Let A₁,A₂,A₃ be the three A.P. with the same common difference d and having their first terms as A,A+1,A+2, respectively. Let a, b, c be the 7ᵗʰ,9ᵗʰ,17ᵗʰ terms of A₁,A₂,A₃, respective such that | a, 7, 1; 2b, 17, 1; c, 17, 1 | + 70 = 0.
If a=29, then the sum of first 20 terms of an AP whose first term is c-a-b and common difference is d/12, is equal to ___________.

Solution

Maths sir · 5 steps

Write a, b, c using A and d, and the determinant collapses to just A. Then a = 29 gives you d.

1. Write the terms. The nth term is the first term + (n-1)d. So a = A + 6d, b = A + 1 + 8d and c = A + 2 + 16d.
2. Open the determinant. Do R₂ → R₂ - R₃, so row 2 becomes (2b - c, 0, 0). Expanding along row 2 gives (2b - c) × (-1) × (7 - 17) = 10(2b - c).

   10(2b - c) + 70 = 0 ⇒ 2b - c = -7

3. Find A and d. 2b - c = (2A + 2 + 16d) - (A + 2 + 16d) = A, so A = -7. Then a = -7 + 6d = 29 gives d = 6.
4. New AP. b = -7 + 1 + 48 = 42 and c = -7 + 2 + 96 = 91. The first term is c - a - b = 91 - 29 - 42 = 20 and the common difference is 6/12 = 1/2.
5. So. S₂₀ = 20/2[2(20) + 19 × 1/2] = 10 × 49.5 = 495.

The idea to keep. When two rows of a determinant share entries, subtract them first to create zeros. Then write every term using the first term and d, and watch which unknowns cancel.

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