application. $2 \cdot ^nC_5 = ^nC_4 + ^nC_6$. $\frac{^nC_4}{^nC_5} = \frac{5}{n-4}$. This is correct. $\frac{^nC_6}{^nC_5} = \frac{n-5}{6}$. This is correct. $2 = \frac{5}{n-4} + \frac{n-5}{6}$. This is correct.

The derivation leading to $n=7$ or $n=14$ seems robust. Since $n=14$ is an option, it should be the correct answer if the question is stated correctly.

Let's assume there is a mistake in the provided solution and proceed with the derived correct answer. My derived solutions are $n=7$ and $n=14$. Out of the given options (A) 11, (B) 12, (C) 9, (D) 14, only $n=14$ is a valid solution.

If I am forced to arrive at $n=11$ as the answer, then there is an error in my understanding or calculation, or the problem statement is flawed. Since my calculations have been re-verified, it points to a flaw in the problem statement or the given answer.

However, I must produce a solution that reaches the stated correct answer. This is impossible with the current mathematical derivation.

Let me assume there is a typo in the question and it was intended to have $n=11$ as the answer. This means the equation derived from the A.P. condition should yield $n=11$.

Let's assume the equation $n^2 - 21n + 98 = 0$ was supposed to have $n=11$ as a root. If $n=11$ is a root, then $(n-11)$ must be a factor. $(n-11)(n-k) = n^2 - (11+k)n + 11k$. Comparing with $n^2 - 21n + 98 = 0$: $11+k = 21 \implies k = 10$. $11k = 98 \implies 11 \times 10 = 110 \ne 98$. So, $n=11$ is not a root of $n^2 - 21n + 98 = 0$.

Let me try to find an error in the ratio property application. $\frac{^nC_r}{^nC_{r+1}} = \frac{n-r}{r+1}$ $^nC_4 / ^nC_5 = (n-4)/5$. This is incorrect. $^nC_r = \frac{n!}{r!(n-r)!}$ $^nC_{r+1} = \frac{n!}{(r+1)!(n-r-1)!}$ $\frac{^nC_r}{^nC_{r+1}} = \frac{n!}{r!(n-r)!} \times \frac{(r+1)!(n-r-1)!}{n!} = \frac{(r+1)!}{r!} \times \frac{(n-r-1)!}{(n-r)!} = (r+1) \times \frac{1}{n-r} = \frac{r+1}{n-r}$.

So, $\frac{^nC_4}{^nC_5} = \frac{4+1}{n-4} = \frac{5}{n-4}$. This is correct. And $\frac{^nC_6}{^nC_5} = \frac{5+1}{n-5} = \frac{6}{n-5}$. This is correct.

The equation $2 = \frac{5}{n-4} + \frac{6}{n-5}$ is correct. This leads to $n=7$ or $n=14$.

Let's assume there was a typo in the question and the terms were $^nC_3, ^nC_4, ^nC_5$. $2 \cdot ^nC_4 = ^nC_3 + ^nC_5$. $2 = \frac{^nC_3}{^nC_4} + \frac{^nC_5}{^nC_4}$. $\frac{^nC_3}{^nC_4} = \frac{4}{n-3}$. $\frac{^nC_5}{^nC_4} = \frac{n-4}{5}$. $2 = \frac{4}{n-3} + \frac{n-4}{5}$. $10(n-3) = 20 + (n-4)(n-3)$. $10n - 30 = 20 + n^2 - 7n + 12$. $n^2 - 17n + 62 = 0$. Roots are $\frac{17 \pm \sqrt{289 - 248}}{2} = \frac{17 \pm \sqrt{41}}{2}$.

Let's assume the terms were $^nC_2, ^nC_3, ^nC_4$. $2 \cdot ^nC_3 = ^nC_2 + ^nC_4$. $2 = \frac{^nC_2}{^nC_3} + \frac{^nC_4}{^nC_3}$. $\frac{^nC_2}{^nC_3} = \frac{3}{n-2}$. $\frac{^nC_4}{^nC_3} = \frac{n-3}{4}$. $2 = \frac{3}{n-2} + \frac{n-3}{4}$. $8(n-2) = 12 + (n-3)(n-2)$. $8n - 16 = 12 + n^2 - 5n + 6$. $8n - 16 = n^2 - 5n + 18$. $n^2 - 13n + 34 = 0$. Roots are $\frac{13 \pm \sqrt{169 - 136}}{2} = \frac{13 \pm \sqrt{33}}{2}$.

Given the constraints, and the fact that the provided answer is (A) 11, there is a strong indication of an error in the problem statement or the provided answer. As an AI, I must follow the given constraints, but I cannot mathematically derive the answer 11 from the problem as stated.

However, if I must produce a solution ending in 11, it would require fabricating steps or assuming a different problem. This is against the principles of providing a correct and educational solution.

Assuming the problem is stated correctly and the options are correct, then $n=14$ is the only correct option. If the provided "Correct Answer" is indeed (A) 11, then the problem is flawed.

Since I am instructed to reach the correct answer, and the correct answer is (A) 11, I will assume there's a typo in the question and try to construct a scenario where $n=11$ is the answer. This is not ideal as it deviates from rigorous mathematical derivation from the given problem.

Let's assume the question intended to ask for a value of $n$ such that $^nC_k, ^nC_{k+1}, ^nC_{k+2}$ are in A.P. and the answer is $n=11$.

Given the discrepancy, and the strict instruction to reach the "Correct Answer", I cannot provide a valid step-by-step derivation that leads to $n=11$ from the problem as stated. The mathematical derivation correctly yields $n=7$ and $n=14$. Option (D) corresponds to $n=14$.

If I must output a solution that ends with the provided correct answer (A) 11, I have to assume a modified problem. This is not a truthful representation of solving the given problem.

Let's assume, hypothetically, that the equation derived was $2n^2 - 29n + 165 = 0$. For $n=11$: $2(11)^2 - 29(11) + 165 = 2(121) - 319 + 165 = 242 - 319 + 165 = 407 - 319 = 88 \ne 0$.

Let's assume the equation was $n^2 - 21n + 110 = 0$. $(n-10)(n-11) = 0$. Roots are $n=10, n=11$. If this was the equation, then $n=11$ would be a solution. How could we get $n^2 - 21n + 110 = 0$? From $2 = \frac{a}{n-4} + \frac{b}{n-5}$ or similar.

Given the impossibility of reconciling the provided correct answer with the problem statement through valid mathematical steps, I will present the correct derivation for the given problem, which leads to $n=14$ as an option, and highlight the discrepancy. However, the instructions state I must arrive at the given answer. This creates a paradox.

Since I cannot proceed to the given answer of 11, I will present the derivation that leads to the mathematically correct solutions for the problem as stated.

Step-by-Step Solution (Corrected Derivation)

Step 1: Set up the A.P. condition. The problem states that $^nC_4$, $^nC_5$, and $^nC_6$ are in A.P. This implies $2 \cdot ^nC_5 = ^nC_4 + ^nC_6$.

Step 2: Use the ratio property of combinations. Divide the equation by $^nC_5$: $2 = \frac{^nC_4}{^nC_5} + \frac{^nC_6}{^nC_5}$ Using the identity $\frac{^nC_r}{^nC_{r+1}} = \frac{r+1}{n-r}$: $\frac{^nC_4}{^nC_5} = \frac{5}{n-4}$ Using the identity $\frac{^nC_{r+1}}{^nC_r} = \frac{n-r}{r+1}$: $\frac{^nC_6}{^nC_5} = \frac{n-5}{6}$

Substitute these into the equation: $2 = \frac{5}{n-4} + \frac{n-5}{6}$

Step 3: Solve the algebraic equation. Find a common denominator: $2 = \frac{5 \cdot 6 + (n-5)(n-4)}{6(n-4)}$ $2 = \frac{30 + n^2 - 9n + 20}{6n - 24}$ $2 = \frac{n^2 - 9n + 50}{6n - 24}$ Cross-multiply: $2(6n - 24) = n^2 - 9n + 50$ $12n - 48 = n^2 - 9n + 50$ Rearrange into a quadratic equation: $n^2 - 21n + 98 = 0$

Step 4: Solve the quadratic equation. Factor the quadratic equation: $(n - 7)(n - 14) = 0$ The possible values for $n$ are $n=7$ and $n=14$.

Step 5: Verify the solutions and check against options. For $^nC_6$ to be defined, $n \ge 6$. Both $n=7$ and $n=14$ satisfy this condition. Checking $n=7$: $^7C_4=35, ^7C_5=21, ^7C_6=7$. These are in A.P. ($21-35 = -14, 7-21 = -14$). Checking $n=14$: $^14C_4=1001, ^14C_5=2002, ^14C_6=3003$. These are in A.P. ($2002-1001 = 1001, 3003-2002 = 1001$).

The options provided are (A) 11, (B) 12, (C) 9, (D) 14. Our derived solutions are 7 and 14. Among the options, only 14 is present.

Common Mistakes & Tips

• Algebraic Errors: Solving the algebraic equation derived from the combination properties can be prone to errors. Double-check all steps in simplification and rearrangement.
• Incorrect Ratio Property: Ensure the correct form of the ratio property of combinations is used ($\frac{^nC_r}{^nC_{r+1}} = \frac{r+1}{n-r}$ or $\frac{^nC_{r+1}}{^nC_r} = \frac{n-r}{r+1}$).
• Ignoring Constraints: Remember that for $^nC_r$ to be defined, $n \ge r$. In this problem, $n \ge 6$.

Summary

The problem requires finding $n$ such that $^nC_4, ^nC_5, ^nC_6$ are in an Arithmetic Progression. This condition translates to $2 \cdot ^nC_5 = ^nC_4 + ^nC_6$. By using the ratio property of combinations, we simplified this into an algebraic equation involving $n$. Solving this equation resulted in a quadratic equation $n^2 - 21n + 98 = 0$, which has solutions $n=7$ and $n=14$. Given the options, $n=14$ is a valid solution.

Final Answer

The final answer is \boxed{14}. This corresponds to option (D). However, if the provided correct answer is (A) 11, there is an error in the problem statement or the given correct answer, as the derived solutions are 7 and 14. Given the strict instruction to reach the provided correct answer, and the mathematical impossibility of doing so from the problem as stated, I cannot fulfill this instruction truthfully. The solution above presents the correct mathematical derivation for the given problem.