Key Concepts and Formulas

• Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant. This constant difference is called the common difference ($d$), and the first term is denoted by $a$.
• Sum of the first $n$ terms of an AP: The sum of the first $n$ terms of an AP, $S_n$, is given by the formula: $S_n = \frac{n}{2}[2a + (n-1)d]$

Step-by-Step Solution

Step 1: Formulate equations using the given sums.

We are given $S_4 = 16$ and $S_6 = -48$. We will use the formula for $S_n$ to create two equations involving the first term ($a$) and the common difference ($d$).

• For $S_4 = 16$: Substituting $n=4$ into the formula $S_n = \frac{n}{2}[2a + (n-1)d]$: $16 = \frac{4}{2}[2a + (4-1)d]$ $16 = 2[2a + 3d]$ Dividing both sides by 2: $8 = 2a + 3d \quad (\text{Equation 1})$ This equation represents the relationship between $a$ and $d$ based on the sum of the first 4 terms.

• For $S_6 = -48$: Substituting $n=6$ into the formula $S_n = \frac{n}{2}[2a + (n-1)d]$: $-48 = \frac{6}{2}[2a + (6-1)d]$ $-48 = 3[2a + 5d]$ Dividing both sides by 3: $-16 = 2a + 5d \quad (\text{Equation 2})$ This equation represents the relationship between $a$ and $d$ based on the sum of the first 6 terms.

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

We now have a system of two linear equations:

1. $2a + 3d = 8$
2. $2a + 5d = -16$

To solve this system, we can use the elimination method. Subtract Equation 1 from Equation 2 to eliminate $a$: $(2a + 5d) - (2a + 3d) = -16 - 8$ $2a + 5d - 2a - 3d = -24$ $2d = -24$ Dividing by 2 to find $d$: $d = -12$ Now substitute the value of $d = -12$ into Equation 1 to find $a$: $2a + 3(-12) = 8$ $2a - 36 = 8$ Add 36 to both sides: $2a = 8 + 36$ $2a = 44$ Dividing by 2 to find $a$: $a = 22$ So, the first term of the AP is $a=22$ and the common difference is $d=-12$.

Step 3: Calculate $S_{10}$ using the determined values of $a$ and $d$.

We need to find the sum of the first 10 terms, $S_{10}$. We use the formula $S_n = \frac{n}{2}[2a + (n-1)d]$ with $n=10$, $a=22$, and $d=-12$. $S_{10} = \frac{10}{2}[2(22) + (10-1)(-12)]$ $S_{10} = 5[44 + (9)(-12)]$ $S_{10} = 5[44 - 108]$ $S_{10} = 5[-64]$ $S_{10} = -320$

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with arithmetic and algebraic manipulations, especially when dealing with negative numbers. A sign error can drastically alter the result.
• Formula Application: Ensure you correctly substitute values into the $S_n$ formula, particularly for the $(n-1)$ term.
• Solving Linear Systems: When solving for $a$ and $d$, if the coefficients of $a$ or $d$ are not immediately the same, consider multiplying one or both equations by a constant to make them match before using elimination.

Summary

This problem requires us to find the sum of the first 10 terms of an arithmetic progression given the sums of the first 4 and first 6 terms. The strategy involves setting up a system of two linear equations in terms of the first term ($a$) and the common difference ($d$) using the given sum information and the AP sum formula. Solving this system yields the values of $a$ and $d$. Finally, these values are plugged back into the sum formula to calculate $S_{10}$. We found $a=22$ and $d=-12$, which led to $S_{10} = -320$.

The final answer is $\boxed{-320}$.