1. Stating the problem: Calculate the sum of all natural numbers: a) even numbers not greater than 250, b) two-digit numbers divisible by 4. 2. Formula and rules: The sum of an arithmetic sequence is given by: $$S_n = \frac{n}{2}(a_1 + a_n)$$ where $n$ is the number of terms, $a_1$ is the first term, and $a_n$ is the last term. 3. Part a) Sum of even natural numbers not greater than 250: - Even numbers start at 2 and go up to 250. - The sequence is: 2, 4, 6, ..., 250. - Number of terms $n = \frac{250}{2} = 125$. - First term $a_1 = 2$, last term $a_n = 250$. - Sum: $$S = \frac{125}{2}(2 + 250) = \frac{125}{2} \times 252 = 125 \times 126 = 15750$$ 4. Part b) Sum of two-digit numbers divisible by 4: - Two-digit numbers range from 10 to 99. - Find first two-digit number divisible by 4: 12. - Find last two-digit number divisible by 4: 96. - Sequence: 12, 16, 20, ..., 96. - Number of terms $n = \frac{96 - 12}{4} + 1 = \frac{84}{4} + 1 = 21 + 1 = 22$. - First term $a_1 = 12$, last term $a_n = 96$. - Sum: $$S = \frac{22}{2}(12 + 96) = 11 \times 108 = 1188$$ Final answers: a) Sum of even numbers not greater than 250 is $15750$. b) Sum of two-digit numbers divisible by 4 is $1188$.