Answer :
To find the sum of the first 47 terms of the arithmetic series that starts with 12 and increases by 4 each time, we can use the formula for the sum of an arithmetic series, which is:
\[ S_n = \frac{n}{2} \times (a_1 + a_n) \]
where:
- \( S_n \) is the sum of the first \( n \) terms of the series,
- \( n \) is the number of terms,
- \( a_1 \) is the first term, and
- \( a_n \) is the \( n \)th term.
Given in the problem:
- The first term (\( a_1 \)) is 12,
- The common difference (\( d \)) is 4, and
- The number of terms (\( n \)) is 47.
First, let's find the 47th term (\( a_{47} \)) of the series. We can use the formula for the \( n \)th term of an arithmetic series:
\[ a_n = a_1 + (n - 1)d \]
Substituting our known values:
\[ a_{47} = 12 + (47 - 1) \times 4 \]
\[ a_{47} = 12 + 46 \times 4 \]
\[ a_{47} = 12 + 184 \]
\[ a_{47} = 196 \]
Now, we can find the sum of the first 47 terms (\( S_{47} \)):
\[ S_{47} = \frac{47}{2} \times (12 + 196) \]
\[ S_{47} = \frac{47}{2} \times 208 \]
\[ S_{47} = 23.5 \times 208 \]
\[ S_{47} = 4892 \]
We don't need to round this answer to the nearest integer because it's already an integer. So, the sum of the first 47 terms of the series is 4892.
