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:
- 390
- 395
- 405
- 410
The Correct Option is B
Approach Solution - 1
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
Approach Solution -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.
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