The number of terms in the series `1+3+5+7+…….+73+75`

Options

A. 28
B. 30
C. 36
D. 38

Answer

To find the number of terms in the series \(1 + 3 + 5 + 7 + \ldots + 73 + 75\), we can recognize that this series is an arithmetic progression (AP) where:

- The first term \(A = 1\)
- The common difference \(D = 2\)

We can use the formula for the \(n\)-th term of an arithmetic progression, which is given by:

\[ A_n = A + (n - 1) \cdot D \]

In our case, we want to find \(n\) such that \(A_n = 75\). Thus, we set up the equation:

\[ 75 = 1 + (n - 1) \cdot 2 \]

Now, let's solve for \(n\):

1. Subtract 1 from both sides:

\[ 75 - 1 = (n - 1) \cdot 2 \]
\[ 74 = (n - 1) \cdot 2 \]

2. Divide both sides by 2:

\[ \frac{74}{2} = n - 1 \]
\[ 37 = n - 1 \]

3. Add 1 to both sides:

\[ n = 37 + 1 \]
\[ n = 38 \]

Thus, the number of terms in the series \(1 + 3 + 5 + 7 + \ldots + 75\) is \(38\).