JEE Main 2025 · 7 April · evening · Single correct · +4 / −1
Let aₙ be the nᵗʰ term of an A.P. If Sₙ = a₁ + a₂ + a₃ + … + aₙ = 700, a₆ = 7 and S₇ = 7, then aₙ is equal to :
- (a) 65
- (b) 56
- (c) 70
- (d) 64
Answer and solution
Answer (d) 64
Solution
Maths sir · 4 steps
S₇ = 7 hands you the middle term a₄ = 1 at once, and with a₆ = 7 the whole A.P. is fixed.
- Use the middle term. For an A.P. with 7 terms, S₇ = 7a₄. So 7a₄ = 7, which gives a₄ = 1.
- Find d and a1. From a₄ to a₆ is two steps: 2d = 7 - 1 = 6, so d = 3. Then a₁ = a₄ - 3d = 1 - 9 = -8.
Sₙ = n/2(2(-8) + (n-1)3) = n(3n-19)/2 = 700
- Solve for n. This gives 3n² - 19n - 1400 = 0. The discriminant is 361 + 16800 = 17161 = 131², so n = (19 + 131)/6 = 25. The other root is negative, so reject it.
- So. a₂₅ = -8 + 24 × 3 = 64. Check: 25/2(-8 + 64) = 25 × 28 = 700. The answer is (d) 64.
The idea to keep. When a sum of an odd number of A.P. terms is given, divide by the count to get the middle term. It turns a messy equation into one clean value.
Answer from the NTA final answer key. Solution by PrepWiser.