Let S₁ be the sum of first 2n terms of an arithmetic progression. Let S₂ be the sum of first 4n terms of the same arithmetic progression. If (S₂ - S₁) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to :

Solution

Maths sir · 4 steps

Terms 2n+1 to 4n are the middle third of the first 6n terms, so they give one third of S₆ₙ.

1. Write the sums. Use Sₖ = k/2[2a + (k-1)d]. Then S₁ = S₂ₙ = n[2a + (2n-1)d] and S₂ = S₄ₙ = 2n[2a + (4n-1)d].
2. Subtract. S₂ - S₁ = 4na + (8n² - 2n)d - 2na - (2n² - n)d = 2na + (6n² - n)d. Take out n to get the form below.

   S₂ - S₁ = n[2a + (6n-1)d] = 1000

3. Compare with S6n. S₆ₙ = 6n/2[2a + (6n-1)d] = 3n[2a + (6n-1)d]. This is exactly 3 times S₂ - S₁.
4. So. S₆ₙ = 3 × 1000 = 3000, which is option (c).

The idea to keep. The middle 2n of the first 6n AP terms pair up to the same sum a₁ + a₆ₙ as the whole run, so that middle block is exactly one third of S₆ₙ.

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