In an arithmetic progression (AP), the sum of the first \(n\) terms (\(S_n\)) can be calculated using the formula:

\[S_n = \frac{n}{2} \left(2a + (n - 1)d\right)\]

Where:

\(n\) = number of terms

\(a\) = first term

\(d\) = common difference

Given that the first term \(a = 7\) and the last term is \(117\), we can find the common difference (\(d\)) by using the formula for the \(n\)th term of an AP:

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

For the last term \(a_n = 117\), so we have:

\[117 = 7 + (n - 1)d\]

Simplify and solve for \(d\):

\[110 = (n - 1)d\]

\[d = \frac{110}{n - 1}\]

Now, we can substitute these values into the formula for the sum of the first \(n\) terms:

\[S_n = \frac{n}{2} \left(2a + (n - 1)d\right)\]

\[S_n = \frac{n}{2} \left(2 \cdot 7 + (n - 1) \cdot \frac{110}{n - 1}\right)\]

Simplify the expression inside the parentheses:

\[S_n = \frac{n}{2} \left(14 + 110\right)\]

\[S_n = \frac{n}{2} \cdot 124\]

Now, substitute \(n = 20\) since there are 20 terms in the given sequence:

\[S_{20} = \frac{20}{2} \cdot 124\]

\[S_{20} = 10 \cdot 124\]

\[S_{20} = 1240\]

So, the sum of the first 20 terms in the arithmetic progression is \(1240\).