Question:medium
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:
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:
Updated On: Jan 13, 2026
- 390
- 395
- 405
- 410
Show Solution
The Correct Option is B
Solution and Explanation
The sum of the first \( n \) terms of an arithmetic progression (AP) is defined as: \[ S_n = \frac{n}{2} [2a + (n - 1)d] \] where \( a \) represents the first term and \( d \) represents the common difference.
- Given \( S_{20} = 790 \): \[ S_{20} = \frac{20}{2} [2a + 19d] = 790 \] This simplifies to: \[ 10[2a + 19d] = 790 \Rightarrow 2a + 19d = 79 \quad \text{(Equation 1)} \]
- Given \( S_{10} = 145 \): \[ S_{10} = \frac{10}{2} [2a + 9d] = 145 \] This simplifies to: \[ 5[2a + 9d] = 145 \Rightarrow 2a +9d = 29 \quad \text{(Equation 2)} \]
- Determining \( a \) and \( d \): Subtract Equation 2 from Equation 1: \[ (2a + 19d) - (2a + 9d) = 79 - 29 \] This yields: \[ 10d = 50\Rightarrow d = 5 \] Substitute \( d = 5 \) into Equation 2: \[ 2a + 9 \times 5 = 29 \] \[ 2a + 45 = 29 \Rightarrow 2a = -16\Rightarrow a = -8 \]
- Calculating \( S_{15} \) and \( S_{5} \):
The difference \( S_{15} - S_{5} \) is: \[ S_{15} - S_{5} = 405 - 10 = 395 \]
Final Answer: (2) 395
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