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Sagot :
To determine the total number of students Mr. Wright has taught over 25 years, we need to sum the number of students he taught each year. We can use the given expression:
[tex]\[ \sum_{n=1}^{25} \left[ 18 + 2(n-1) \right] \][/tex]
This represents the number of students taught each year where he starts with 18 students in the first year, and the number of students increases by 2 each subsequent year.
The number of students Mr. Wright teaches in the [tex]\(n\)[/tex]-th year is:
[tex]\[ 18 + 2(n-1) \][/tex]
Therefore, the sequence of the number of students he teaches per year is:
[tex]\[ 18, 20, 22, \ldots \][/tex]
This sequence is an arithmetic series where the first term [tex]\(a\)[/tex] is 18 and the common difference [tex]\(d\)[/tex] is 2. The sum of the first [tex]\(n\)[/tex] terms of an arithmetic series can be expressed as:
[tex]\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \][/tex]
Here, [tex]\(n\)[/tex] is 25 (the number of years), [tex]\(a\)[/tex] is 18, and [tex]\(d\)[/tex] is 2. Plugging in these values, we get:
[tex]\[ S_{25} = \frac{25}{2} \times (2 \cdot 18 + (25-1) \cdot 2) \][/tex]
Simplify the expression inside the parentheses first:
[tex]\[ 2 \cdot 18 + (25-1) \cdot 2 = 36 + 48 = 84 \][/tex]
Now plug this back into the formula:
[tex]\[ S_{25} = \frac{25}{2} \times 84 \][/tex]
[tex]\[ S_{25} = 25 \times 42 \][/tex]
[tex]\[ S_{25} = 1050 \][/tex]
Therefore, the total number of students that Mr. Wright has taught over 25 years is:
[tex]\[ \boxed{1050} \][/tex]
[tex]\[ \sum_{n=1}^{25} \left[ 18 + 2(n-1) \right] \][/tex]
This represents the number of students taught each year where he starts with 18 students in the first year, and the number of students increases by 2 each subsequent year.
The number of students Mr. Wright teaches in the [tex]\(n\)[/tex]-th year is:
[tex]\[ 18 + 2(n-1) \][/tex]
Therefore, the sequence of the number of students he teaches per year is:
[tex]\[ 18, 20, 22, \ldots \][/tex]
This sequence is an arithmetic series where the first term [tex]\(a\)[/tex] is 18 and the common difference [tex]\(d\)[/tex] is 2. The sum of the first [tex]\(n\)[/tex] terms of an arithmetic series can be expressed as:
[tex]\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \][/tex]
Here, [tex]\(n\)[/tex] is 25 (the number of years), [tex]\(a\)[/tex] is 18, and [tex]\(d\)[/tex] is 2. Plugging in these values, we get:
[tex]\[ S_{25} = \frac{25}{2} \times (2 \cdot 18 + (25-1) \cdot 2) \][/tex]
Simplify the expression inside the parentheses first:
[tex]\[ 2 \cdot 18 + (25-1) \cdot 2 = 36 + 48 = 84 \][/tex]
Now plug this back into the formula:
[tex]\[ S_{25} = \frac{25}{2} \times 84 \][/tex]
[tex]\[ S_{25} = 25 \times 42 \][/tex]
[tex]\[ S_{25} = 1050 \][/tex]
Therefore, the total number of students that Mr. Wright has taught over 25 years is:
[tex]\[ \boxed{1050} \][/tex]
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