Even numbers from 1 to 100: 2, 4, 6, ..., 100. This is an AP with first term 2, last term 100, common difference 2. Number of terms = (100-2)/2 + 1 = 50.

We need a number of form LCM(12,15,20) + 7. LCM(12,15,20): 12=2²×3, 15=3×5, 20=2²×5. LCM = 2²×3×5 = 60. So the number = 60k + 7. For k=2: 120+7 = 127.

Let the integers be x and x+1. Sum = x + (x+1) = 51. 2x + 1 = 51. 2x = 50. x = 25. The larger integer = 26.

Using formula for sum of first n natural numbers: S = n(n+1)/2 where n=99. S = 99(100)/2 = 9900/2 = 4950.

Let the number be 7k + 3. When divided by 14: if k is even (k=2m), number = 14m + 3 (remainder 3); if k is odd (k=2m+1), number = 14m + 10 (remainder 10). So remainder is either 3 or 10.

For divisibility by both 6 and 8, the number must be divisible by LCM(6,8) = 24. Check: 48 ÷ 24 = 2 ✓. Option A is correct.

Using the formula: HCF × LCM = Product of two numbers. Therefore, 13 × LCM = 2028. LCM = 2028 ÷ 13 = 156.

Let the number be x. According to problem: 3x - 4 = 17. Therefore, 3x = 21, x = 7.

Let the three consecutive odd numbers be (x-2), x, and (x+2). Their sum: (x-2) + x + (x+2) = 51, so 3x = 51, x = 17.

Let smaller number be x. Then x(x+12) = 189. x² + 12x - 189 = 0. Factoring: (x+21)(x-9) = 0. Since x must be positive, x = 9.