GATE Overflow Test Series | Quantitative Aptitude | Shallow Test 1: 13

Question

If \( ^{n}C_{3}, ^{n}C_{4} \), and \( ^{n}C_{5} \) are in A.P., then the possible values of \( n \) are ___________

• A. \(\dfrac{19 + \sqrt{41}}{2}, \dfrac{19 - \sqrt{41}}{2}\)
• B. \(\dfrac{17 + \sqrt{41}}{2}, \dfrac{17 - \sqrt{41}}{2}\)
• C. \(\dfrac{23 + \sqrt{41}}{2}, \dfrac{23 - \sqrt{41}}{2}\)
• D. \(\dfrac{25 + \sqrt{41}}{2}, \dfrac{25 - \sqrt{41}}{2}\)

Answer 1

Given: \[ ^{n}C_{3}, ^{n}C_{4}, \] and \[ ^{n}C_{5} \] are in arithmetic progression (A.P.).

For these to be in A.P., the following condition must hold:
\[ 2 \cdot ^{n}C_{4} = ^{n}C_{3} + ^{n}C_{5} \]

First, let's express the binomial coefficients in terms of \( n \):

\[ ^{n}C_{k} = \frac{n!}{k!(n-k)!} \]

Thus,
\[ ^{n}C_{3} = \frac{n(n-1)(n-2)}{6} \]

\[ ^{n}C_{4} = \frac{n(n-1)(n-2)(n-3)}{24} \]

\[ ^{n}C_{5} = \frac{n(n-1)(n-2)(n-3)(n-4)}{120} \]

Given that these are in A.P., we have:
\[ 2 \cdot \frac{n(n-1)(n-2)(n-3)}{24} = \frac{n(n-1)(n-2)}{6} + \frac{n(n-1)(n-2)(n-3)(n-4)}{120} \]

Simplifying this:
\[ \frac{n(n-1)(n-2)(n-3)}{12} = \frac{n(n-1)(n-2)}{6} + \frac{n(n-1)(n-2)(n-3)(n-4)}{120} \]

Multiply through by 120 to clear the denominators:
\[ 10n(n-1)(n-2)(n-3) = 20n(n-1)(n-2) + n(n-1)(n-2)(n-3)(n-4) \]

Divide through by \( n(n-1)(n-2) \) (assuming \( n > 3 \)):
\[ 10(n-3) = 20 + (n-3)(n-4) \]
\[ 10n - 30 = 20 + n^2 - 7n + 12 \]
\[ n^2 - 17n + 62 = 0 \]

Solving the quadratic equation:
\[ n = \frac{17 \pm \sqrt{17^2 - 4 \cdot 1 \cdot 62}}{2 \cdot 1} \]
\[ n = \frac{17 \pm \sqrt{289 - 248}}{2} \]
\[ n = \frac{17 \pm \sqrt{41}}{2} \]

Hence, the possible values for \( n \) are:
\[ \boxed{\dfrac{17 + \sqrt{41}}{2}, \dfrac{17 - \sqrt{41}}{2}} \]

Thus, the valid final values are:
\[ \dfrac{17 + \sqrt{41}}{2}, \dfrac{17 - \sqrt{41}}{2} \]

These values correspond to option (B).