Let Tᵣ be the rᵗʰ term of an A.P. If for some m, Tₘ=1/25, T₂₅=1/20, and 20 Σᵣ₌₁²⁵ Tᵣ=13, then 5 m Σᵣ₌ₘ²ᵐ Tᵣ is equal to

Solution

Maths sir · 4 steps

Use the sum of 25 terms to find T₁, and you will see the whole A.P. is just Tᵣ = r/500.

1. Find the first term. From 20 Σ Tᵣ = 13, the sum of the first 25 terms is 13/20. Using S₂₅ = 25/2(T₁ + T₂₅) gives T₁ + T₂₅ = 26/500. Since T₂₅ = 25/500, we get T₁ = 1/500.
2. Common difference. d = (T₂₅ - T₁)/24 = (24/500)/24 = 1/500. So Tᵣ = 1/500 + (r-1)1/500 = r/500.
3. Find m. Tₘ = m/500 = 1/25 = 20/500, so m = 20 and 2m = 40.

   Σᵣ₌₂₀⁴⁰ Tᵣ = 1/500Σᵣ₌₂₀⁴⁰ r = 1/500 × (21 × (20 + 40))/2 = 630/500

4. So. There are 21 terms from r = 20 to r = 40. Then 5m Σ Tᵣ = 100 × 630/500 = 126, which is option (b).

The idea to keep. When an A.P. question gives you a term and a sum, use Sₙ = n/2(first + last) to get the first term. Then pin down d and write Tᵣ in closed form.

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