Question

In an arithmetic progression, if \( S_{40} = 1030 \) and \( S_{12} = 57 \), then \( S_{30} - S_{10} \) is equal to:

• A. \( 515 \)
• B. \( 525 \)
• C. \( 510 \)
• D. \( 505 \)

Hint:

For arithmetic progressions, use the sum formula \( S_n = \frac{n}{2} \left( 2a + (n - 1)d \right) \) to relate the sum of terms to the first term and common difference. Solve for the unknowns and calculate the desired sum.

Answer 1

The Correct Option is A

The sum of the first \( n \) terms of an arithmetic progression is \( S_n = \frac{n}{2} (2a + (n - 1) d) \), where \( a \) is the first term and \( d \) is the common difference. Given \( S_{40} = 1030 \) and \( S_{12} = 57 \), we can determine \( a \) and \( d \). Subsequently, we compute \( S_{30} - S_{10} \) using the same formula.
Final Answer: \( S_{30} - S_{10} = 515 \).