Question

Find the sum of the series \( 1 + 3 + 5 + \ldots + 99 \).

• A. \( 2500 \)
• B. \( 2400 \)
• C. \( 2600 \)
• D. 2300

Hint:

For an AP, use the formula \( S_n = \frac{n}{2}(a + l) \) to find the sum.

Answer 1

The Correct Option is A

The given series is an arithmetic series with the first term \( a = 1 \) and the last term \( l = 99 \). The common difference \( d \) is \( 2 \).

To find the sum of the series, we first need to determine the number of terms \( n \) in the series using the formula for the \( n \)-th term of an arithmetic series:

n-th term: \( a_n = a + (n-1)d \)

Setting \( a_n = 99 \), we get:

\( 99 = 1 + (n-1) \cdot 2 \)

\( 99 = 1 + 2n - 2 \)

\( 99 = 2n - 1 \)

\( 100 = 2n \)

\( n = 50 \)

The number of terms is \( 50 \). The sum \( S_n \) of an arithmetic series is given by:

\( S_n = \frac{n}{2} \times (a + l) \)

Substituting the values, we have:

\( S_{50} = \frac{50}{2} \times (1 + 99) \)

\( S_{50} = 25 \times 100 \)

\( S_{50} = 2500 \)

Therefore, the sum of the series \( 1 + 3 + 5 + \ldots + 99 \) is \( 2500 \).