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: Problems involving two unknown variables (like $a$ and $d$) often require solving a system of two linear equations.

Step-by-Step Solution

Step 1: Formulate equations using the given sums. We are given $S_{10} = 530$ and $S_5 = 140$. We will use the formula for the sum of an AP to create two equations with the first term ($a$) and the common difference ($d$).

• For $S_{10} = 530$: $S_{10} = \frac{10}{2}[2a + (10-1)d]$ $530 = 5[2a + 9d]$ Divide by 5: $106 = 2a + 9d \quad \text{(Equation 1)}$ This equation relates $a$ and $d$ based on the sum of the first 10 terms.

• For $S_5 = 140$: $S_5 = \frac{5}{2}[2a + (5-1)d]$ $140 = \frac{5}{2}[2a + 4d]$ Multiply by $\frac{2}{5}$: $140 \times \frac{2}{5} = 2a + 4d$ $56 = 2a + 4d$ Divide by 2: $28 = a + 2d \quad \text{(Equation 2)}$ This equation relates $a$ and $d$ based on the sum of the first 5 terms.

Step 2: Solve the system of linear equations for $a$ and $d$. We have the system:

1. $2a + 9d = 106$
2. $a + 2d = 28$

We can use the elimination method. Multiply Equation 2 by 2: $2(a + 2d) = 2(28)$ $2a + 4d = 56 \quad \text{(Equation 3)}$

Now, subtract Equation 3 from Equation 1: $(2a + 9d) - (2a + 4d) = 106 - 56$ $5d = 50$ $d = \frac{50}{5}$ $d = 10$

Substitute $d=10$ into Equation 2: $a + 2(10) = 28$ $a + 20 = 28$ $a = 28 - 20$ $a = 8$ We have found that the first term $a=8$ and the common difference $d=10$.

Step 3: Calculate $S_{20}$ and $S_6$ using the found values of $a$ and $d$. Now we apply the sum formula with $a=8$ and $d=10$ for $n=20$ and $n=6$.

• Calculate $S_{20}$: $S_{20} = \frac{20}{2}[2(8) + (20-1)10]$ $S_{20} = 10[16 + (19)10]$ $S_{20} = 10[16 + 190]$ $S_{20} = 10[206]$ $S_{20} = 2060$

• Calculate $S_6$: $S_6 = \frac{6}{2}[2(8) + (6-1)10]$ $S_6 = 3[16 + (5)10]$ $S_6 = 3[16 + 50]$ $S_6 = 3[66]$ $S_6 = 198$

Step 4: Compute the required value $S_{20} - S_6$. $S_{20} - S_6 = 2060 - 198$ $S_{20} - S_6 = 1862$

Common Mistakes & Tips

• Algebraic Errors: Be meticulous when solving the system of linear equations. Small arithmetic mistakes can lead to incorrect values for $a$ and $d$.
• Formula Misapplication: Ensure you use the correct formula for $S_n$ and substitute the values of $n$, $a$, and $d$ accurately.
• Order of Operations: Follow the order of operations (PEMDAS/BODMAS) carefully when calculating the sums to avoid errors.

Summary The problem requires us to find the difference between the sum of the first 20 terms and the sum of the first 6 terms of an arithmetic progression. We first used the given information ($S_{10}=530$ and $S_5=140$) to set up and solve a system of two linear equations for the first term ($a$) and the common difference ($d$). Once $a$ and $d$ were determined, we calculated $S_{20}$ and $S_6$ using the sum formula and then found their difference.

The final answer is $\boxed{1862}$.