Question
Download Solution PDFIf C(n, 4), C(n, 5) and C(n, 6) are in AP, then what is the value of n?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFFormula 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.
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