College

Calculate: [tex]S = 27 + 30 + 33 + \ldots + 333[/tex]

A. 10050
B. 18540
C. 30000
D. 5040
E. 3700

Answer :

Sure, let's solve the problem step-by-step!

The sequence given is [tex]\( 27, 30, 33, \ldots, 333 \)[/tex]. This is an arithmetic sequence because the difference between consecutive terms is constant.

### Step 1: Identify the first term ([tex]\(a\)[/tex]) and the common difference ([tex]\(d\)[/tex])
- The first term [tex]\(a\)[/tex] of the series is 27.
- The common difference [tex]\(d\)[/tex] between each term is 3 (i.e., [tex]\( 30 - 27 = 3 \)[/tex]).

### Step 2: Find the total number of terms ([tex]\(n\)[/tex])
To find the number of terms in the sequence, we use the formula for the [tex]\(n\)[/tex]-th term of an arithmetic sequence:
[tex]\[ l = a + (n - 1) \cdot d \][/tex]
where [tex]\(l\)[/tex] is the last term and [tex]\(n\)[/tex] is the number of terms.

Here:
- [tex]\( l = 333 \)[/tex]
- [tex]\( a = 27 \)[/tex]
- [tex]\( d = 3 \)[/tex]

Substitute these values into the formula:
[tex]\[ 333 = 27 + (n - 1) \cdot 3 \][/tex]
First, isolate [tex]\( (n - 1) \cdot 3 \)[/tex]:
[tex]\[ 333 - 27 = (n - 1) \cdot 3 \][/tex]
[tex]\[ 306 = (n - 1) \cdot 3 \][/tex]

Next, solve for [tex]\(n - 1\)[/tex]:
[tex]\[ n - 1 = \frac{306}{3} \][/tex]
[tex]\[ n - 1 = 102 \][/tex]

So, the number of terms [tex]\(n\)[/tex] is:
[tex]\[ n = 102 + 1 \][/tex]
[tex]\[ n = 103 \][/tex]

### Step 3: Calculate the sum of the arithmetic sequence ([tex]\(S\)[/tex])
The sum [tex]\(S\)[/tex] of the first [tex]\(n\)[/tex] terms of an arithmetic sequence is given by:
[tex]\[ S = \frac{n}{2} \cdot (a + l) \][/tex]

Substitute the known values:
- [tex]\( n = 103 \)[/tex]
- [tex]\( a = 27 \)[/tex]
- [tex]\( l = 333 \)[/tex]

[tex]\[ S = \frac{103}{2} \cdot (27 + 333) \][/tex]
[tex]\[ S = \frac{103}{2} \cdot 360 \][/tex]
[tex]\[ S = 103 \cdot 180 \][/tex]
[tex]\[ S = 18540 \][/tex]

Thus, the sum of the sequence [tex]\( S = 27 + 30 + 33 + \ldots + 333 \)[/tex] is [tex]\( 18540 \)[/tex].

### Answer
The correct answer is [tex]\( B) 18540 \)[/tex].