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

To find the sum of the series \(1 + 3 + 5 + \ldots + 99\), we observe that this is an arithmetic series where the first term \(a = 1\) and the common difference \(d = 2\). The general form of an arithmetic series is expressed as:

\(S_n = \frac{n}{2} \times (2a + (n-1)d)\)

First, we need to determine the number of terms \(n\). For an arithmetic series, the \(n\)-th term is given by:

\(a_n = a + (n-1)d\)

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

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

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

\(98 = 2(n-1)\)

\(49 = n-1\)

\(n = 50\)

Now, substitute \(n = 50\), \(a = 1\), and \(d = 2\) into the formula for the sum \(S_n\):

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

\(S_{50} = 25 \times (2 + 98)\)

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

\(S_{50} = 2500\)

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

Answer 2

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 \( 2500 \).