If , and are in AP, then can be
- A.9
- B.11
- C.14
- D.12
Show answer and solution
Correct option: C
Solution
For three terms in AP, 2·(middle) = first + last, so 2·ⁿC₅ = ⁿC₄ + ⁿC₆.
Dividing through by ⁿC₅:
2 = ⁿC₄/ⁿC₅ + ⁿC₆/ⁿC₅ = 5/(n-4) + (n-5)/6
Multiply through by 6(n-4):
12(n-4) = 30 + (n-5)(n-4)
12n - 48 = 30 + n² - 9n + 20
n² - 21n + 98 = 0
(n-14)(n-7) = 0, giving n = 14 or n = 7.
Among the options, n = 14 is present, so the answer is option 3.
Dividing through by ⁿC₅:
2 = ⁿC₄/ⁿC₅ + ⁿC₆/ⁿC₅ = 5/(n-4) + (n-5)/6
Multiply through by 6(n-4):
12(n-4) = 30 + (n-5)(n-4)
12n - 48 = 30 + n² - 9n + 20
n² - 21n + 98 = 0
(n-14)(n-7) = 0, giving n = 14 or n = 7.
Among the options, n = 14 is present, so the answer is option 3.
[JEE Main 2019 · 12 Jan · Shift II]