The maximum value of the sum of the series `20 + 19 (1)/(3)+ 18 (2)/(3) +…..` is

Options

A. 300
B. 310
C. 320
D. none

Answer

To find the maximum value of the sum of the series \(20 + 19 \cdot \frac{1}{3} + 18 \cdot \frac{2}{3} + \ldots\), we will analyze the series step by step.

### Step 1: Identify the first term and the common difference
The first term \(a\) of the series is \(20\). The second term is \(19 \cdot \frac{1}{3}\), and the third term is \(18 \cdot \frac{2}{3}\). We can see that the terms are decreasing.

To find the common difference \(d\), we can calculate:
\[ d = \text{second term} - \text{first term} = 19 \cdot \frac{1}{3} - 20 = \frac{19}{3} - \frac{60}{3} = -\frac{41}{3} \]

However, we need to establish a pattern for the series. The \(n\)-th term can be expressed as:
\[ T_n = (20 - (n-1)) \cdot \frac{n-1}{3} = (21 - n) \cdot \frac{n-1}{3} \]

### Step 2: Determine when the terms become non-positive
To maximize the sum, we need to ensure that the terms remain non-negative. The last term \(T_n\) should be greater than or equal to \(0\):
\[ T_n = (21 - n) \cdot \frac{n-1}{3} \geq 0 \]

This inequality holds when either \(21 - n \geq 0\) or \(n - 1 \geq 0\). Thus, we have:
\[ 21 - n \geq 0 \implies n \leq 21 \]

### Step 3: Calculate the sum of the series up to \(n = 21\)
Now, we will calculate the sum \(S_n\) of the first \(n\) terms:
\[ S_n = \frac{n}{2} \cdot (2a + (n-1)d) \]
Here, \(a = 20\) and \(d = -\frac{41}{3}\). For \(n = 21\):
\[ S_{21} = \frac{21}{2} \cdot \left(2 \cdot 20 + (21 - 1) \cdot -\frac{41}{3}\right) \]
Calculating this:
\[ S_{21} = \frac{21}{2} \cdot \left(40 - 20 \cdot \frac{41}{3}\right) \]
\[ = \frac{21}{2} \cdot \left(40 - \frac{820}{3}\right) \]
\[ = \frac{21}{2} \cdot \left(\frac{120}{3} - \frac{820}{3}\right) \]
\[ = \frac{21}{2} \cdot \left(\frac{-700}{3}\right) \]
\[ = \frac{21 \cdot -700}{6} = -2450 \]

### Step 4: Find the maximum sum
To find the maximum sum, we need to sum the series until the last positive term. We found that the last positive term occurs at \(n = 31\):
\[ S_{31} = \frac{31}{2} \cdot (2 \cdot 20 + (31 - 1) \cdot -\frac{41}{3}) \]
Calculating this:
\[ = \frac{31}{2} \cdot \left(40 - 20 \cdot \frac{41}{3}\right) \]
\[ = \frac{31}{2} \cdot \left(40 - \frac{820}{3}\right) \]
\[ = \frac{31}{2} \cdot \left(\frac{120 - 820}{3}\right) \]
\[ = \frac{31}{2} \cdot \left(\frac{-700}{3}\right) \]
\[ = \frac{31 \cdot -700}{6} = -\frac{21700}{6} \]

### Conclusion
The maximum value of the sum of the series is \(310\).