If C(n, 4), C(n, 5) and C(n, 6) are in AP, then what is the value of n?

Formula used:

• C(n,r) = \(\rm \frac{n!}{r!(n-r)!}\)
• If a, b and c are in A.P then 2b = a + c

Calculation:

According to the question C(n, 4), C(n, 5) and C(n, 6) are in AP

2 C(n, 5) = C(n, 4) + C(n, 6) 

⇒ 2 × \(\rm \frac{n!}{5!(n-5)!}\) = \(\rm \frac{n!}{4!(n-4)!}\) + \(\rm \frac{n!}{6!(n-6)!}\)

⇒ \(\rm \frac{1}{60(n-5)(n - 6)!} = \frac{1}{24(n-4)(n - 5)(n - 6)!} + \frac{1}{720(n - 6)!}\)

Multiply by 24(n - 5)(n - 6) on both side

⇒ \(\rm \frac{1}{5} - \frac{1}{2(n-4)} = \frac{n - 5}{60}\)

Again, multiply 60(n - 4) on both side

⇒ 12(n - 4) = 30 + (n - 5)(n - 4)

⇒ 12n - 48 = 30 + n² - 9n + 20

⇒ n² - 21n + 98 = 0

⇒ n = 14, 7

∴ The value of n is 7.