Question:

Step 1: Find the sum $S_A$ by identifying it as an arithmetic progression. $S_A$ is the sum of consecutive integers from $2m$ to $4m$ with common difference $1$. The number of terms is: $$n_A = 4m - 2m + 1 = 2m + 1$$ Using the arithmetic progression sum formula $S = \frac{n(\text{first term} + \text{last term})}{2}$: $$S_A = \frac{(2m+1)(2m + 4m)}{2} = \frac{(2m+1)(6m)}{2} = 3m(2m+1)$$ Step 2: Find the sum $S_B$ by identifying it as an arithmetic progression of odd numbers. $S_B$ is the sum of odd numbers from $2m+1$ to $4m-1$ with common difference $2$. The number of terms is: $$n_B = \frac{(4m-1)-(2m+1)}{2} + 1 = \frac{2m-2}{2} + 1 = m - 1 + 1 = m$$ Using the arithmetic progression sum formula: $$S_B = \frac{m(2m+1 + 4m-1)}{2} = \frac{m(6m)}{2} = 3m^2$$ Step 3: Calculate the ratio $\frac{S_A}{S_B}$ and express it in the form $k + \frac{1}{l}$. $$\frac{S_A}{S_B} = \frac{3m(2m+1)}{3m^2} = \frac{2m+1}{m}$$ Dividing the numerator by the denominator: $$\frac{2m+1}{m} = \frac{2m + 1}{m} = 2 + \frac{1}{m}$$ Step 4: Identify the values of $k$ and $l$, then find $k + l$. From the expression $\frac{S_A}{S_B} = 2 + \frac{1}{m}$, we can match it with the form $k + \frac{1}{l}$: $$k = 2 \quad \text{and} \quad l = m$$ Therefore: $$k + l = 2 + m$$ The answer is **Option 1: $2 + m$**