In an AP, a = 3 and the sum from the 11th term to the 20th term is 465. Find d.

Correct answer: 3

The governing concept is that a range sum can be found by subtracting partial sums. The sum from the 11th through the 20th term is S₂₀ − S₁₀. Using Sₙ = n/2[2a + (n − 1)d] with a = 3, we get S₂₀ = 10(6 + 19d) and S₁₀ = 5(6 + 9d). Hence 10(6 + 19d) − 5(6 + 9d) = 465. Expanding gives 60 + 190d − 30 − 45d = 465, so 30 + 145d = 465. Therefore 145d = 435 and d = 3. Option C is correct. A common mistake is to subtract S₁₁ rather than S₁₀, because the terms before the 11th term are precisely the first ten terms. Other options also arise from counting the interval incorrectly or using the wrong coefficient in the sum formula.