Question

If \( a_1, a_2, a_3, \dots \) is an arithmetic progression with the common difference of 1 and \( a_2 + a_4 + a_6 + \dots + a_{98} = 93 \), then \( \sum_{i=1}^{98} a_i \) is equal to \( k \). The sum of the digits of \( k \) is:

• A. 9
• B. 11
• C. 12
• D. 13

Hint:

For arithmetic progressions, use the sum formula and remember that the sum of an even number of terms can be simplified by pairing terms symmetrically.

Answer 1

The Correct Option is C

Given the arithmetic progression, the sum of the terms is given by the formula \[ S = \frac{n}{2} (2a + (n - 1)d) \] where \( a \) is the first term and \( d \) is the common difference. First, calculate the sum of the terms using the given condition for \( a_2 + a_4 + a_6 + \dots + a_{98} \), and then sum the first 98 terms to find \( k \).