If the coefficients of x⁴, x⁵ and x⁶ in the expansion of (1+x)ⁿ are in the arithmetic progression, then the maximum value of n is:

Solution

Maths sir · 4 steps

Divide the AP condition by the middle coefficient: the factorials cancel and a quadratic in n is left.

1. AP condition. The coefficient of xʳ in (1+x)ⁿ is C(n, r). Three terms are in AP when twice the middle one equals the sum of the outer two, so 2C(n, 5) = C(n, 4) + C(n, 6).
2. Divide by the middle. Use the ratios C(n, 4)/C(n, 5) = 5/(n-4) and C(n, 6)/C(n, 5) = (n-5)/6. Dividing the condition by C(n, 5) gives 2 = 5/(n-4) + (n-5)/6.

   12(n-4) = 30 + (n-5)(n-4) ⇒ n² - 21n + 98 = 0

3. Factorise. n² - 21n + 98 = (n-7)(n-14) = 0, so n = 7 or n = 14. Both work: for n = 7 the coefficients are 35, 21, 7, and for n = 14 they are 1001, 2002, 3003.
4. So. The maximum value of n is 14, option (d).

The idea to keep. For consecutive binomial coefficients, divide by the middle one. The ratio C(n, r)/C(n, r-1) = (n-r+1)/r turns factorials into a simple equation in n.

Answer from the NTA final answer key. Solution by PrepWiser.