Question

The sum of the first 10 terms of an arithmetic progression is 145, and the first term is 1. Find the common difference.

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

For an arithmetic progression, use the sum formula \( S_n = \frac{n}{2} [2a + (n-1)d] \) to solve for the common difference when given the sum and first term.

Answer 1

The Correct Option is B

The sum of the first \( n \) terms of an arithmetic progression is: \[ S_n = \frac{n}{2} [2a + (n-1)d] \] Given \( n = 10 \), \( S_{10} = 145 \), \( a = 1 \): \[ 145 = \frac{10}{2} [2 \cdot 1 + (10-1)d] = 5 [2 + 9d] \] \[ 145 = 10 + 45d \] \[ 45d = 135 \implies d = \frac{135}{45} = 3 \] Thus, the common difference is: \[ \boxed{3} \]