Let a₁, a₂, a₃, ..... be an A.P. If (a₁ + a₂ + .... + a₁₀)/(a₁ + a₂ + .... + aₚ) = 100/p², p ≠ 10, then a₁₁/a₁₀ is equal to :

Solution

Maths sir · 4 steps

Write both sums with the AP formula, cross-multiply, and the condition forces d = 2a.

1. Write both sums. Let the first term be a and the common difference d. Then Sₙ = n/2(2a + (n-1)d), so the ratio becomes 10(2a + 9d)/p(2a + (p-1)d) = 100/p².
2. Cross-multiply. Multiplying out gives p(2a + 9d) = 10(2a + (p-1)d), that is 2ap + 9dp = 20a + 10pd - 10d. Collecting terms gives 2a(p - 10) = d(p - 10).

   p ≠ 10 ⇒ d = 2a

3. General term. With d = 2a, aₙ = a + (n-1)(2a) = (2n - 1)a. So a₁₁ = 21a and a₁₀ = 19a.
4. So. a₁₁/a₁₀ = 21a/19a = 21/19, which is option (c).

The idea to keep. If Sₘ : Sₙ = m² : n² for an AP, then d = 2a and the terms go a, 3a, 5a, …. So aₘ : aₙ = (2m-1) : (2n-1).

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