Let `S_n` denote sum of first n-terms of an A.P where `S_(10) =530 , S_(5) =140` then find `S_(20)` -`S_(6)`

Options

A. 1872
B. 1842
C. 1852
D. 1862

Answer

To solve the problem, we need to find \( S_{20} - S_{6} \) given that \( S_{10} = 530 \) and \( S_{5} = 140 \).

### Step 1: Use the formula for the sum of the first \( n \) terms of an A.P.
The sum of the first \( n \) terms of an arithmetic progression (A.P.) is given by the formula:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Write equations for \( S_{10} \) and \( S_{5} \)
Using the formula, we can write:
1. For \( S_{10} \):
\[ S_{10} = \frac{10}{2} \left(2a + (10-1)d\right) = 5(2a + 9d) = 530 \]
Simplifying this gives:
\[ 2a + 9d = \frac{530}{5} = 106 \quad \text{(Equation 1)} \]

2. For \( S_{5} \):
\[ S_{5} = \frac{5}{2} \left(2a + (5-1)d\right) = \frac{5}{2}(2a + 4d) = 140 \]
Simplifying this gives:
\[ 2a + 4d = \frac{140 \times 2}{5} = 56 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have the two equations:
1. \( 2a + 9d = 106 \)
2. \( 2a + 4d = 56 \)

We can subtract Equation 2 from Equation 1:
\[ (2a + 9d) - (2a + 4d) = 106 - 56 \]
This simplifies to:
\[ 5d = 50 \implies d = 10 \]

### Step 4: Substitute \( d \) back to find \( a \)
Now substitute \( d = 10 \) back into Equation 2:
\[ 2a + 4(10) = 56 \]
This simplifies to:
\[ 2a + 40 = 56 \implies 2a = 16 \implies a = 8 \]

### Step 5: Calculate \( S_{20} \) and \( S_{6} \)
Now we can find \( S_{20} \) and \( S_{6} \).

1. For \( S_{20} \):
\[ S_{20} = \frac{20}{2} \left(2a + (20-1)d\right) = 10(2 \cdot 8 + 19 \cdot 10) = 10(16 + 190) = 10 \cdot 206 = 2060 \]

2. For \( S_{6} \):
\[ S_{6} = \frac{6}{2} \left(2a + (6-1)d\right) = 3(2 \cdot 8 + 5 \cdot 10) = 3(16 + 50) = 3 \cdot 66 = 198 \]

### Step 6: Find \( S_{20} - S_{6} \)
Now we can find the final result:
\[ S_{20} - S_{6} = 2060 - 198 = 1862 \]

Thus, the final answer is:
\[ \boxed{1862} \]