Let `S_(n)` denote the sum of the first `n` terms of an `A.P..` If `S_(4)=16` and `S_(6)=-48`, then `S_(10)` is equal to :

Options

A. `-260`
B. `-410`
C. `-320`
D. `-380`

Answer

To solve the problem, we need to find the sum of the first 10 terms of an arithmetic progression (A.P.) given that \( S_4 = 16 \) and \( S_6 = -48 \).

### Step 1: Write the formula for the sum of the first n terms of an A.P.
The sum of the first \( n \) terms of an A.P. can be expressed as:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Set up equations using the given sums
Using the formula, we can write:
1. For \( S_4 \):
\[ S_4 = \frac{4}{2} \times (2a + 3d) = 16 \]
This simplifies to:
\[ 2(2a + 3d) = 16 \implies 2a + 3d = 8 \quad \text{(Equation 1)} \]

2. For \( S_6 \):
\[ S_6 = \frac{6}{2} \times (2a + 5d) = -48 \]
This simplifies to:
\[ 3(2a + 5d) = -48 \implies 2a + 5d = -16 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have two equations:
1. \( 2a + 3d = 8 \) (Equation 1)
2. \( 2a + 5d = -16 \) (Equation 2)

We can subtract Equation 1 from Equation 2:
\[ (2a + 5d) - (2a + 3d) = -16 - 8 \]
This simplifies to:
\[ 2d = -24 \implies d = -12 \]

### Step 4: Substitute \( d \) back to find \( a \)
Now substitute \( d = -12 \) back into Equation 1:
\[ 2a + 3(-12) = 8 \]
This simplifies to:
\[ 2a - 36 = 8 \implies 2a = 44 \implies a = 22 \]

### Step 5: Find \( S_{10} \)
Now we can find \( S_{10} \):
\[ S_{10} = \frac{10}{2} \times (2a + 9d) \]
Substituting \( a = 22 \) and \( d = -12 \):
\[ S_{10} = 5 \times (2(22) + 9(-12)) \]
Calculating inside the parentheses:
\[ = 5 \times (44 - 108) = 5 \times (-64) = -320 \]

### Final Answer
Thus, \( S_{10} = -320 \).