If `S_n` denotes sum of first n terms of an A.P. such that, `S_(20) = 790, S_(10) = 145` then `S_(15) - S_5`

Options

A. 540
B. 395
C. 355
D. 475

Answer

To solve the problem, we need to find \( S_{15} - S_5 \) given that \( S_{20} = 790 \) and \( S_{10} = 145 \).

1. **Understanding 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:
\[ S_n = \frac{n}{2} \left(2A + (n-1)D\right) \]
where \( A \) is the first term and \( D \) is the common difference.

2. **Setting up the equations**:
- For \( n = 20 \):
\[ S_{20} = \frac{20}{2} \left(2A + 19D\right) = 790 \]
Simplifying this:
\[ 10(2A + 19D) = 790 \implies 2A + 19D = 79 \quad \text{(Equation 1)} \]

- For \( n = 10 \):
\[ S_{10} = \frac{10}{2} \left(2A + 9D\right) = 145 \]
Simplifying this:
\[ 5(2A + 9D) = 145 \implies 2A + 9D = 29 \quad \text{(Equation 2)} \]

3. **Solving the equations**:
Now we have two equations:
- \( 2A + 19D = 79 \) (Equation 1)
- \( 2A + 9D = 29 \) (Equation 2)

We can subtract Equation 2 from Equation 1:
\[ (2A + 19D) - (2A + 9D) = 79 - 29 \]
This simplifies to:
\[ 10D = 50 \implies D = 5 \]

4. **Finding \( A \)**:
Substitute \( D = 5 \) back into Equation 2:
\[ 2A + 9(5) = 29 \implies 2A + 45 = 29 \implies 2A = 29 - 45 \implies 2A = -16 \implies A = -8 \]

5. **Calculating \( S_{15} \) and \( S_5 \)**:
Now we can find \( S_{15} \) and \( S_5 \):
- For \( S_{15} \):
\[ S_{15} = \frac{15}{2} \left(2A + 14D\right) = \frac{15}{2} \left(2(-8) + 14(5)\right) \]
\[ = \frac{15}{2} \left(-16 + 70\right) = \frac{15}{2} \times 54 = 15 \times 27 = 405 \]

- For \( S_5 \):
\[ S_5 = \frac{5}{2} \left(2A + 4D\right) = \frac{5}{2} \left(2(-8) + 4(5)\right) \]
\[ = \frac{5}{2} \left(-16 + 20\right) = \frac{5}{2} \times 4 = 5 \times 2 = 10 \]

6. **Finding \( S_{15} - S_5 \)**:
Now, we can calculate:
\[ S_{15} - S_5 = 405 - 10 = 395 \]

Thus, the final answer is:
\[ \boxed{395} \]