Question

Let \( S_n \) denote the sum of the first \( n \) terms of an arithmetic progression. If \( S_{20} = 790 \) and \( S_{10} = 145 \), then \( S_{15} - S_5 \) is:

• A. 390
• B. 395
• C. 405
• D. 410

Answer 1

The Correct Option is B

The sum of the first \( n \) terms in an arithmetic progression (AP) is given by: \[ S_n = \frac{n}{2} [2a + (n - 1)d] \] where \( a \) is the first term and \( d \) is the common difference.

• Using \( S_{20} = 790 \): \[ S_{20} = \frac{20}{2} [2a + 19d] = 790 \] Simplifying, we get: \[ 10[2a + 19d] = 790 \Rightarrow 2a + 19d = 79 \quad \text{(Equation 1)} \]
• Using \( S_{10} = 145 \): \[ S_{10} = \frac{10}{2} [2a + 9d] = 145 \] Simplifying, we get: \[ 5[2a + 9d] = 145 \Rightarrow 2a +9d = 29 \quad \text{(Equation 2)} \]
• Solving for \( a \) and \( d \): Subtract Equation 2 from Equation 1: \[ (2a + 19d) - (2a + 9d) = 79 - 29 \] \[ 10d = 50\Rightarrow d = 5 \] Substitute \( d = 5 \) back into Equation 2: \[ 2a + 9 \times 5 = 29 \] \[ 2a + 45 = 29 \Rightarrow 2a = -16\Rightarrow a = -8 \]
• Calculating \( S_{15} \) and \( S_{5} \):

Finding \( S_{15} - S_{5} \): \[ S_{15} - S_{5} = 405 - 10 = 395 \]

Final Answer: (2) 395

Answer 2

To solve this problem, we need to find \( S_{15} - S_5 \) for an arithmetic progression (AP) given that \( S_{20} = 790 \) and \( S_{10} = 145 \). Let's denote the first term of the AP as \( a \) and the common difference as \( d \).

The formula for the sum of the first \( n \) terms of an AP is given by:

\(S_n = \frac{n}{2} (2a + (n - 1)d)\)

Given:

• \(S_{10} = 145\)
• \(S_{20} = 790\)

Substituting into the formula for \( S_{10} \):

\(\frac{10}{2} (2a + 9d) = 145\)

\(5(2a + 9d) = 145\)

\(2a + 9d = 29 \quad \text{(Equation 1)}\)

Substituting into the formula for \( S_{20} \):

\(\frac{20}{2} (2a + 19d) = 790\)

\(10(2a + 19d) = 790\)

\(2a + 19d = 79 \quad \text{(Equation 2)}\)

Now, we solve the two simultaneous equations:

• Equation 1: \(2a + 9d = 29\)
• Equation 2: \(2a + 19d = 79\)

Subtract Equation 1 from Equation 2:

\((2a + 19d) - (2a + 9d) = 79 - 29\)

\(10d = 50\)

\(d = 5\)

Substitute \( d = 5 \) in Equation 1:

\(2a + 9 \times 5 = 29\)

\(2a + 45 = 29\)

\(2a = -16\)

\(a = -8\)

Now, let's calculate \( S_{15} \) and \( S_5 \) using the values of \( a \) and \( d \):

For \( S_{15} \):

\(S_{15} = \frac{15}{2} (2 \times (-8) + (15-1) \times 5)\)

\(S_{15} = \frac{15}{2} (-16 + 70)\)

\(S_{15} = \frac{15}{2} \times 54\)

\(S_{15} = 15 \times 27 = 405\)

For \( S_5 \):

\(S_5 = \frac{5}{2} (2 \times (-8) + (5-1) \times 5)\)

\(S_5 = \frac{5}{2} (-16 + 20)\)

\(S_5 = \frac{5}{2} \times 4\)

\(S_5 = 10\)

Finally, find \( S_{15} - S_5 \):

\(S_{15} - S_5 = 405 - 10 = 395\)

Therefore, the answer is 395.