Sum of first n natural numbers

Use the governing formula Sₙ = n(n + 1)/2 for the sum of the first n natural numbers. For n = 200, S₂₀₀ = 200 × 201 / 2 = 100 × 201 = 20100. Convert 30% to the fraction 30/100 = 3/10. Therefore, 30% of the sum is (3/10) × 20100 = 6030. Hence option C is correct. A quick check is that 10% of 20100 is 2010, so three such portions equal 6030. Option A is too small and corresponds to an incorrect treatment of the sum, while B and D differ from the correct result by 15 and 15 respectively. The percentage must be applied after calculating the complete sum.

Use the governing formula for the sum of the first n natural numbers: S_n = n(n + 1)/2. With S_n = 10,440, we obtain n(n + 1) = 20,880. Testing n = 144 gives 144 × 145 = 20,880, so 144 × 145/2 = 10,440. Therefore option C is correct. The nearby alternatives do not satisfy the equation: n = 141 gives 10,011, n = 142 gives 10,153, and n = 145 gives 10,585. Since the sum increases strictly as n increases through positive natural numbers, there cannot be another positive natural-number answer. Direct substitution therefore confirms n = 144 uniquely.

Treat the two statements as simultaneous equations. Add them to eliminate Sₙ₋₁: (Sₙ + Sₙ₋₁) + (Sₙ − Sₙ₋₁) = 12544 + 112. The Sₙ₋₁ terms cancel, leaving 2Sₙ = 12656. Dividing by 2 gives Sₙ = 6328, so option C is correct. The result can be checked by finding the other partial sum: Sₙ₋₁ = 12544 − 6328 = 6216. Then 6328 + 6216 = 12544 and 6328 − 6216 = 112, so both original conditions hold. Option A is the value of Sₙ₋₁, a likely distractor. Options B and D arise from an arithmetic or division error and fail the verification.

First calculate the sum of the first 175 natural numbers using Sₙ = n(n + 1)/2. Thus S₁₇₅ = 175 × 176 / 2 = 175 × 88 = 15400. Next, express 40% as 40/100 = 2/5. Therefore, 40% of the required sum is (2/5) × 15400 = 2 × 3080 = 6160. Hence option C is correct. Another check is that 20% of 15400 is 3080, so 40% must be twice that amount, or 6160. Options A, B, and D do not result from applying 2/5 to 15400; they may come from an incorrect sum or percentage calculation. The order of operations matters: find the complete natural-number sum first, then calculate its percentage.

For a consecutive interval from a to b, use the difference of partial sums: sum(a to b) = S_b − Sₐ₋₁, where Sₖ = k(k + 1)/2. Here the required sum is S₃₅₀ − S₃₀₀. Calculate S₃₅₀ = 350 × 351 / 2 = 613? More directly, 175 × 351 = 61425. Also, S₃₀₀ = 300 × 301 / 2 = 150 × 301 = 45150. Therefore, the interval sum is 61425 − 45150 = 16275, so option B is correct. Subtracting S₃₀₀ is necessary because the range begins at 301. Using S₃₅₀ alone would include numbers 1 through 300 and give an incorrect result; the other options reflect nearby arithmetic errors.

Use the partial-sum difference rule for a consecutive range: the sum from a to b equals S_b − Sₐ₋₁, with Sₖ = k(k + 1)/2. Since the range is 251 through 300, calculate S₃₀₀ − S₂₅₀. We have S₃₀₀ = 300 × 301 / 2 = 150 × 301 = 45150, and S₂₅₀ = 250 × 251 / 2 = 125 × 251 = 31375. Subtracting gives 45150 − 31375 = 13775. Thus option B is correct. This method excludes all numbers through 250 and retains exactly 251 to 300. Adding the 50 terms directly would also work, but the partial-sum method is quicker and less error-prone for a large interval. The neighboring options result from arithmetic mistakes.