Question

If in the binomial expansion of (1 + x)ⁿ where n is a natural number, the coefficients of the 5^th, 6^th and 7^th terms are in A.P., then n is equal to:

Options

• 7 or 13

• 7 or 14

• 7 or 15

• 7 or 17

Answer 1

7 or 14

Explanation:

In the binomial expansion of (1 + x)ⁿ,

`{:(T_r = ""^nC_(r - 1)*(x)^(y - 1), "For"  r = 5",", T_5 = ""^nC_4 x^4):}`

`{:(r = 6",", T_6 = ""nC_5 x^5";", "and" r = 7",", T_7 = ""^nC_6 x^6):}`

Since, the coefficients of these terms are in A.P.

⇒ T₅ + T₇ = ₂T₆

⇒ ⁿC₆ + ⁿC₆ = 2 × ⁿC₅ 

⇒ `(n!)/((n - 4)4!) + (n!)/((n - 6)!6!) = (2 xx n!)/((n - 5)!5!)`

⇒ `(n(n - 1)(n - 2)(n - 3))/4 + (n(n - 1)(n - 2)(n - 3)(n - 4)(n - 5))/6 = (2n(n - 1)(n - 2)(n - 3)(n - 4))/(5!)`

⇒ `1/(4!) + ((n - 4)(n - 5))/(6!) = (2(n - 4))/(5!)`

⇒ `1/1 + ((n - 4)(n - 5))/(5 xx 6) = (2(n - 4))/(5!)`

⇒ `(30 + n^2 - 9n + 20)/(5 xx 6) = (2n - 8)/5`

⇒ `n^2 - 9n + 50 = 6(2n - 8)`

⇒ `n^2 - 9n + 50 - 12n + 48` = 0

⇒ `n^2 - 21n + 98` = 0

⇒ `(n - 7)(n - 14)` = 0

⇒ n = 7 or n = 14.