Finding the sum of the first $n$ terms of an AP

The governing concept is the sum of the first n terms of an arithmetic progression. The first n even natural numbers are 2, 4, 6, ..., 2n. Factoring out 2 gives their sum as 2(1+2+...+n). Since 1+2+...+n = n(n+1)/2, the required sum is n(n+1). Therefore n(n+1)=420. We need two consecutive positive integers whose product is 420; 20×21=420, so n=20. A direct check also confirms that the twentieth even number is 40 and the sum of the first 20 even numbers is 20×21=420. Thus option C is correct. Options A, B, and D give 342, 380, and 462 respectively, so none of them satisfies the condition.

Direct answer: 528, so option B is correct. The first 32 natural numbers are 1, 2, 3, ..., 32. Their sum can be found with Sₙ = n(n + 1)/2, where n is the last natural number when the sequence begins at 1. Thus S₃₂ = 32×33/2 = 16×33 = 528. A second explanation uses pairing: pair the first and last terms, 1 + 32 = 33; then 2 + 31 = 33, and so on. There are 16 pairs, so the total is 16×33 = 528. Option A, 520, results from an arithmetic or formula error. Option C, 536, and option D, 544, likewise do not agree with the formula; one common mistake is using the wrong neighbouring number or forgetting the division by 2. The phrase ‘first 32 natural numbers’ starts at 1 here, so zero is not included. Memory cue: the sum 1 through n is n(n+1)/2.