Key Concepts and Formulas

• Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant.
  • $n$-th term: $T_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.
  • Sum of the first $n$ terms: $S_n = \frac{n}{2}(2a + (n-1)d)$.
• Difference of Sums: $S_m - S_n$ represents the sum of terms from $(n+1)$ to $m$, i.e., $T_{n+1} + T_{n+2} + \dots + T_m$.

Step-by-Step Solution

Step 1: Translate the given information into algebraic equations. We are given $S_{10} = 390$ and $\frac{T_{10}}{T_5} = \frac{15}{7}$. We will use the formulas for $S_n$ and $T_n$ to form equations involving the first term ($a$) and the common difference ($d$).

• Using $S_{10} = 390$: $S_{10} = \frac{10}{2}(2a + (10-1)d)$ $390 = 5(2a + 9d)$ Dividing by 5, we get: $78 = 2a + 9d \quad \text{.........(Equation 1)}$

• Using $\frac{T_{10}}{T_5} = \frac{15}{7}$: We know $T_{10} = a + (10-1)d = a + 9d$ and $T_5 = a + (5-1)d = a + 4d$. $\frac{a + 9d}{a + 4d} = \frac{15}{7}$ Cross-multiplying gives: $7(a + 9d) = 15(a + 4d)$ $7a + 63d = 15a + 60d$ Rearranging terms to one side: $63d - 60d = 15a - 7a$ $3d = 8a \quad \text{.........(Equation 2)}$

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

1. $2a + 9d = 78$
2. $3d = 8a$

From Equation 2, we can express $d$ in terms of $a$: $d = \frac{8}{3}a$. Substitute this expression for $d$ into Equation 1: $2a + 9\left(\frac{8}{3}a\right) = 78$ $2a + 3 \cdot 8a = 78$ $2a + 24a = 78$ $26a = 78$ $a = \frac{78}{26} = 3$ Now, substitute the value of $a$ back into the expression for $d$: $d = \frac{8}{3}a = \frac{8}{3}(3) = 8$ So, the first term is $a=3$ and the common difference is $d=8$.

Step 3: Calculate $S_{15}$ and $S_5$ using the found values of $a$ and $d$. We need to find $S_{15} - S_5$. Let's calculate each sum individually.

• Calculating $S_{15}$: $S_{15} = \frac{15}{2}(2a + (15-1)d)$ $S_{15} = \frac{15}{2}(2(3) + 14(8))$ $S_{15} = \frac{15}{2}(6 + 112)$ $S_{15} = \frac{15}{2}(118)$ $S_{15} = 15 \times 59 = 885$

• Calculating $S_5$: $S_5 = \frac{5}{2}(2a + (5-1)d)$ $S_5 = \frac{5}{2}(2(3) + 4(8))$ $S_5 = \frac{5}{2}(6 + 32)$ $S_5 = \frac{5}{2}(38)$ $S_5 = 5 \times 19 = 95$

Step 4: Compute the difference $S_{15} - S_5$. $S_{15} - S_5 = 885 - 95$ $S_{15} - S_5 = 790$

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful when solving the system of equations and simplifying expressions. A small mistake in calculation can lead to an incorrect final answer.
• Formula Misapplication: Ensure you are using the correct formulas for the $n$-th term and the sum of terms. The formulas for $S_n$ involving $a$ and $d$ are crucial here.
• Understanding the Question: The question asks for $S_{15} - S_5$. This is the sum of terms from the 6th term to the 15th term ($T_6 + T_7 + \dots + T_{15}$). While direct calculation of this sum is possible, calculating $S_{15}$ and $S_5$ separately and then subtracting is often more straightforward.

Summary

The problem required us to find the first term ($a$) and common difference ($d$) of an arithmetic progression using the given sum of the first 10 terms and the ratio of the 10th to the 5th term. We translated these conditions into two linear equations, solved them to find $a=3$ and $d=8$. Finally, we used these values to compute $S_{15}$ and $S_5$ and found their difference to be 790.

The final answer is $\boxed{790}$, which corresponds to option (C).