Let a₁, a₂, a₃, … be an A.P. If a₇=3, the product a₁ a₄ is minimum and the sum of its first n terms is zero, then n !-4 aₙ₍ₙ₊₂₎ is equal to :

Solution

Maths sir · 5 steps

Write everything in terms of d using a₇ = 3. The product then becomes a quadratic you can minimise.

1. Use the given term. Let the first term be a and the common difference be d. From a₇ = a + 6d = 3 you get a = 3 - 6d, so a₁ = 3 - 6d and a₄ = a + 3d = 3 - 3d.

   a₁ a₄ = (3-6d)(3-3d) = 18d² - 27d + 9

2. Minimise the product. This is an upward parabola in d, so its minimum is at the vertex d = 27/(2 × 18) = 3/4. Then a = 3 - 6 × 3/4 = -3/2.
3. Sum equals zero. Sₙ = n/2[2a + (n-1)d] = 0 with n ≠ 0 gives -3 + (n-1) × 3/4 = 0, so n - 1 = 4 and n = 5.
4. Find the term. n(n+2) = 5 × 7 = 35. So a₃₅ = -3/2 + 34 × 3/4 = -3/2 + 51/2 = 24.
5. So. n! - 4a₃₅ = 120 - 4 × 24 = 120 - 96 = 24, which is option (a).

The idea to keep. When one term of an A.P. is fixed, write a in terms of d. Any product of two terms then becomes a quadratic in d, and its vertex gives the minimum.

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