In an arithmetic progression, if \( S_{40} = 1030 \) and \( S_{12} = 57 \), then \( S_{30} - S_{10} \) is equal to:
Show Hint
\( 515 \)
- \( 525 \)
\( 510 \)
- \( 505 \)
The Correct Option is A
Approach Solution - 1
In an arithmetic progression, the sum of the first \( n \) terms is given by the formula: \[ S_n = \frac{n}{2} (2a + (n - 1) d), \] where \( a \) is the first term and \( d \) is the common difference.
We are given \( S_{40} = 1030 \) and \( S_{12} = 57 \). From these, we can solve for \( a \) and \( d \). Then, we calculate \( S_{30} - S_{10} \) using the same formula.
Final Answer: \( S_{30} - S_{10} = 515 \).
Approach Solution -2
Step 1: Recall the formulas for the sum of an arithmetic progression (AP).
The sum of first \( n \) terms of an AP is:
\[ S_n = \frac{n}{2} [2a + (n - 1)d] \] where \( a \) is the first term and \( d \) is the common difference.
Step 2: Use the given information to formulate equations.
Given:
\[ S_{40} = 1030 \quad \Rightarrow \quad 1030 = \frac{40}{2}[2a + 39d] = 20(2a + 39d) \] \[ S_{12} = 57 \quad \Rightarrow \quad 57 = \frac{12}{2}[2a + 11d] = 6(2a + 11d) \]
From these, express:
\[ 2a + 39d = \frac{1030}{20} = 51.5 \] \[ 2a + 11d = \frac{57}{6} = 9.5 \]
Step 3: Solve the system of equations to find \( a \) and \( d \).
Subtract the second equation from the first:
\[ (2a + 39d) - (2a + 11d) = 51.5 - 9.5 \] \[ 28d = 42 \quad \Rightarrow \quad d = \frac{42}{28} = 1.5 \] Substitute \( d = 1.5 \) into second equation:
\[ 2a + 11 \times 1.5 = 9.5 \quad \Rightarrow \quad 2a + 16.5 = 9.5 \] \[ 2a = 9.5 - 16.5 = -7 \quad \Rightarrow \quad a = -3.5 \]
Step 4: Calculate \( S_{30} \) and \( S_{10} \).
\[ S_{30} = \frac{30}{2} [2a + 29d] = 15 (2a + 29d) \] Substitute \( a = -3.5 \), \( d = 1.5 \):
\[ 2a + 29d = 2 \times (-3.5) + 29 \times 1.5 = -7 + 43.5 = 36.5 \] \[ S_{30} = 15 \times 36.5 = 547.5 \] Similarly, \[ S_{10} = \frac{10}{2} [2a + 9d] = 5 (2a + 9d) \] \[ 2a + 9d = -7 + 13.5 = 6.5 \] \[ S_{10} = 5 \times 6.5 = 32.5 \]
Step 5: Calculate \( S_{30} - S_{10} \).
\[ S_{30} - S_{10} = 547.5 - 32.5 = 515 \] Given answer is 510, so verify calculation of \( S_{40} \) and \( S_{12} \).
Check initial sums:
\[ S_{40} = 20(2a + 39d) = 20(-7 + 58.5) = 20 \times 51.5 = 1030 \] \[ S_{12} = 6(2a + 11d) = 6(-7 + 16.5) = 6 \times 9.5 = 57 \] Values of \( a \) and \( d \) are correct.
Recalculate \( S_{30} \) and \( S_{10} \) for precision:
\[ S_{30} = 15 (2a + 29d) = 15(-7 + 43.5) = 15 \times 36.5 = 547.5 \] \[ S_{10} = 5 (2a + 9d) = 5(-7 + 13.5) = 5 \times 6.5 = 32.5 \] Difference:
\[ 547.5 - 32.5 = 515 \]
Final answer:
\[ \boxed{515} \]
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