Let a₁, a₂, a₃, … be in an arithmetic progression of positive terms.
Let Aₖ=a₁²-a₂²+a₃²-a₄²+…+a₂ₖ₋₁²-a₂ₖ².
If A₃=-153, A₅=-435 and a₁²+a₂²+a₃²=66, then a₁₇-A₇ is equal to ________.

Solution

Maths sir · 5 steps

Pair the squares as differences of squares and each pair gives -d times a sum of terms.

1. Pair the terms. Each pair is a₂ᵢ₋₁² - a₂ᵢ² = (a₂ᵢ₋₁-a₂ᵢ)(a₂ᵢ₋₁+a₂ᵢ) = -d (a₂ᵢ₋₁+a₂ᵢ). Adding all k pairs gives -d times the sum of the first 2k terms.

   Aₖ = -d S₂ₖ = -d k(2a + (2k-1)d)

2. Find a and d. A₃ = -153 gives d(2a+5d) = 51, and A₅ = -435 gives d(2a+9d) = 87. Subtracting gives 4d² = 36, so d = ± 3. With d = -3 you get a = -1, which is not positive, so d = 3 and a = 1.
3. Check. The terms are 1, 4, 7, … and 1 + 16 + 49 = 66, which matches the given condition.
4. Compute. a₁₇ = 1 + 16 × 3 = 49. Also A₇ = -3 × 7 × (2 + 13 × 3) = -21 × 41 = -861.
5. So. a₁₇ - A₇ = 49 - (-861) = 910.

The common slip: -812. Dropping the minus sign of A₇ gives 49 - 861 = -812. Remember that A₇ = -861 is negative.

The idea to keep. In an AP, aₘ² - aₘ₊₁² = -d(aₘ + aₘ₊₁). So alternating sums of squares turn into -d times an ordinary AP sum.

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