Question

If \(^nC_4,^nC_5\) and \(^nC_6\) are in arithmetic progression (A.P.), then the value of n is...

• A. \(5\) or \(11\)
• B. \(7\) or \(14\)
• C. \(8\) or \(15\)
• D. \(6\) or \(13\)

Hint:

Use the AP condition 2 nC5 = nC4 + nC6 and divide by nC5.

Answer 1

The Correct Option is B

Step 1: Write the AP condition
Three terms are in A.P. when the middle term is the mean: \(2\,{}^nC_5 = {}^nC_4 + {}^nC_6\).

Step 2: Divide by \({}^nC_5\)
Use \({}^nC_4/{}^nC_5 = \frac{5}{n-4}\) and \({}^nC_6/{}^nC_5 = \frac{n-5}{6}\). Then
\[ 2 = \frac{5}{n-4} + \frac{n-5}{6} \]

Step 3: Clear the fractions
Multiply by \(6(n-4)\): \(12(n-4) = 30 + (n-5)(n-4)\). So \(12n - 48 = n^2 - 9n + 50\), which gives
\[ n^2 - 21n + 98 = 0 \]

Step 4: Solve
\((n - 7)(n - 14) = 0\), so \(n = 7\) or \(n = 14\). Check \(n = 7\): \({}^7C_4 = 35\), \({}^7C_5 = 21\), \({}^7C_6 = 7\), and \(35 + 7 = 42 = 2 \times 21\). Option (B) is correct.

Final Answer:
The values of n are 7 and 14. This is option (B). \[ \boxed{\text{(B) }7 \text{ or } 14} \]