Key Concepts and Formulas

• Sum of an Arithmetic Progression (AP): The sum of the first $n$ terms of an AP is given by $S_n = \frac{n}{2}[2a + (n-1)d]$, where $a$ is the first term and $d$ is the common difference.
• System of Linear Equations: A set of two or more linear equations that can be solved simultaneously to find the values of the variables.
• Sum of a Range of Terms: The sum of terms from the $(m+1)$-th term to the $n$-th term in an AP can be expressed as $S_n - S_m$.

Step-by-Step Solution

Step 1: Understand the Problem and Recall the Formula The problem asks us to find the difference between the sum of the first 30 terms and the sum of the first 10 terms ($S_{30} - S_{10}$) of an arithmetic progression, given the sums of the first 40 terms ($S_{40}$) and the first 12 terms ($S_{12}$). The fundamental formula for the sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}[2a + (n-1)d]$.

Step 2: Formulate Equations from the Given Information We are given $S_{40} = 1030$ and $S_{12} = 57$. We use the AP sum formula to translate these into equations involving the first term ($a$) and the common difference ($d$).

For $S_{40} = 1030$: $S_{40} = \frac{40}{2}[2a + (40-1)d] = 1030$ $20[2a + 39d] = 1030$ Dividing both sides by 20: $2a + 39d = \frac{1030}{20} = \frac{103}{2} \quad \text{..... (1)}$

For $S_{12} = 57$: $S_{12} = \frac{12}{2}[2a + (12-1)d] = 57$ $6[2a + 11d] = 57$ Dividing both sides by 6: $2a + 11d = \frac{57}{6} = \frac{19}{2} \quad \text{..... (2)}$

Step 3: Solve the System of Linear Equations for a and d We now have a system of two linear equations with two unknowns ($a$ and $d$). We can solve this system using the elimination method. Subtract equation (2) from equation (1):

$(2a + 39d) - (2a + 11d) = \frac{103}{2} - \frac{19}{2}$ $28d = \frac{103 - 19}{2}$ $28d = \frac{84}{2}$ $28d = 42$ Solving for $d$: $d = \frac{42}{28} = \frac{3 \times 14}{2 \times 14} = \frac{3}{2}$

Now, substitute the value of $d = \frac{3}{2}$ into equation (2) to find $a$: $2a + 11\left(\frac{3}{2}\right) = \frac{19}{2}$ $2a + \frac{33}{2} = \frac{19}{2}$ $2a = \frac{19}{2} - \frac{33}{2}$ $2a = \frac{19 - 33}{2}$ $2a = \frac{-14}{2}$ $2a = -7$ Solving for $a$: $a = -\frac{7}{2}$

Step 4: Calculate $S_{30} - S_{10}$ We need to find the value of $S_{30} - S_{10}$. We can express this directly using the AP sum formula for $S_{30}$ and $S_{10}$.

$S_{30} = \frac{30}{2}[2a + (30-1)d] = 15[2a + 29d]$ $S_{10} = \frac{10}{2}[2a + (10-1)d] = 5[2a + 9d]$

Now, find the difference: $S_{30} - S_{10} = 15[2a + 29d] - 5[2a + 9d]$ Distribute the coefficients: $S_{30} - S_{10} = (30a + 435d) - (10a + 45d)$ Combine like terms: $S_{30} - S_{10} = (30a - 10a) + (435d - 45d)$ $S_{30} - S_{10} = 20a + 390d$

Step 5: Substitute the Values of a and d Substitute the calculated values of $a = -\frac{7}{2}$ and $d = \frac{3}{2}$ into the expression for $S_{30} - S_{10}$:

$S_{30} - S_{10} = 20\left(-\frac{7}{2}\right) + 390\left(\frac{3}{2}\right)$ $S_{30} - S_{10} = \left(\frac{20}{2}\right)(-7) + \left(\frac{390}{2}\right)(3)$ $S_{30} - S_{10} = 10(-7) + 195(3)$ $S_{30} - S_{10} = -70 + 585$ $S_{30} - S_{10} = 515$

Common Mistakes & Tips

• Algebraic Accuracy: Carefully check all arithmetic calculations, especially when dealing with fractions and solving the system of equations. A small error in finding $a$ or $d$ will lead to an incorrect final answer.
• Formula Application: Ensure the correct formula for $S_n$ is used and applied consistently.
• Interpreting $S_{30} - S_{10}$: Remember that $S_{30} - S_{10}$ represents the sum of the terms from the 11th term to the 30th term, inclusive. This can sometimes be used as an alternative method by forming an AP of these 20 terms.

Summary The problem was solved by first setting up a system of two linear equations using the given information about the sums of terms in an arithmetic progression. By solving these equations, we found the first term ($a$) and the common difference ($d$). Subsequently, we used these values to calculate the required difference $S_{30} - S_{10}$ by first deriving a general expression for it in terms of $a$ and $d$, and then substituting the found values. The calculation resulted in $515$.

The final answer is $\boxed{515}$, which corresponds to option (D).